Post on 26-Jan-2023
Engineering Mathematics -II
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ENGINEERING MATHEMATICS -II
SUBJECT CODE: MA8251
(Regulation 2017) Common to all branches of B.E
UNIT โ I
MATRICES
Engineering Mathematics -II
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UNIT โ I
MATRICES
1.1 INTRODUCTION
The definition of matrices and the basic operations related to them originated in a memoir of
Cayley in 1958.This grew out of a simple observation on bilinear transformation and theory of Algebraic
invariants .Subsequently a number of eminent mathematicians like Sylvester, Hamilton contributed to the
development of the theory. Heisenberg used matrices in quantum theory as early as1925 AD. The matrix
algebra is a kind of shorthand technique for translating complicated mathematical relationship in to compact
forms .It has high degree of applicability in most of the quantitative applied sciences and has become
indispensable tool. In Physical, Social and Biological Sciences, application of matrix algebra has become
highly wide โspread. In this chapter we shall discuss and solve the Eigen value problem ๐ด๐ = X
Definition
An arrangement of mn elements in a rectangular form having an ordered set of m rows and n
columns is called an m ร n matrix A
๐ด = [(
๐11 ๐12 ๐13 โฏ ๐1๐
๐21๐22 ๐23 โฑ ๐2๐
โฆ โฏ โฆ .
)]
In short A=[aij] ,i = 1,2,โฆโฆm
j = 1,2, โฆโฆn
Here each aij is called an element of the matrix in, i th row and j th column.
Special Types of Matrices
Let A be Square Matrix of order n,
1. A = AT,then it is called a symmetric matrix (๐. ๐) aij = aji for all i, j.
2. If the square matrix A = โAT, then it is called skew โsymmetric matrix
(๐. ๐) aij = โaji for all i, j.
3. An ๐ ร 1 matrix is called a column matrix.
4. An 1 ร ๐ matrix is called a row matrix.
5. In the square matrix A=(aij) the elements ๐11, ๐22, ๐33, โฆ , ๐๐๐ are called the diagonal elements of A and
the diagonal elements constitute the principal diagonal of the matrix A .
6. A square matrix is called a diagonal matrix if all the elements except the leading diagonal are zero (๐. ๐) In
a diagonal matrix aij = 0 if i โ j.
7. A square matrix A=(aij) is called an upper triangular matrix if all the entries below the leading diagonal
are zero.
8. A square matrix A=(aij) is called an lower triangular matrix if all the entries above the leading diagonal
are zero.
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9. A square matrix A is said to be singular if |A| = 0, otherwise it is a non โsingular matrix.
1.2 CHARACTERISTIC EQUATION
If A is any square matrix of order n, the matrix A โ ฮปI where I is the unit matrix and ฮป be scaler
of order n can be formed as
|A โ ฮปI| = |๐11 โ ๐ ๐12 โฏ ๐1๐
โฎ โฑ โฎ๐๐1 ๐๐2 โฏ ๐๐๐ โ ๐
| = 0 is called the characteristic equation of A.
Working Rule for Characteristic Equation
Type I: For 2 ร 2 matrix
If ๐ด = (๐11 ๐12
๐21 ๐22), then the characteristic equation of A is ๐2 โ ๐ 1๐ + ๐ 2 = 0
Where ๐ 1 =Sum of the leading diagonal elements =๐11 + ๐22
๐ 2 = |๐ด| =Determinant of a matrix A.
Type II: For 3 ร 3 matrix
If ๐ด = (
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
), then the characteristic equation of A is ๐3 โ ๐ 1๐2 + ๐ 2 ๐ โ ๐ 3 = 0
Where ๐ 1 =Sum of the leading diagonal elements =๐11 + ๐22 + ๐33
๐ 2 =Sum of minors of leading diagonal elements
= |๐22 ๐23
๐32 ๐33| + |
๐11 ๐13
๐31 ๐33| + |
๐11 ๐12
๐21 ๐22|
๐ 3 = |๐ด| = Determinant of a matrix A.
Example: 1.1 Find the characteristic equation of the matrix (๐ ๐๐ ๐
)
Solution:
The characteristic equation is 2 โ s1+ s2 = 0
s1 = sum of the main diagonal element
= 1 + 2 = 3
s2 = |A| = |1 20 2
| = 2
Characteristic equation is 2 โ 3+ 2 = 0
Example: 1.2 Find the characteristic equation of the matrix (๐ โ๐
โ๐ ๐)
Solution:
The characteristic equation is 2 โ s1 + s2 = 0
s1 = sum of the main diagonal element
= 1 + 4 = 5
s2 = |A| = |1 โ2
โ5 4| = 4 โ 10 = โ6
Characteristic equation is 2 โ 5โ 6 = 0
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Example: 1.3 Find the characteristic equation of the matrix (๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
)
Solution:
The characteristic equation is 3 โ s1
2 + s2 โ s3 = 0
s1 = sum of the main diagonal element
= 2 + 2 + 2 = 6
s2 = sum of the minors of the main diagonalelement
= |2 00 2
| + |2 11 2
| + |2 00 2
| = 11
s3 = |A| = |2 0 10 2 01 0 2
| = 2(4 โ 0) โ 0 + 1(0 โ 2)
= 8 โ 2 = 6
Characteristic equation is 3 โ 62 + 11โ 6 = 0
Example: 1.4 Find the characteristic equation of the matrix (๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
)
Solution:
The characteristic equation is 3 โ s1
2 + s2โ s3 = 0
s1 = sum of the main diagonal element
= 2 + 2 + 1 = 5
s2 = sum of the minors of the main diagonalelement
= |2 10 1
| + |2 10 1
| + |2 11 2
| = 7
s3 = |A| = |2 1 11 2 10 0 1
| = 2(2 โ 0) โ 1(1 โ 0) + 1(0 โ 0)
= 4 โ 1 = 3
Characteristic equation is 3 โ 52 + 7โ 3 = 0
Example: 1.5 Find the characteristic polynomial of the matrix (๐ โ๐ โ๐
โ๐ ๐ ๐โ๐ โ๐ ๐
)
Solution:
The characteristic polynomial is 3 โ s1
2 + s2 โ s3
s1 = sum of the main diagonal element
= 0 + 1 + 2 = 3
s2 = sum of the minors of the main diagonalelement
= |1 2
โ1 2| + |
0 โ2โ1 2
| + |0 โ2
โ1 1| =0
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s3 = |A| = |0 โ2 โ2
โ1 1 2โ1 โ1 2
| = 0 + 2(โ2 + 2) โ 2(1 + 1)
= โ4
Characteristic polynomial is 3 โ 32 + 4
1.3 EIGEN VALUES AND EIGEN VECTORS
Definition
The values of ฮป obtained from the characteristic equation |A โ ฮปI| = 0 are called Eigenvalues of
โAโ.[or Latent values of A or characteristic values of A]
Definition
Let A be square matrix of order 3 and ฮป be scaler. The column matrix ๐ = (
๐ฅ1
๐ฅ2
๐ฅ3
) which satisfies
(A โ ฮปI)X = 0 is called Eigen vector or Latent vector or characteristic vector.
Linearly Dependent and Independent Eigenvectors
Let A be the matrix whose columns are Eigen vectors of A
* If |A| = 0 then the eigenvectors are linearly dependent.
* If |A| โ 0 then the eigenvectors are linearly independent.
Example: 1.6 Find the Eigen values for the matrix (๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
)
Solution:
The characteristic equation is 3 โ s1
2 + s2 โ s3=0
s1 = sum of the main diagonal element
= 2 + 3 + 2 = 7
s2 = sum of the minors of the main diagonalelement
= |3 12 2
| + |2 11 2
| + |2 21 3
| = 4 + 3 + 4 = 11
s3 = |A| = |2 2 11 3 11 2 2
| = 2(6 โ 2) โ 2(2 โ 1) + 1(2 โ 3)
= 8 โ 2 โ 1 = 5
Characteristic equation is 3 โ 72 + 11โ 5 = 0
โ = 1 ,2 โ 6+ 5 = 0
โ = 1, ( โ 1)(โ 5) = 0
The Eigen values are = 1, 1,5
Example: 1.7 Determine the Eigen values for the matrix (โ๐ ๐ โ๐๐ ๐ โ๐
โ๐ โ๐ ๐)
Solution:
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The characteristic equation is 3 โ s1
2 + s2 โ s3=0
s1 = sum of the main diagonal element
= โ2 + 1 + 0 = โ1
s2 = sum of the minors of the main diagonalelement
= |1 โ6
โ2 0| + |
โ2 โ3โ1 0
| + |โ2 22 1
| = โ12 โ 3 โ 6 = โ21
s3 = |A| = |โ2 2 โ32 1 โ6
โ1 โ2 0| = โ2(0 โ 12) โ 2(0 โ 6) โ 3(โ4 + 1)
= 24 + 12 + 9 = 45
Characteristic equation is 3 +
2 โ 21โ 45 = 0
โ = โ3 , 2 โ 2โ 15 = 0
โ = โ3, (+ 3)(โ 5) = 0
The Eigen values are = โ3,โ3,5
Exercise: 1.1
Find the Eigen values for the following matrices:
1.(3 44 โ3
) Ans: = 5; โ5
2.(2 โ1
โ8 4) Ans: = 0,6
3.(โ15 4 310 โ12 620 โ4 2
) Ans: = โ10,โ20,5
4.(2 0 00 2 01 0 2
) Ans: = 1,2,3
5.(1 0 โ11 2 12 2 3
) Ans: = 1,2,3
Eigen values and Eigen vectors for Non โ Symmetric matrix
Example: 1.8 Find the Eigen values and Eigen vectors for the matrix (๐ โ๐ ๐
โ๐ ๐ โ๐๐ โ๐ ๐
)
Solution:
The characteristic equation is 3 โ s1
2 + s2 โ s3=0
s1 = sum of the main diagonal element
= 8 + 7 + 3 = 18
s2 = sum of the minors of the main diagonalelement
= |7 โ4
โ4 3| + |
8 22 3
| + |8 โ6
โ6 7| = 5 + 20 + 20 = 45
s3 = |A| = |8 โ6 2
โ6 7 โ42 โ4 3
| = 8(21 โ 16) + 6(โ18 + 8) + 2(24 โ 14)
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= 40 โ 40 + 20 = 0
Characteristic equation is 3 โ 182 + 45 = 0
โ (2 โ 18+ 45) = 0
โ = 0, (โ 15)(โ 3) = 0
โ = 0,3,15
To find the Eigen vectors:
Case( i) When = 0 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (8 โ 0 โ6 2โ6 7 โ 0 โ42 โ4 3 โ 0
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
8x1 โ 6x2 + 2x3 = 0โฆ (1)
โ6x1 + 7x2 โ 4x3 = 0โฆ (2)
2x1 โ 4x2 + 3x3 = 0โฆ (3)
From (1) and (2)
x1
24โ14=
x2
โ12+32=
x3
56โ36
x1
10=
x2
20=
x3
20
x1
1=
x2
2=
x3
2
X1 = (123)
Case (ii) When = 3 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (8 โ 3 โ6 2โ6 7 โ 3 โ42 โ4 3 โ 3
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
5x1 โ 6x2 + 2x3 = 0โฆ (4)
โ6x1 + 4x2 โ 4x3 = 0โฆ (5)
2x1 โ 4x2 + 0x3 = 0โฆ (6)
From (4) and (5)
x1
24โ8=
x2
โ12+20=
x3
20โ36
x1
16=
x2
8=
x3
โ16
x1
2=
x2
1=
x3
โ2
X2 = (21
โ2)
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Case (iii) When = 15 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (8 โ 15 โ6 2
โ6 7 โ 15 โ42 โ4 3 โ 15
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
7x1 โ 6x2 + 2x3 = 0โฆ (7)
โ6x1 โ 8x2 โ 4x3 = 0โฆ (8)
2x1 โ 4x2 โ 12x3 = 0โฆ(9)
From (7) and (8)
x1
24+16=
x2
โ12โ28=
x3
56โ36
x1
40=
x2
โ40=
x3
20
x1
2=
x2
โ2=
x3
1
X3 = (2
โ21
)
Hence the corresponding Eigen vectors are X1 = (123) ; X2 = (
21
โ2) ; X3 = (
2โ21
)
Example: 1.9 Determine the Eigen values and Eigen vectors of the matrix (๐ โ๐ ๐
โ๐ ๐ โ๐๐ โ๐ ๐
)
Solution:
The characteristic equation is 3 โ s1
2 + s2 โ s3= 0
s1 = sum of the main diagonal element
= 7 + 6 + 5 = 18
s2 = sum of the minors of the main diagonalelement
= |6 โ2
โ2 5| + |
7 00 5
| + |7 โ2
โ2 6| = 26 + 35 + 38 = 99
s3 = |A| = |7 โ2 0
โ2 6 โ20 โ2 5
| = 182 โ 20 + 0 = 162
Characteristic equation is 3 โ 182 + 99โ 162 = 0
โ = 3, (2 โ 15+ 54) = 0
โ = 3, ( โ 9)(โ 6) = 0
โ = 3, 6, 9
To find the Eigen vectors:
Case (i) When = 3 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
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โ (7 โ 3 โ2 0โ2 6 โ 3 โ20 โ2 5 โ 3
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
4x1 โ 2x2 + 0x3 = 0โฆ (1)
โ2x1 + 3x2 โ 2x3 = 0โฆ (2)
0x1 โ 2x2 + 2x3 = 0โฆ (3)
From (1) and (2)
x1
4โ0=
x2
0+8=
x3
12โ4
x1
4=
x2
8=
x3
8
x1
1=
x2
2=
x3
2
X1 = (122)
Case (ii) When = 6 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (7 โ 6 โ2 0โ2 6 โ 6 โ20 โ2 5 โ 6
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
x1 โ 2x2 + 0x3 = 0โฆ (4)
โ2x1 + 0x2 โ 2x3 = 0โฆ (5)
0x1 โ 2x2 โ x3 = 0โฆ (6)
From (4)and (5)
x1
4โ0=
x2
0+2=
x3
0โ4
x1
4=
x2
2=
x3
โ4
x1
2=
x2
1=
x3
โ2
X2 = (21
โ2)
Case (iii) When = 9 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (7 โ 9 โ2 0โ2 6 โ 9 โ20 โ2 5 โ 9
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โ2x1 โ 2x2 + 0x3 = 0โฆ (7)
โ2x1 โ 3x2 โ 2x3 = 0โฆ (8)
0x1 โ 2x2 โ 4x3 = 0โฆ (9)
From (7) and (8)
x1
4โ0=
x2
0โ4=
x3
6โ4
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x1
4=
x2
โ4=
x3
2
x1
2=
x2
โ2=
x3
1
X3 = (2
โ21
)
Hence the corresponding Eigen vectors are X1 = (122) ; X2 = (
21
โ2) ; X3 = (
2โ21
)
Example: 1.10 Determine the Eigen values and Eigen vectors of the matrix (๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
)
Solution:
The characteristic equation is 3 โ s1
2 + s2 โ s3=0
s1 = sum of the main diagonal element
= 3 + 2 + 5 = 10
s2 = sum of the minors of the main diagonalelement
= |2 60 5
| + |3 40 5
| + |3 10 2
| = 10 + 15 + 6 = 31
s3 = |A| = |3 1 40 2 60 0 5
| = 30
Characteristic equation is 3 โ 102 + 31โ 30 = 0
โ = 2, (2 โ 8+ 15) = 0
โ = 2, ( โ 5)(โ 3) = 0
โ = 2,3,5
To find the Eigen vectors:
Case( i) When = 2 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (3 โ 2 1 4
0 2 โ 2 60 0 5 โ 2
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
x1 + x2 + 4x3 = 0 โฆ(1)
0x1 + 0x2 + 6x3 = 0โฆ (2)
0x1 + 0x2 + 3x3 = 0โฆ (3)
From (1) and (2)
x1
6โ0=
x2
0โ6=
x3
0โ0
x1
6=
x2
โ6=
x3
0
x1
1=
x2
โ1=
x3
0
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X1 = (1
โ10
)
Case (ii) When = 3 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (3 โ 3 1 4
0 2 โ 3 60 0 5 โ 3
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
0x1 + x2 + 4x3 = 0โฆ (4)
0x1 โ x2 + 6x3 = 0โฆ (5)
0x1 + 0x2 + 2x3 = 0โฆ (6)
From (4) and (5)
x1
4+6=
x2
0โ0=
x3
0โ0
x1
10=
x2
0=
x3
0
x1
1=
x2
0=
x3
0
X2 = (100)
Case (iii) When = 5 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (3 โ 5 1 4
0 2 โ 5 60 0 5 โ 5
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โ2x1 + x2 + 4x3 = 0โฆ (7)
0x1 โ 3x2 + 6x3 = 0โฆ (8)
0x1 + 0x2 + 0x3 = 0โฆ (9)
From (7) and (8)
x1
6+12=
x2
0+12=
x3
6โ0
x1
18=
x2
12=
x3
6
x1
3=
x2
2=
x3
1
X3 = (321)
Hence the corresponding Eigen vectors are X1 = (1
โ10
) ; X2 = (100) ; X3 = (
321)
Example: 1.11 Find the Eigen values and Eigen vectors of the matrix (๐ ๐
โ๐ ๐)
Solution:
The characteristic equation is 2 โ s1 + s2 = 0
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s1 = sum of the main diagonal element
= a+ a = 2a
s2 = |A| = | a bโb a
| = a2 + b2
Characteristic equation is 2 โ 2a+ a2 + b2 = 0
=โbยฑโb2โ4ac
2a
=2aยฑโ4a2โ4(a2+b2)
2
=2aยฑโโ4b2
2
=2aยฑi2b
2
= a ยฑ ib
The Eigen values are = a ยฑ ib
To find the Eigen vectors:
Case (i) When = a + ib the eigen vector is given by(A โ I)X = 0 where X = (x1
x2)
(a โ (a + ib) b
โb a โ (a + ib)) (
x1
x2)=(
00)
โibx1 + bx2 = 0โฆ (1)
โbx1 โ ibx2 = 0โฆ (2)
โด (1) โ โ๐๐x1 = โbx2
x1
b=
x2
ibโ
x1
1=
x2
i
X1 = (1i)
Case (ii) When = a โ ib the eigen vector is given by(A โ I)X = 0 where X = (x1
x2)
(a โ (a โ ib) b
โb a โ (a โ ib)) (
x1
x2)=(
00)
ibx1 + bx2 = 0โฆ (3)
โbx1 + ibx2 = 0โฆ (4)
โด (3) โ ๐๐x1 = โbx2
x1
โb=
x2
ibโ
x1
โ1=
x2
i
X2 = (โ1i
)
Exercise: 1.2
Find the Eigen values and Eigenvectors of the following matrices:
1.(1 0 00 3 โ10 โ1 3
) Ans: = 1; (1,0,0); = 2; (0,1,1); = 4; (0,โ1,1)
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2.(2 4 โ64 2 โ6
โ6 โ6 โ15) Ans: = โ2; (1,โ1,0); = 9; (2,2,โ1); = โ18; (1,1,4)
3.(2 2 02 1 1
โ7 2 โ3) Ans: = 1; (2,โ1, โ4); = 3; (2,1,โ2); = โ4; (1,โ3,13)
4.(4 โ20 โ10
โ2 10 46 โ30 โ13
) Ans: = 0; (5,1,0); = โ1; (2,0,1); = 2; (0,1,โ2)
5.(2 2 02 2 00 0 1
) Ans: = 0; (1,โ1,0); = 1; (0,0,1); = 4; (1,1,0)
Problems on Symmetric matrices with repeated Eigen values
Example: 1.12 Determine the Eigen values and Eigen vectors of the matrix (๐ โ๐ ๐
โ๐ ๐ โ๐๐ โ๐ ๐
)
Solution:
The characteristic equation is 3 โ s1
2 + s2 โ s3=0
s1 = sum of the main diagonal element
= 6 + 3 + 3 = 12
s2 = sum of the minors of the main diagonalelement
= |3 โ1
โ1 3| + |
6 22 3
| + |6 โ2
โ2 3| = 8 + 14 + 14 = 36
s3 = |A| = |6 โ2 2
โ2 3 โ12 โ1 3
| = 32
Characteristic equation is 3 โ 122 + 36โ 32 = 0
โ = 2, (2 โ 10+ 16) = 0
โ = 2, ( โ 2)(โ 8) = 0
โ = 2,2,8
To find the Eigen vectors:
Case (i) When = 8 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (6 โ 8 โ2 2โ2 3 โ 8 โ12 โ1 3 โ 8
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โ2x1 โ 2x2 + 2x3 = 0โฆ (1)
โ2x1 โ 5x2 โ x3 = 0โฆ (2)
2x1 โ x2 โ 5x3 = 0โฆ (3)
From (1) and (2)
x1
2+10=
x2
โ4โ2=
x3
10โ4
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x1
12=
x2
โ6=
x3
6
x1
2=
x2
โ1=
x3
1
X1 = (2
โ11
)
Case (ii) When = 2 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (6 โ 2 โ2 2โ2 3 โ 2 โ12 โ1 3 โ 2
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
4x1 โ 2x2 + 2x3 = 0โฆ (4)
โ2x1 + x2 โ x3 = 0โฆ (5)
2x1 โ x2 + x3 = 0 โฆ(6)
In (1) put x1 = 0 โ โ2x2 = โ2x3
โx2
1=
x3
1 โ ๐2 = (
011)
In (1) put x2 = 0 โ 4x1 + 2x3 = 0
โ 4x1 = โ2x3
โx1
โ1=
x3
2 โ ๐3 = (
โ102
)
Hence the corresponding Eigen vectors are X1 = (2
โ11
) ; X2 = (011) ; X3 = (
โ102
)
Example: 1.13 Find the Eigen values and Eigen vectors of the matrix (๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
)
Solution:
The characteristic equation is 3 โ s1
2 + s2 โ s3=0
s1 = sum of the main diagonal element
= 2 + 3 + 2 = 7
s2 = sum of the minors of the main diagonalelement
= |3 12 2
| + |2 11 2
| + |2 21 3
| = 4 + 3 + 4 = 1
s3 = |A| = |2 2 11 3 11 2 2
| = 5
Characteristic equation is 3 โ 72 + 11โ 5 = 0
โ = 1, (2 โ 6+ 5) = 0
โ = 1, ( โ 1)(โ 5) = 0
โ = 1, 1, 5
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15
To find the Eigen vectors:
Case (i) When = 5 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (2 โ 5 2 1
1 3 โ 5 11 2 2 โ 5
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โ3x1 + 2x2 + x3 = 0โฆ (1)
x1 โ 2x2 + x3 = 0โฆ (2)
x1 + 2x2 โ 3x3 = 0โฆ (3)
From (1) and (2)
x1
2+2=
x2
1+3=
x3
6โ2
x1
4=
x2
4=
x3
4
x1
1=
x2
1=
x3
1
X1 = (111)
Case (ii) When = 1 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (2 โ 1 2 1
1 3 โ 1 11 2 2 โ 1
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
x1 + 2x2 + x3 = 0โฆ (4)
x1 + 2x2 + x3 = 0 โฆ(5)
x1 + 2x2 + x3 = 0 โฆ(6)
In (1) put x1 = 0 โ 2x2 = โx3
โx2
โ1=
x3
2 โ ๐2 = (โ
012)
In (1) put x2 = 0 => x1 = โx3
โx1
โ1=
x3
1 โ ๐3 = (
โ101
)
Hence the corresponding Eigen vectors are X1 = (111) ; X2 = (โ
012) ; X3 = (
โ101
)
Example: 1.14 Find the Eigen values and Eigen vectors of the matrix (๐ โ๐ ๐๐๐ โ๐๐ ๐๐๐ โ๐ ๐
)
Solution:
The characteristic equation is 3 โ s1
2 + s2 โ s3=0
s1 = sum of the main diagonal element
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= 6 โ 13 + 4 = โ3
s2 = sum of the minors of the main diagonalelement
= |โ13 10โ6 4
| + |6 57 4
| + |6 โ614 โ13
| = 8 โ 11 + 6 = 3
s3 = |A| = |6 โ6 514 โ13 107 โ6 4
| = โ1
Characteristic equation is 3 + 32 + 3+ 1 = 0
โ = โ1, (2 + 2+ 1) = 0
โ = 1, ( + 1)(+ 1) = 0
โ = โ1,โ1, โ1
To find the Eigen vectors:
When = โ1 the Eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (6 + 1 โ6 514 โ13 + 1 107 โ6 4 + 1
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
7x1 โ 6x2 + 5x3 = 0โฆ (1)
14x1 โ 12x2 + 10x3 = 0โฆ (2)
7x1 โ 6x2 + 5x3 = 0โฆ (3)
In (1) put x1 = 0 โ โ6x2 = โ5x3
โx2
5=
x3
6 โ ๐1 = (
056)
In (1) put x2 = 0 โ 7x1 = โ5x3
โx1
โ5=
x3
7 โ ๐2 = (
โ507
)
In (1) put x3 = 0 โ 7x1 = 6x2
โx1
6=
x2
7 โ ๐3 = (
670)
Hence the corresponding Eigen vectors are X1 = (056) ; X2 = (
โ507
) ; X3 = (670)
Example: 1.15 Find the Eigen values and Eigen vectors of the matrix (๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
)
Solution:
The characteristic equation s 3 โ s1
2 + s2 โ s3=0
s1 = sum of the main diagonal element
= 2 + 2 + 2 = 6
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s2 = sum of the minors of the main diagonalelement
= |2 10 2
| + |2 00 2
| + |2 10 2
| = 4 + 4 + 4 = 12
s3 = |A| = |2 1 00 2 10 0 2
| = 8
Characteristic equation is 3 โ 62 + 12โ 8 = 0
โ = 2, (2 โ 4+ 4) = 0
โ = 2, ( โ 2)(โ 2) = 0
โ = 2 ,2, 2
To find the Eigen vectors:
When = 2 the Eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (2 โ 2 1 0
0 2 โ 2 10 0 2 โ 2
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
0x1 + x2 + 0x3 = 0โฆ (1)
0x1 + 0x2 + x3 = 0โฆ (2)
0x1 + 0x2 + 0x3 = 0โฆ (3)
From (1) and (2)
x1
1 โ 0=
x2
0 โ 0=
x3
0 โ 0
x1
1=
x2
0=
x3
0
Hence the corresponding Eigen vectors are X1 = X2 = X3 = (100)
Exercise: 1.3
Find the Eigen values and Eigenvectors of the following matrices:
1.(2 2 11 3 11 2 2
) Ans: = 1; (2,โ1,0); = 1; (1,0,โ1); = 5; (1,1,1)
2.(2 1 12 3 23 3 4
) Ans: = 1; (0,1,โ1); = 1; (1,0,โ1); = 7; (1,2,3)
3.(1 2 20 2 1
โ1 2 2) Ans: = 1; (1,1,โ1); = 1; (1,1,โ1); = 2; (2,1,0)
4.(โ3 โ9 โ121 3 40 0 1
) Ans: = 0; (3,โ1,0); = 0; (3,โ1,0); = 1; (12,โ4,โ1)
5. (โ2 2 โ32 1 โ6
โ1 โ2 0) Ans: = โ3; (3,0,1); = โ3; (โ2,1,0); = 5; (1,2,โ1)
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1.4 PROPERTIES OF EIGEN VALUES AND EIGEN VECTORS
Property: 1(a) The sum of the Eigen values of a matrix is equal to the sum of the elements of the
principal (main) diagonal.
(or)
The sum of the Eigen values of a matrix is equal to the trace of the matrix.
1. (b) product of the Eigen values is equal to the determinant of the matrix.
Proof:
Let A be a square matrix of order ๐.
The characteristic equation of A is |๐ด โ ๐๐ผ| = 0
(๐. ๐. )๐๐ โ ๐1๐๐โ1 + ๐2๐
๐โ2 โ โฏ+ (โ1)๐๐ = 0 ... (1)
where S1 = Sum of the diagonal elements of A.
. . .
. . .
. . .
Sn = determinant of A.
We know the roots of the characteristic equation are called Eigen values of the given matrix.
Solving (1) we get ๐ roots.
Let the ๐ be ๐1, ๐2, โฆ ๐๐ .
i.e., ๐1, ๐2, โฆ ๐๐ . are the Eignvalues of A.
We know already,
ฮปn โ (Sum of the roots ฮปnโ1 + [sum of the product of the roots taken two at a time] ฮปnโ2 โ
โฆ + (โ1)n(Product of the roots) = 0 ... (2)
Sum of the roots = ๐1๐๐ฆ (1)&(2)
(๐. ๐. ) ๐1 + ๐2 + โฏ + ๐๐ = ๐1
(๐. ๐. ) ๐1 + ๐2 + โฏ + ๐๐ = ๐11 + ๐22 + โฏ + ๐๐๐
Product of the roots = Snby (1)&(2)
(๐. ๐. )ฮป1ฮป2 โฆ ฮปn = det of A
Property: 2 A square matrix A and its transpose ๐๐ have the same Eigenvalues.
(or)
A square matrix A and its transpose ๐๐ have the same characteristic values.
Proof:
Let A be a square matrix of order ๐.
The characteristic equation of A and AT are
Sum of the Eigen values = Sum of the main diagonal elements
Product of the Eigen values = |A|
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19
|A โ ฮปI| = 0 โฆโฆ . (1)
and |AT โ ฮปI| = 0 โฆโฆ . (2)
Since, the determinant value is unaltered by the interchange of rows and columns.
We know |A| = |AT|
Hence, (1) and (2) are identical.
โด The Eigenvalues of A and AT are the same.
Property: 3 The characteristic roots of a triangular matrix are just the diagonal elements of the
matrix.
(or)
The Eigen values of a triangular matrix are just the diagonal elements of the matrix.
Proof: Let us consider the triangular Characteristic equation of is
matrix. |A โ ฮปI| = 0
A = [๐11 0 0๐21 ๐22 0๐31 ๐32 ๐33
] i.e., |๐11 โ ๐ 0 0
๐21 ๐22 โ ๐ 0๐31 ๐32 ๐33 โ ๐
| = 0
On expansion it gives (๐11 โ ๐)(๐22 โ ๐)(๐33 โ ๐) = 0
i.e., ๐ = ๐11, ๐22, ๐33
which are diagonal elements of the matrix A.
Property: 4 If ๐ is an Eigenvalue of a matrix A, then ๐
๐, (๐ โ ๐) is the Eignvalue of ๐โ๐.
(or)
If ๐ is an Eigenvalue of a matrix A, what can you say about the Eigenvalue of matrix ๐โ๐. Prove your
statement.
Proof:
If X be the Eigenvector corresponding to ๐,
then ๐ด๐ = ๐๐ ... (i)
Pre multiplying both sides by Aโ1, we get
Aโ1AX = Aโ1ฮปX
(1) โ X = ฮปAโ1X
X = ฮปAโ1X
รท ฮป โ 1
ฮปX = Aโ1X
(๐. ๐. ) Aโ1X =1
ฮปX
This being of the same form as (i), shows that 1
๐ is an Eigenvalue of the inverse matrix Aโ1.
Property: 5 If ๐ is an Eigenvalue of an orthogonal matrix, then ๐
๐ is an Eigenvalue.
Proof:
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20
Definition: Orthogonal matrix.
A square matrix A is said to be orthogonal if AAT = ATA = I
i.e., AT = Aโ1
Let A be an orthogonal matrix.
Given ๐ is an Eignevalue of A.
โ1
๐is an Eigenvalue of Aโ1
Since, AT = Aโ1
โด1
๐ is an Eigenvalue of AT
But, the matrices A and AT have the same Eigenvalues, since the determinants
|A โ ฮปI|and |AT โ ฮปI| are the same.
Hence, 1
๐ is also an Eigenvalue of A.
Property: 6 If ๐๐, ๐๐, โฆ ๐๐. are the Eignvalues of a matrix A, then ๐๐ฆ has the Eigenvalues
๐๐๐ฆ, ๐๐
๐ฆ, โฆ . , ๐๐ง๐ฆ (๐ฆ ๐๐๐ข๐ง๐ ๐ ๐ฉ๐จ๐ฌ๐ข๐ญ๐ข๐ฏ๐ ๐ข๐ง๐ญ๐๐ ๐๐ซ)
Proof:
Let ๐ด๐ be the Eigenvalue of A and ๐๐ the corresponding Eigenvector.
Then ๐ด๐๐ = ๐๐๐๐ โฆ (1)
We have A2Xi = A(AXi)
= A(ฮปiXi)
= ฮปiA(Xi)
= ฮปi(ฮปiXi)
= ฮปi2Xi
|โฃ| 1y A3Xi = ฮปi3Xi
In general, AmXi = ฮปimXi โฆ . (2)
Hence, ๐๐๐ is an Eigenvalue of Am.
The corresponding Eigenvector is the same ๐๐.
Note: If ฮป is the Eigenvalue of the matrix A then ฮป2 is the Eigenvalue of A2
Property: 7 The Eigen values of a real symmetric matrix are real numbers.
Proof:
Let ฮป be an Eigenvalue (may be complex) of the real symmetric matrix A. Let the corresponding
Eigenvector be X. Let A denote the transpose of A.
We have AX = ฮปX
Pre-multiplying this equation by 1 ร ๐ matrix ๐โฒฬ , where the bar denoted that all elements of ๐โฒฬ are the
complex conjugate of those of ๐โฒ, we get
XโฒAX = ฮปXโฒX โฆ . (1)
Taking the conjugate complex of this we get Xโฒ AX = ฮปXโฒX or
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XโฒA X = ฮป Xโฒ X since, A = A for A is real.
Taking the transpose on both sides, we get
(XโฒAX)โฒ = (ฮป Xโฒ X)โฒ(๐. ๐. , )Xโฒ Aโฒ X = ฮป Xโฒ ๐
(๐. ๐. )Xโฒ Aโฒ X = ฮป Xโฒ ๐ since Aโฒ = A for A is symmetric.
But, from (1), Xโฒ A X = ฮป Xโฒ ๐ Hence ฮป Xโฒ ๐ = ฮป Xโฒ ๐
Since, Xโฒ X is an 1 ร 1 matrix whose only element is a positive value, ฮป = ฮป (๐. ๐. ) ๐ is real).
Property: 8 The Eigen vectors corresponding to distinct Eigen values of a real symmetric matrix are
orthogonal.
Proof:
For a real symmetric matrix A, the Eigen values are real.
Let ๐1, ๐2 be Eigenvectors corresponding to two distinct eigen values ๐1 , ๐2 [๐1 , ๐2 are real]
AX1 = ฮป1X1 โฆ . (1)
AX2 = ฮป2X2 โฆ . (2)
Pre multiplying (1) by X2โฒ, we get
X2โฒAX1 = X2โฒฮป1X1
= ฮป1X2โฒX1
Pre-multiplying (2) by X1โฒ, we get
X1โฒAX2 = ฮป2X1โฒX2 โฆ . (3)
But(X2โฒAX1)โฒ = (ฮป1X2โฒX1)โฒ
X1โฒA X2 = ฮป1X1โฒX2
(๐. ๐) X1โฒA X2 = ฮป1X1โฒX2 ..... (4) [โต Aโฒ = A]
From (3) and (4)
ฮป1X1 โฒ X2 = ฮป2X1โฒ X2
(i.e.,)(ฮป1 โ ฮป2)X1 โฒ X2 = O
ฮป1 โ ฮป2, X1 โฒ X2 = O
โด X1 , X2 are orthogonal.
Property: 9 The similar matrices have same Eigen values.
Proof:
Let A, B be two similar matrices.
Then, there exists an non-singular matrix P such that B = Pโ1 AP
B โ ฮปI = Pโ1AP โ ฮปI
= Pโ1AP โ Pโ1 ฮปIP
= Pโ1(A โ ฮปI)P
|B โ ฮปI| = |Pโ1| |A โ ฮปI| |P|
= |A โ ฮปI| |Pโ1P|
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= |A โ ฮปI| |I|
= |A โ ฮปI|
Therefore, A, B have the same characteristic polynomial and hence characteristic roots.
โด They have same Eigen values.
Property: 10 If a real symmetric matrix of order 2 has equal Eigen values, then the matrix is a scalar
matrix.
Proof :
Rule 1 : A real symmetric matrix of order ๐ can always be diagonalised.
Rule 2 : If any diagonalized matrix with their diagonal elements are equal, then the matrix is a scalar matrix.
Given A real symmetric matrix โAโ of order 2 has equal Eigen values.
By Rule: 1 A can always be diagonalized, let ฮป1 and ฮป2 be their Eigenvalues then
we get the diagonlized matrix = [ฮป1 00 ฮป2
]
Givenฮป1 = ฮป2
Therefore, we get = [ฮป1 00 ฮป2
]
By Rule: 2 The given matrix is a scalar matrix.
Property: 11 The Eigen vector X of a matrix A is not unique.
Proof :
Let ฮป be the Eigenvalue of A, then the corresponding Eigenvector X such that A X = ฮป X .
Multiply both sides by non-zero K,
K (AX) = K (ฮปX)
โ A (KX) = ฮป (KX)
(๐. ๐. )an Eigenvector is determined by a multiplicative scalar.
(๐. ๐. ) Eigenvector is not unique.
Property: 12 ๐๐, ๐๐, โฆ ๐๐ be distinct Eigenvalues of an ๐ ร ๐ matrix, then the corresponding
Eigenvectors ๐๐, ๐๐ , โฆ ๐๐ง form a linearly independent set.
Proof:
Let ๐1, ๐2, โฆ ๐๐(๐ โค ๐) be the distinct Eigen values of a square matrix A of order ๐.
Let X1, X2 , โฆ Xm be their corresponding Eigenvectors we have to prove โ ๐ผ๐๐๐=1 Xi =
0 implies each ๐ผ๐ = 0, ๐ = 1,2, โฆ ,๐
Multiplying โ ๐ผ๐๐๐=1 Xi = 0 by (A โ ๐1I),we get
(A โ ๐1I)๐ผ1X1 = ๐ผ1(๐ดX1 โ ๐1X1 ) = ๐ผ1(0) = 0
When โ ๐ผ๐๐๐=1 Xi = 0 Multiplied by
(A โ ๐2I)(A โ ๐2I)โฆ (A โ ๐๐โ1I)(A โ ๐๐I) (A โ ๐๐+1I)โฆ (A โ ๐๐I)
We get, ๐ผ๐(๐๐ โ ๐1)(๐๐ โ ๐2)โฆ (๐๐ โ ๐๐โ1)(๐๐ โ ๐๐+1)โฆ (๐๐ โ ๐๐) = 0
Since, ฮปโs are distinct, ๐ผ๐ = 0
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Since, ๐ is arbitrary, each ๐ผ๐ = 0, ๐ = 1, 2,โฆ ,๐
โ ๐ผ๐๐๐=1 Xi = 0 implies each ๐ผ๐ = 0, ๐ = 1,2, โฆ , ๐
Hence, X1, X2 , โฆ Xm are linearly independent.
Property: 13 If two or more Eigen values are equal it may or may not be possible to get linearly
independent Eigenvectors corresponding to the equal roots.
Property: 14 Two Eigenvectors ๐๐and ๐๐ are called orthogonal vectors if ๐ฟ๐ ๐ป ๐๐ = ๐
Property: 15 If A and B are ๐ ร ๐ matrices and B is a non singular matrix, then A and ๐ฉโ๐ ๐๐ have
same eigenvalues.
Proof:
Characteristic polynomial of ๐ตโ1 AB
= |Bโ1 AB โ ฮปI| = |Bโ1 AB โ Bโ1(ฮปI)B|
= |Bโ1 (A โ ฮปI)B| = |Bโ1||A โ ฮปI||B|
= |Bโ1| |B| |A โ ฮปI| = |Bโ1B||A โ ฮปI|
= Characterisstisc polynomial of A
Hence, A and Bโ1 AB have same Eigenvalues.
Problems based on properties
Example: 1.16 Find the sum and product of the Eigen values of the matrix[โ๐ ๐ โ๐๐ ๐ โ๐
โ๐ โ๐ ๐]
Solution:
Sum of the Eigen values =Sum of the main diagonal elements
= (โ2) + (1) + (0)
= โ1
Product of the Eigen values = |โ2 2 โ32 1 โ6
โ1 โ2 0|
= โ2(0 โ 12) โ 2(0 โ 6) โ 3(โ4 + 1)
= 24 + 12 + 9 = 45
Example: 1.17 Find the sum and product of the Eigen values of the matrix A= [๐ ๐ ๐
โ๐ ๐ ๐๐ ๐ ๐
]
Solution:
Sum of the Eigen values = Sum of its diagonal elements = 1 + 2 + 1 = 4
Product of Eigen values = |C| = |1 2 3
โ1 2 11 1 1
|
= 1(2 โ 1) โ 2(โ1 โ 1) + 3(โ1 โ 2)
= 1(1) โ 2(โ2) + 3(โ3)
= 1 + 4 โ 9 = โ4
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Example: 1.18 The product of two Eigen values of the matrix ๐จ = [๐ โ๐ ๐
โ๐ ๐ โ๐๐ โ๐ ๐
]is 16. Find the third
Eigenvalue.
Solution:
Let Eigen values of the matrix A be ๐1, ๐2, ๐3.
Given ๐1๐2 = 16
We know that, ๐1๐2๐3 = |A|
[Product of the Eigen values is equal to the determinant of the matrix]
โด ๐1๐2๐3 = | 6 โ2 2โ2 3 โ1 2 โ1 3
|
= 6(9 โ 1) + (โ6 + 2) + 2(2 โ 6)
= 6(8) + 2(โ4) + 2(โ4)
= 48 โ 8 โ 8
โ ๐1๐2๐3 = 32
โ 16 ๐3 = 32
โ ๐3 =32
16= 2
Example: 1.19 Two of the Eigen values of [ ๐ โ๐ ๐โ๐ ๐ โ๐ ๐ โ๐ ๐
]are 2 and 8. Find the third Eigen value.
Solution:
We know that, Sum of the Eigen values = Sum of its diagonal elements
= 6 + 3 + 3 = 12
Given ๐1 = 2, ๐2 = 8, ๐3 = ?
We get, ๐1 + ๐2 + ๐3 = 12
2 + 8 + ๐3 = 12
๐3 = 12 โ 10
๐3 = 2
โด The third Eigenvalue = 2
Example: 1.20 If 3 and 15 are the two Eigen values of ๐จ = [ ๐ โ๐ ๐โ๐ ๐ โ๐ ๐ โ๐ ๐
] ๐๐ข๐ง๐ |๐|, without
expanding the determinant.
Solution:
Given ๐1 = 3, ๐2 = 15, ๐3 = ?
We know that, Sum of the Eigen values = Sum of the main diagonal elements
โ ๐1 + ๐2 + ๐3 = 8 + 7 + 3
3 + 15 + ๐3 = 18
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25
โ ๐3 = 0
We know that, Product of the Eigen values = |A|
โ (3)(15)(0) = |A|
โ |A| = (3)(15)(0)
โ |A| = 0
Example: 1.21 If 2, 2, 3 are the Eigen values of ๐จ = [ ๐ ๐๐ ๐โ๐ โ๐ โ๐ ๐ ๐ ๐
] find the Eigen values of ๐๐.
Solution:
By Property โA square matrix A and its transpose AT have the same Eigen valuesโ.
Hence, Eigen values of AT are 2, 2, 3
Example: 1.22 If the Eigen values of the matrix ๐จ = [๐ ๐๐ โ๐
] are ๐, โ๐ then find the Eigen values of
๐๐.
Solution:
Eigen values of ๐ด = Eigen values ๐๐ AT
โด Eigen values ๐๐ AT are 2, โ2.
Example: 1.23 Find the Eigen values of ๐ = [๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
]without using the characteristic equation idea.
Solution:
Given A = [2 1 00 2 10 0 2
] clearly given matrix A is an upper triangular matrix. Then by property, the
characteristic roots of a triangular matrix are just the diagonal elements of the matrix.
Hence, the Eigen values are 2, 2, 2.
Example: 1.24 Find the Eigen values of ๐ = [๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
]
Solution:
Given A = [2 0 01 3 00 4 4
]
Clearly given matrix A is a lower triangular matrix
Hence, by property the Eigen values of A are 2, 3, 4.
Example: 1.25 Two of the Eigen values of ๐จ = [๐ โ๐ ๐
โ๐ ๐ โ๐๐ โ๐ ๐
] are 3 and 6. Find the Eigen values of
๐โ๐.
Solution:
Sum of the Eigen values = Sum of the main diagonal elements
= 3 + 5 + 3 = 11
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26
Let K be the third Eigen value
โด 3 + 6 + ๐ = 11
โ 9 + ๐ = 11
โ ๐ = 2
โด The Eigenvalues of Aโ1 are 1
2 ,
1
3 ,
1
6
Example: 1.26 Two Eigen values of the matrix ๐ = [๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
] are equal to 1 each. Find the
Eigenvalues of ๐โ๐.
Solution:
Given A = [2 2 11 3 11 2 2
]
Let the Eigen values of the matrix A be ๐1, ๐2, ๐3
Given condition is ๐2 = ๐3 = 1
We have, Sum of the Eigen values = Sum of the main diagonal elements
โ ๐1 + ๐2 + ๐3 = 2 + 3 + 2
โ ๐1 + 1 + 1 = 7
โ ๐1 + 2 = 7
โ ๐1 = 5
Hence, the Eigen values of A are 1, 1, 5
Eigen values of Aโ1 are 1
1 ,
1
1 ,
1
5 , i. e. , 1, 1,
1
5
Example: 1.27 Find the Eigen values of the inverse of the matrix ๐ = [๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
]
Solution:
We know that, the given matrix is a upper triangular matrix.
Therefore, the Eigen values of A are 2, 3, 4
Example: 1.28 Find the Eigen values of ๐๐given ๐ = [๐ ๐ ๐๐ ๐ โ๐๐ ๐ ๐
]
Solution:
Given A = [1 2 30 2 โ70 0 3
]
Clearly given A is an upper triangular matrix
Hence, the Eigen values are 1, 2, 3
(๐. ๐. ) The Eigen values of the given matrix A are 1, 2, 3.
By the property, the Eigen values of the matrix A3 are 13, 23 , 33. i.e., 1, 8, 27.
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27
Example: 1.29 If 1 & 2 are the Eigen values of a ๐ ร ๐ matrix A, what are the Eigenvalues of ๐๐ and
๐โ๐?
Solution:
We know that, if ๐1, ๐2, . โฆ๐๐ are the Eigenvalues of A, then ๐1๐ , ๐2
๐ โฆ ๐๐๐ are the Eigenvalues of Am.
Given 1 and 2 are the Eigen values of A.
โด 12 and 22 i.e., 1 and 4 are the Eigen values of ๐ด2 ; 1 ๐๐๐ 1
2 are the Eigenvalues of Aโ1.
Example: 1.30 If ๐ผ ๐๐ง๐ ๐ฝ are the Eigen values of [๐ โ๐
โ๐ ๐], form the matrix whose Eigenvalues are
๐ผ๐ ๐๐๐ ๐ฝ๐.
Note: If ๐1, ๐2, . โฆ ๐๐ are the Eigenvalues of A, then ๐1๐, ๐2
๐ , ๐3๐ โฆ ๐๐
๐ are the Eigenvalues of Ak for any
positive integer.
Solution:
Let ๐ด = (3 โ1
โ1 5), Let ๐ผ , ๐ฝ be the Eigen values of A
โ A2 = A.A
= (3 โ1
โ1 5) (
3 โ1โ1 5
) = (10 โ8โ8 26
)
โ A3 = A2A
= (10 โ8โ8 26
) (3 โ1
โ1 5) = (
38 โ50โ50 138
)
Hence, the required matrix is (38 โ50
โ50 138)
Example: 1.31 Form the matrix whose Eigen values are ๐ถ โ ๐ , ๐ท โ ๐, ๐ธ โ ๐ where ๐ถ , ๐ท, ๐ธ are the
Eigenvalues of ๐จ = [โ๐ โ๐ โ๐๐ ๐ โ๐๐ โ๐ ๐
]
Solution:
Note: The matrix A โ KI has the Eigen values ๐1 โ ๐, ๐2 โ ๐, . โฆ ๐๐ โ ๐.
Hence, the required matrix is A โ 5I = [โ1 โ2 โ34 5 โ67 โ8 9
] โ 5 [1 0 00 1 00 0 1
]
= [โ6 โ2 โ34 0 โ67 โ8 4
]
Example: 1.32 Two Eigen values of ๐ = [๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
] are equal and they are ๐
๐ times to the third. Find
them.
Solution:
Let the third Eigen value be ๐3.
We know that, ๐1 + ๐2 + ๐3 = 2 + 3 + 2 = 7
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28
Given ๐1 = ๐2 โฆ(2) ๐1 =1
5๐3 โฆ(3) ๐2 =
1
5๐3 โฆ(4)
(1) โ 2๐1 + ๐3 = 7 by (2)
2
5๐3 + ๐3 = 7 by (3)
โ 7
5๐3 = 7
(3) โ ๐1 =1
5(5) = 1
(2) โ [โต ๐1 = ๐2]
Hence, the Eigen values are 1, 1, 5.
Example: 1.33 If 2, 3 are the Eigen values of [๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
], find the value of a.
Solution:
Let A = [2 0 00 2 0๐ 0 2
]
Let ๐1, ๐2, ๐3 by the Eigenvalues of A.
Given ๐1 = 2, ๐2 = 3
We know that, ๐1 + ๐2 + ๐3 = Sum to the main diagonal elements
โ 2 + 3 + ๐3 = 2 + 2 + 2
โ 5 + ๐3 = 6
โ ๐3 = 1
We know that, ๐1 ๐2 ๐3 = |A|
โ (2)(3)(1) = |2 0 00 2 0๐ 0 2
|
โ 6 = 2(4 โ 0) โ 0(0 โ 0) + 1(0 โ 2๐)
โ 6 = 8 โ 2๐
โ 2๐ = 8 โ 6 = 2
โด ๐ = 1
Example: 1.34 If the Eigen values of A of order ๐ ร ๐ are 2, 3 and 1, then find the Eigen values of
adjoint of A.
Solution:
Given the Eigen values of A are 2, 3, 1
The Eigen values of Aโ1 are 1
2 ,
1
3 , 1,
Formula: adj A = |A| Aโ1
|A| = Product of the Eigenvalues of ๐ด = (2)(3)(1) = 6
โด adj A = 6 Aโ1
โ ๐3 = 5
โ ๐3 = 5
๐2 = 1
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29
โด The Eivenvalues of adj A are 6 (1
2) , 6 (
1
3) , 6 (1), i. e. , 3, 2, 6
Example: 1.35 If ๐,โ๐, โ๐ are the Eigenvalues of the matrix A, then find the Eigenvalues of the matrix
๐๐ โ ๐๐.
Solution:
Given the Eigen values of A are 2,โ1,โ3
The Eigen values of A2 โ 2I are 2,โ1, 7
Since 22 โ 2(1) = 2, (โ1)2 โ 2(1) = โ1, (โ3)2 โ 2(1) = 7
Example: 1.36 What are the Eigen values of the matrix ๐ + ๐๐ if the Eigenvalues of the matrix ๐จ =
[๐ โ๐
โ๐ ๐] are 6 and โ๐? why ?
Solution:
The Eigen values of A are 6 and โ1
The Eigen values of A + 3I are 6 + 3and โ1 + 3
i.e., The Eigen values of A + 3I are 9 and 2
Reason:
If ๐1, ๐2, โฆ ๐๐ are Eigenvalues of A, then
๐1 + ๐, ๐2 + ๐, ๐๐ + ๐ are the Eigenvalues of A + kI, then
Example: 1.37 Find the Eigen values of ๐๐ + ๐๐, where ๐จ = [๐ ๐๐ ๐
]
Solution:
The Eigen values of A are 5 and 2.
The Eigen values of 3A + 2I are 3(5) + 2 and 3(2) + 2
(๐. ๐. ) The Eigen values of 3A +2๐ผ are 17 and 8
Exercise: 1.4
1. If A = [1 2 โ21 0 3
โ2 โ1 3] , then find the sum and product of all Eigenvalues of A.
Ans: Sum = 4; Product = โ13
2. Find the product of the Eigen values of
(a) A = [3 โ4 41 โ2 41 โ1 3
] Ans: Product = โ6
(b) A = [ 1 2 โ2 1 0 3โ2 โ1 โ3
] Ans: Product = โ1
3. If the Eigen values of the matrix ๐ด = [1 1 31 5 13 1 1
] are โ2, 3, 6, then find the Eigenvalues of AT.
Ans: 2, 3, 6,
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30
4. Find the Eigen values of A = [1 3 40 2 50 0 3
] Ans: 1, 2, 3
5. Find the Eigen values of the inverse of the matrix A = [1 3 40 2 50 0 3
] Ans: 1,1
2,1
3,
6. Find the Eigen values of A2 if A = [2 1 00 3 40 0 4
] Ans: 4, 9, 16
7. Obtain the Eigen values of A3where A = [3 21 2
] Ans: 1, 64
1.5 DIAGONALISATION OF A MATRIX BY ORTHOGONAL TRANSFORMATION
Orthogonal matrix
Definition
A matrix โAโ is said to be orthogonal if ๐ด๐ด๐ = ๐ด๐๐ด = ๐ผ
Example: 1.38 Show that the following matrix is orthogonal (๐๐จ๐ฌ๐ ๐ฌ๐ข๐ง๐โ๐ฌ๐ข๐ง๐ ๐๐จ๐ฌ๐
)
Solution:
Let A = (cosฮธ sinฮธโsinฮธ cosฮธ
)
โ AT = (cosฮธ โsinฮธ sinฮธ cosฮธ
)
AAT = (cosฮธ sinฮธโsinฮธ cosฮธ
) (cosฮธ โsinฮธ sinฮธ cosฮธ
)
= ( cos2ฮธ + sin2ฮธ โcosฮธsinฮธ + cosฮธsinฮธโsinฮธcosฮธ + sinฮธcosฮธ sin2ฮธ + cos2ฮธ
)
= (1 00 1
) = I
โด A is orthogonal.
Modal Matrix
Modal matrix is a matrix in which each column specifies the eigenvectors of a matrix .It is
denoted by N.
A square matrix A with linearity independent Eigen vectors can be diagonalized by a similarly
transformation, ๐ท = ๐โ1๐ด๐, where N is the modal matrix .The diagonal matrix D has as its diagonal
elements, the Eigen values of A.
Normalized vector
Eigen vector ๐๐ is said to be normalized if each element of ๐๐ is being divided by the square root
of the sum of the squares of all the elements of ๐๐ .i.e.,the normalized vector is ๐
|๐|
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31
๐๐ = [
๐ฅ1
๐ฅ2
๐ฅ3
] ; Normalized vector of ๐๐ = [
๐ฅ1/โ๐ฅ12 + ๐ฅ2
2 + ๐ฅ32
๐ฅ2/โ๐ฅ12 + ๐ฅ2
2 + ๐ฅ32
๐ฅ3/โ๐ฅ12 + ๐ฅ2
2 + ๐ฅ32
]
Working rule for diagonalization of a square matrix A using orthogonal reduction:
i) Find all the Eigen values of the symmetric matrix A.
ii) Find the Eigen vectors corresponding to each Eigen value.
iii) Find the normalized modal matrix N having normalized Eigen vectors as its column vectors.
iv) Find the diagonal matrix ๐ท = ๐๐๐ด๐.The diagonal matrix D has Eigen values of A as its
diagonal elements.
Example: 1.39 Diagonalize the matrix (๐ ๐ โ๐๐ ๐ โ๐
โ๐ โ๐ ๐)
Solution:
The characteristic equation is 3 โ s1
2 + s2 โ s3=0
s1 = sum of the main diagonal element
= 2 + 1 + 1 = 4
s2 = sum of the minors of the main diagonalelement
= |1 โ2
โ2 1| + |
2 โ1โ1 1
| + |2 11 1
| = โ3 + 1 + 1 = โ1
s3 = |A| = |2 1 โ11 1 โ2
โ1 โ2 1| = โ4
Characteristic equation is 3 โ 42 โ + 4 = 0
โ = 1, (2 โ 3โ 4) = 0
โ = 1, ( + 1)(โ 4) = 0
โ = โ1, 1, 4
To find the Eigen vectors:
Case (i) When = 1 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (2 โ 1 1 โ1
1 1 โ 1 โ2โ1 โ2 1 โ 1
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
x1 + x2 โ x3 = 0โฆ (1)
x1 + 0x2 โ 2x3 = 0โฆ (2)
โx1 โ 2x2 + 0x3 = 0โฆ(3)
From (1) and (2)
x1
โ2โ0=
x2
โ1+2=
x3
0โ1
x1
โ2=
x2
1=
x3
โ1
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32
X1 = (โ21
โ1)
Case( ii) When = โ1 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (2 + 1 1 โ1
1 1 + 1 โ2โ1 โ2 1 + 1
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
3x1 + x2 โ x3 = 0โฆ (4)
x1 + 2x2 โ 2x3 = 0โฆ (5)
โx1 โ 2x2 + 2x3 = 0โฆ(6)
From (4) and (5)
x1
โ2+2=
x2
โ1+6=
x3
6โ1
x1
0=
x2
5=
x3
5
x1
0=
x2
1=
x3
1
X2 = (011)
Case (iii) When = 4 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (2 โ 4 1 โ1
1 1 โ 4 โ2โ1 โ2 1 โ 4
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โ2x1 + x2 โ x3 = 0โฆ (7)
x1 โ 3x2 โ 2x3 = 0โฆ (8)
โx1 โ 2x2 โ 3x3 = 0โฆ (9)
From (7) and (8)
x1
โ2โ3=
x2
โ1โ4=
x3
6โ1
x1
โ5=
x2
โ5=
x3
5
x1
โ1=
x2
โ1=
x3
1
X3 = (โ1โ11
)
Hence the corresponding Eigen vectors are X1 = (โ21
โ1) ; X2 = (
011) ; X3 = (
โ1โ11
)
To check X1, X2& X3 are orthogonal
X1TX2 = (โ2 1 โ1)(
011) = 0 + 1 โ 1 = 0
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33
X2TX3 = (0 1 1) (
โ1โ11
) = 0 โ 1 + 1 = 0
X3TX1 = (โ1 โ1 1)(
โ21
โ1) = 2 โ 1 โ 1 = 0
Normalized Eigen vectors are
(
โ2
โ61
โ6โ1
โ6)
(
01
โ21
โ2
)
(
โ1
โ3โ1
โ31
โ3)
Normalized modal matrix
N =
(
โ2
โ60
โ1
โ31
โ6
1
โ2
โ1
โ3โ1
โ6
1
โ2
1
โ3)
NT =
(
โ2
โ6
1
โ6
โ1
โ6
01
โ2
1
โ2โ1
โ3
โ1
โ3
1
โ3)
Thus the diagonal matrix D = NTAN
=
(
โ2
โ6
1
โ6
โ1
โ6
01
โ2
1
โ2โ1
โ3
โ1
โ3
1
โ3)
(2 1 โ11 1 โ2
โ1 โ2 1)
(
โ2
โ60
โ1
โ31
โ6
1
โ2
โ1
โ3โ1
โ6
1
โ2
1
โ3)
= (1 0 00 โ1 00 0 4
)
Example: 1.40 Diagonalize the matrix (๐๐ โ๐ โ๐โ๐ ๐ ๐โ๐ โ๐ ๐
)
Solution:
The characteristic equation is 3 โ s1
2 + s2โ s3=0
s1 = sum of the main diagonal element
= 10 + 2 + 5 = 17
s2 = sum of the minors of the main diagonalelement
= |2 3
โ3 5| + |
10 โ5โ5 5
| + |10 โ2โ2 2
| = 1 + 25 + 16 = 42
s3 = |A| = |10 โ2 โ5โ2 2 3โ5 โ3 5
| = 0
Characteristic equation is 3 โ 172 + 42 = 0
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34
โ (2 โ 17+ 42) = 0
โ = 0, 3, 14
To find the Eigen vectors:
Case (i) When = 0 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (10 โ 0 โ2 โ5
โ2 2 โ 0 3โ5 โ3 5 โ 0
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
10x1 โ 2x2 โ 5x3 = 0โฆ (1)
โ2x1 + 2x2 + 3x3 = 0โฆ (2)
โ5x1 + 3x2 + 5x3 = 0โฆ (3)
From (1) and (2)
x1
โ6+10=
x2
10โ30=
x3
20โ4
x1
4=
x2
โ20=
x3
16
x1
1=
x2
โ5=
x3
4
X1 = (1
โ54
)
Case (ii) When = 3 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (10 โ 3 โ2 โ5
โ2 2 โ 3 3โ5 โ3 5 โ 3
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
7x1 โ 2x2 โ 5x3 = 0โฆ (4)
โ2x1 โ x2 + 3x3 = 0โฆ (5)
โ5x1 + 3x2 + 2x3 = 0โฆ (6)
From (4) and (5)
x1
โ6โ5=
x2
10โ21=
x3
โ7โ4
x1
โ11=
x2
โ11=
x3
โ11
x1
1=
x2
1=
x3
1
X2 = (111)
Case (iii) When = 14 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (10 โ 14 โ2 โ5
โ2 2 โ 14 3โ5 โ3 5 โ 14
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
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35
โ4x1 โ 2x2 โ 5x3 = 0โฆ (7)
โ2x1 โ 12x2 + 3x3 = 0โฆ (8)
โ5x1 + 3x2 โ 9x3 = 0โฆ (9)
From (7) and (8)
x1
โ6โ60=
x2
10+12=
x3
48โ4
x1
โ66=
x2
22=
x3
44
x1
โ6=
x2
2=
x3
4
X3 = (โ312
)
Hence the corresponding Eigen vectors are X1 = (1
โ54
) ; X2 = (111) ; X3 = (
โ312
)
To check X1, X2& X3 are orthogonal
X1TX2 = (1 โ5 4)(
111) = 1 โ 5 + 4 = 0
X2TX3 = (1 1 1) (
โ312
) = โ3 + 1 + 2 = 0
X3TX1 = (โ3 1 2)(
1โ54
) = โ3 โ 5 + 8 = 0
Normalized Eigen vectors are
(
1
โ42โ5
โ424
โ42)
(
1
โ31
โ31
โ3)
(
โ3
โ141
โ142
โ14)
Normalized modal matrix
N =
(
1
โ42
1
โ3
โ3
โ14
โ5
โ42
1
โ3
1
โ144
โ42
1
โ3
2
โ14)
NT =
(
1
โ42
โ5
โ42
4
โ421
โ3
1
โ3
1
โ3โ3
โ14
1
โ14
2
โ14)
Thus the diagonal matrix D = NTAN
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36
=
(
1
โ42
โ5
โ42
4
โ421
โ3
1
โ3
1
โ3โ3
โ14
1
โ14
2
โ14)
(10 โ2 โ5โ2 2 3โ5 โ3 5
)
(
1
โ42
1
โ3
โ3
โ14
โ5
โ42
1
โ3
1
โ144
โ42
1
โ3
2
โ14)
= (0 0 00 3 00 0 14
)
Example: 1.41 Diagonalize the matrix (๐ โ๐ ๐
โ๐ ๐ โ๐๐ โ๐ ๐
)
Solution:
The characteristic equation is 3 โ s1
2 + s2โ s3=0
s1 = sum of the main diagonal element
= 6 + 3 + 3 = 12
s2 = sum of the minors of the main diagonalelement
= |3 โ1
โ1 3| + |
6 22 3
| + |6 โ2
โ2 3| = 8 + 14 + 14 = 36
s3 = |A| = |6 โ2 2
โ2 3 โ12 โ1 3
| = 32
Characteristic equation is 3 โ 122 + 36โ 32 = 0
โ = 2, (2 โ 10+ 16) = 0
โ = 2,2,8
To find the Eigen vectors:
Case (i) When = 8 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (6 โ 8 โ2 2โ2 3 โ 8 โ12 โ1 3 โ 8
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โ2x1 โ 2x2 + 2x3 = 0โฆ (1)
โ2x1 โ 5x2 โ x3 = 0โฆ(2)
2x1 โ x2 โ 5x3 = 0โฆ (3)
From (1) and (2)
x1
2+10=
x2
โ4โ2=
x3
10โ4
x1
12=
x2
โ6=
x3
6
x1
2=
x2
โ1=
x3
1
X1 = (2
โ11
)
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37
Case (ii) When = 2 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (6 โ 2 โ2 2โ2 3 โ 2 โ12 โ1 3 โ 2
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
4x1 โ 2x2 + 2x3 = 0โฆ (4)
โ2x1 + x2 โ x3 = 0โฆ (5)
2x1 โ x2 + x3 = 0โฆ (6)
Put x1 = 0 โ โ2x2 = โ2x3
x2
1=
x3
1
X2 = (011)
Case (iii) Let X3 = (abc) be a new vector orthogonal to both X1and X2
(๐. ๐) X1TX3 = 0 & X2
TX3 = 0
(2 โ1 1) (abc) = 0 & (0 1 1) (
abc) = 0
2a โ b + c = 0โฆ (7)
a + b + c = 0โฆ (8)
From (7) and (8)
a
โ1โ1=
b
0โ2=
c
2
a
โ2=
b
โ2=
c
2
a
โ1=
b
โ1=
c
1
X3 = (โ1โ11
)
Hence the corresponding Eigen vectors are X1 = (โ211) ; X2 = (
011) ; X3 = (
โ1โ11
)
Normalized Eigen vectors are
(
2
โ6โ1
โ61
โ6)
(
01
โ21
โ2
)
(
โ1
โ3โ1
โ31
โ3)
Normalized modal matrix
Engineering Mathematics -II
38
N =
(
2
โ60
โ1
โ3โ1
โ6
1
โ2
โ1
โ31
โ6
1
โ2
1
โ3)
NT =
(
2
โ6
โ1
โ6
1
โ6
01
โ2
1
โ2โ1
โ3
โ1
โ3
1
โ3)
Thus the diagonal matrix D = NTAN
=
(
2
โ6
โ1
โ6
1
โ6
01
โ2
1
โ2โ1
โ3
โ1
โ3
1
โ3)
(6 โ2 2
โ2 3 โ12 โ1 3
)
(
2
โ60
โ1
โ3โ1
โ6
1
โ2
โ1
โ31
โ6
1
โ2
1
โ3)
= (8 0 00 2 00 0 2
)
Example: 1.42 Diagonalize the matrix (๐ ๐ ๐๐ ๐ โ๐๐ โ๐ ๐
)
Solution:
The characteristic equation is 3 โ s1
2 + s2โ s3=0
s1 = sum of the main diagonal element
= 3 + 3 + 3 = 9
s2 = sum of the minors of the main diagonalelement
= |3 โ1
โ1 3| + |
3 11 3
| + |3 11 3
| = 8 + 8 + 8 = 24
s3 = |A| = |3 1 11 3 โ11 โ1 3
| = 16
Characteristic equation is 3 โ 92 + 24โ 16 = 0
โ = 1, (2 โ 8+ 16) = 0
โ = 1, 4, 4
To find the Eigen vectors:
Case (i) When = 1 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (3 โ 1 1 1
1 3 โ 1 โ11 โ1 3 โ 1
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
2x1 + x2 + x3 = 0โฆ (1)
x1 + 2x2 โ x3 = 0 โฆ(2)
Engineering Mathematics -II
39
x1 โ x2 + 2x3 = 0 โฆ(3)
From (1) and (2)
x1
โ1โ2=
x2
1+2=
x3
4โ1
x1
โ3=
x2
3=
x3
3
x1
โ1=
x2
1=
x3
1
X1 = (โ111
)
Case (ii) When = 4 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (3 โ 4 1 1
1 3 โ 4 โ11 โ1 3 โ 4
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โx1 + x2 + x3 = 0โฆ (4)
x1 โ x2 โ x3 = 0โฆ (5)
x1 โ x2 โ x3 = 0โฆ (6)
put x1 = 0 โ x2 = โx3
x2
1=
x3
โ1
X2 = (0
โ11
)
Case (iii) Let X3 = (abc) be a new vector orthogonal to both X1and X2
(๐. ๐) X1TX3 = 0 & X2
TX3 = 0
(โ1 1 1)(abc) = 0 & (0 โ1 1) (
abc) = 0
โa + b + c = 0โฆ (7)
0a โ b + c = 0โฆ (8)
From (7) and (8)
a
1+1=
b
0+1=
c
1+0
a
2=
b
1=
c
1
X3 = (211)
Hence the corresponding Eigen vectors are X1 = (โ111
) ; X2 = (0
โ1โ1
) ; X3 = (211)
Normalized Eigen vectors are
Engineering Mathematics -II
40
(
โ1
โ31
โ31
โ3)
(
0โ1
โ2โ1
โ2
)
(
2
โ61
โ61
โ6)
Normalized modal matrix
N =
(
โ1
โ30
2
โ61
โ3
โ1
โ2
1
โ61
โ3
โ1
โ2
1
โ6)
NT =
(
โ1
โ3
1
โ3
1
โ3
0โ1
โ2
โ1
โ22
โ6
1
โ6
1
โ6)
Thus the diagonal matrix D = NTAN
=
(
โ1
โ3
1
โ3
1
โ3
0โ1
โ2
โ1
โ22
โ6
1
โ6
1
โ6)
(3 1 11 3 โ11 โ1 3
)
(
โ1
โ30
2
โ61
โ3
โ1
โ2
1
โ61
โ3
โ1
โ2
1
โ6)
= (1 0 00 4 00 0 4
)
Example: 1.43 Reduce the matrix A=(๐ โ๐
โ๐ ๐) to the diagonal form, find ๐๐.
Solution:
The characteristic equation is 2 โ s1+ s2 = 0
s1 = sum of the main diagonal element
= 3+3=6
s2 = |A| = |3 โ1
โ1 3| = 9 โ 1 = 8
Characteristic equation is 2 โ 6+ 8 = 0
= 2,4
To find the Eigen vectors:
Case (i) When = 2 the eigen vector is given by(A โ I)X = 0 where X = (x1
x2)
โ (3 โ 2 โ1โ1 3 โ 2
)(x1
x2)=0
x1 โ x2 = 0
โx1 + x2 = 0
x1
1=
x2
1
Engineering Mathematics -II
41
X1 = (11)
Case (ii)When = 4 the eigen vector is given by(A โ I)X = 0 where X = (x1
x2)
โ (3 โ 4 โ1โ1 3 โ 4
)(x1
x2)=0
โx1 โ x2 = 0
โx1 โ x2 = 0
โ โx1 = x2
x1
1=
x2
โ1
X2 = (1
โ1)
Hence the corresponding Eigen vectors are X1 = (11) ; X2 = (
1โ1
)
To check X1& X2 are orthogonal
X1TX2 = (1 1) (
1โ1
) = 0
X2TX1 = (1 โ1)(
11) = 0
Normalized Eigen vectors are
(
1
โ2โ1
โ2
) (
1
โ21
โ2
)
Normalized modal matrix
N = (
1
โ2
1
โ21
โ2
โ1
โ2
)
NT = (
1
โ2
1
โ21
โ2
โ1
โ2
)
Thus the diagonal matrix D = NTAN
= (
1
โ2
1
โ21
โ2
โ1
โ2
) (3 โ1
โ1 3)(
1
โ2
1
โ21
โ2
โ1
โ2
)
= (2 00 4
)
To find A3: A3 = ND3NT
= (
1
โ2
1
โ21
โ2
โ1
โ2
) (8 00 64
)(
1
โ2
1
โ21
โ2
โ1
โ2
)
= (36 โ28
โ28 36)
Engineering Mathematics -II
42
Example: 1.44 Find the 3ร3 square symmetric matrix A having Eigen values 2,3,6 and corresponding
Eigen vectors(๐ ๐ โ๐)๐, (๐ ๐ ๐)๐, ๐๐ง๐(๐ โ๐ ๐)๐.
Solution:
Given ฮป = 2,3,6; X1 = (10
โ1) ; X2 = (
111) ; X3 = (
1โ21
)
Normalized modal matrix
N =
(
1
โ2
1
โ3
1
โ6
01
โ3
โ2
โ6โ1
โ2
1
โ3
1
โ6)
; NT =
(
1
โ20
โ1
โ21
โ3
1
โ3
1
โ31
โ6
โ2
โ6
1
โ6)
D = (2 0 00 3 00 0 6
)
D = NTAN โ ๐ด = ๐๐ทNT
DNT = (2 0 00 3 00 0 6
)
(
1
โ20
โ1
โ21
โ3
1
โ3
1
โ31
โ6
โ2
โ6
1
โ6)
=
(
2
โ20
โ2
โ23
โ3
3
โ3
3
โ36
โ6
โ12
โ6
6
โ6)
A = NDNT =
(
1
โ2
1
โ3
1
โ6
01
โ3
โ2
โ6โ1
โ2
1
โ3
1
โ6)
(
2
โ20
โ2
โ23
โ3
3
โ3
3
โ36
โ6
โ12
โ6
6
โ6)
= (1 + 1 + 1 0 + 1 โ 2 โ1 + 1 + 10 + 1 โ 2 0 + 1 + 4 0 + 1 โ 2
โ1 + 1 + 1 1 + 0 โ 2 1 + 1 + 1)
= (3 โ1 1
โ1 5 โ11 โ1 3
)
Example: 1.45 Reduce the matrix A = (๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
) to the diagonal form, find ๐๐.
Solution:
The characteristic equation is 3 โ s1
2 + s2 โ s3=0
s1 = sum of the main diagonal element
= 1 + 5 + 1 = 7
s2 = sum of the minors of the main diagonalelement
= |5 11 1
| + |1 33 1
| + |1 11 5
| = 4 โ 8 + 4 = 0
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43
s3 = |A| = |1 1 31 5 13 1 1
| = โ36
Characteristic equation is 3 โ 72 + 36 = 0
โ = 3 ; (2 โ 4โ 12) = 0
โ = โ2,6,3
To find the Eigen vectors:
Case (i) When = โ2 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (1 + 2 1 3
1 5 + 2 13 1 1 + 2
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
3x1 + x2 + 3x3 = 0โฆ (1)
x1 + 7x2 + x3 = 0โฆ (2)
3x1 + x2 + 3x3 = 0โฆ (3)
From (1) and (2)
x1
1โ21=
x2
3โ3=
x3
21โ1
x1
โ20=
x2
0=
x3
20
x1
โ1=
x2
0=
x3
1
X1 = (โ101
)
Case (ii) When = 3 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (1 โ 3 1 3
1 5 โ 3 13 1 1 โ 3
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โ2x1 + x2 + 3x3 = 0โฆ (4)
x1 + 2x2 + x3 = 0โฆ (5)
3x1 + x2 โ 2x3 = 0โฆ (6)
From (4) and (5)
x1
1โ6=
x2
3+2=
x3
โ4โ1
x1
โ5=
x2
5=
x3
โ5
x1
1=
x2
โ1=
x3
1
X2 = (โ111)
Engineering Mathematics -II
44
Case (iii) When = 6 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (1 โ 6 1 3
1 5 โ 6 13 1 1 โ 6
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โ5x1 + x2 + 3x3 = 0โฆ (7)
x1 โ x2 + x3 = 0โฆ (8)
3x1 + x2 โ 5x3 = 0โฆ (9)
From (7) and (8)
x1
1+3=
x2
3+5=
x3
5โ1
x1
4=
x2
8=
x3
4
x1
1=
x2
2=
x3
1
X3 = (121)
Hence the corresponding Eigen vectors are X1 = (โ101
) ; X2 = (โ111) ; X3 = (
121)
To check X1, X2& X3 are orthogonal
X1TX2 = (โ1 0 1)(โ
111) = โ1 + 0 + 1 = 0
X2TX3 = (1 โ1 1)(
121) = 1 โ 2 + 1 = 0
X3TX1 = (1 2 1)(
โ101
) = โ1 + 0 + 1 = 0
Normalized Eigen vectors are
(
โ1
โ2
01
โ2
)
(
1
โ3โ1
โ31
โ3)
(
1
โ62
โ61
โ6)
Normalized modal matrix
N =
(
โ1
โ2
1
โ3
1
โ6
0โ1
โ3
2
โ61
โ2
1
โ3
1
โ6)
NT =
(
โ1
โ20
1
โ21
โ3
โ1
โ3
1
โ31
โ6
2
โ6
1
โ6)
Engineering Mathematics -II
45
Thus the diagonal matrix D = NTAN
=
(
โ1
โ20
1
โ21
โ3
โ1
โ3
1
โ31
โ6
2
โ6
1
โ6)
(1 1 31 5 13 1 1
)
(
โ1
โ2
1
โ3
1
โ6
0โ1
โ3
2
โ61
โ2
1
โ3
1
โ6)
โ D = (โ2 0 00 3 00 0 6
)
โ D3 = (โ8 0 00 27 00 0 216
)
A3 = ND3NT
=
(
โ1
โ2
1
โ3
1
โ6
0โ1
โ3
2
โ61
โ2
1
โ3
1
โ6)
(โ8 0 00 27 00 0 216
)
(
โ1
โ20
1
โ21
โ3
โ1
โ3
1
โ31
โ6
2
โ6
1
โ6)
= (41 63 4963 153 6349 63 41
)
Exercise: 1.5
1. Diagonalise the following using orthogonal transformation
a. A = (5 1 โ11 3 โ1
โ1 โ1 3
) Ans:๐ท(2,3,6)
b. A = (2 2 02 5 00 0 3
) Ans:๐ท(1,3,6)
c. A = (6 โ2 2
โ2 3 โ12 โ1 3
) Ans:๐ท(8,2,2)
d. A = (1 โ1 โ12
โ1 1 โ1โ1 โ1 1
) Ans:๐ท(2,2,โ1)
2. a Reduce A = (3 โ1 1
โ1 5 โ11 โ1 3
) to diagonal form by an orthogonal transformation and also findA4 .
Ans: A4 = (251 โ405 235
โ405 891 โ405235 โ405 251
)
b. Reduce the matrix A=(2 0 40 6 04 0 2
) to the diagonal form, find A3.
Ans: A3 = (104 0 1120 216 0
112 0 104)
Engineering Mathematics -II
46
c. Reduce the matrix A=(1 1 10 2 1
โ4 4 3) to the diagonal form, find A4.
Ans: A4 = (โ99 115 65โ100 116 65โ160 160 81
)
3. Find the 3x3 square symmetric matrix A having Eigen values -2,1, 1 and corresponding Eigen
vectors(โ111
) , (101)and (
12
โ1) . Ans: A = (
0 1 11 0 โ11 โ1 0
)
1.6 REDUCTION OF QUADRATIC FORM TO CANON ICAL FORM BY
ORTHOGONAL TRANSFORMATION
Linear Form
The expression of the form ๐1๐ฅ1 + ๐2๐ฅ2 + โฏ+๐๐๐ฅ๐, where ๐1, ๐2 โฆ๐๐ are constants is called
a linear form in the variables ๐ฅ1, ๐ฅ2 โฆ๐ฅ๐. This linear form can also be written as ๐1๐ฅ1 + ๐2๐ฅ2 + โฏ+๐๐๐ฅ๐ =
โ ๐๐๐ฅ๐๐๐=1
Bilinear Form
Any expression which is linear and homogeneous in each of the sets of variables
{๐ฅ1, ๐ฅ2 โฆโฆโฆโฆ . . ๐ฅ๐}, {๐ฆ1, ๐ฆ2, โฆโฆ . ๐ฆ๐} is called a bilinear form in these variables.
The general bilinear form of the two sets of variables {๐ฅ1, ๐ฅ2, ๐ฅ3}๐๐๐ {๐ฆ1, ๐ฆ2, ๐ฆ3} can be written as
๐(๐ฅ, ๐ฆ) = ๐11๐ฅ1๐ฆ1 + ๐12๐ฅ1๐ฆ2 + ๐13๐ฅ1๐ฆ3+
๐21๐ฅ2๐ฆ1 + ๐22๐ฅ2๐ฆ2 + ๐23๐ฅ2๐ฆ3 + ๐31๐ฅ3๐ฆ1 + ๐32๐ฅ3๐ฆ2 + ๐33๐ฅ3๐ฆ3 โฆ (1)
This bilinear form can written as ๐(๐ฅ, ๐ฆ) = โ โ ๐๐๐๐ฅ๐๐ฆ๐3๐=1
3๐=1
Equation (1) can be written in matrix form as
๐(๐ฅ, ๐ฆ) = (๐ฅ1 ๐ฅ2 ๐ฅ3) (
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
)(
๐ฆ1
๐ฆ2
๐ฆ3
) = ๐โฒ๐ด๐
Where ๐โฒ is the transpose of ๐ = (
๐ฅ1
๐ฅ2
๐ฅ3
) ๐๐๐ ๐ = (
๐ฆ1
๐ฆ2
๐ฆ3
)
The matrix ๐ด = (
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
) is called the matrix of the bilinear form.
Quadratic Form
A homogeneous polynomial of second degree in any number of variables is called a quadratic form.
The general Quadratic form in three variables {๐ฅ1, ๐ฅ2, ๐ฅ3} is given by
๐(๐ฅ1, ๐ฅ2, ๐ฅ3) = ๐11๐ฅ12 + ๐12๐ฅ1๐ฅ2 + ๐13๐ฅ1๐ฅ3+
๐21๐ฅ1๐ฅ2 + ๐22๐ฅ22 + ๐23๐ฅ2๐ฅ3 + ๐31๐ฅ3๐ฅ1 + ๐32๐ฅ2๐ฅ2 + ๐33๐ฅ3
2
This Quadratic form can written as ๐(๐ฅ1, ๐ฅ2, ๐ฅ3) = โ โ ๐๐๐๐ฅ๐๐ฆ๐3๐=1
3๐=1
Engineering Mathematics -II
47
๐(๐ฅ1, ๐ฅ2, ๐ฅ3) = (๐ฅ1 ๐ฅ2 ๐ฅ3) (
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
)(
๐ฅ1
๐ฅ2
๐ฅ3
)
= ๐โฒ๐ด๐
Where ๐ = (
๐ฅ1
๐ฅ2
๐ฅ3
) and A is called the matrix of the Quadratic form.
Note: To write the matrix of a quadratic form as
A = (
coeff. ofx2 1/2coeff. ofxy 1/2coeff. ofxz
1/2coeff. ofxy coeff. ofy2 1/2coeff. ofyz
1/2coeff. of xz 1/2coeff. ofyz coeff. of z2
)
Example: 1.46 Write down the Quadratic form in to matrix form
(i)๐๐ฑ๐ + ๐๐ฒ๐ + ๐๐ฑ๐ฒ
Solution:
A = (coeff. of x2 1/2coeff. of xy
1/2coeff. of xy coeff. ofy2 )
= (2 33 3
)
(ii) ๐๐ฑ๐ + ๐๐ฒ๐ โ ๐๐ณ๐ โ ๐๐ฑ๐ฒ โ ๐ฒ๐ณ + ๐๐ณ๐ฑ
Solution:
A = (
coeff. ofx2 1/2coeff. ofxy 1/2coeff. ofxz
1/2coeff. ofxy coeff. ofy2 1/2coeff. ofyz
1/2coeff. of xz 1/2coeff. ofyz coeff. of z2
)
= (2 โ1 4
โ1 5 โ1/24 โ1/2 โ6
)
(iii) ๐๐ ๐ฑ๐๐ + ๐๐ฑ๐
๐ + ๐๐ฑ๐๐ โ ๐๐ฑ๐๐ฑ๐ + ๐๐ฑ๐๐ฑ๐ โ ๐๐ฑ๐๐ฑ๐
Solution:
A = (
coeff. of x12 1/2coeff. of x1x2 1/2coeff. of x1x3
1/2coeff. of x1x2 coeff. of x22 1/2coeff. of x2x3
1/2coeff. of x1x3 1/2coeff. of x2x3 coeff. of x32
)
= (12 โ4 3โ4 4 โ23 โ2 5
)
Example: 1.47 Write down the matrix form in to Quadratic form
(i)(๐ ๐ โ๐๐ โ๐ ๐
โ๐ โ๐ ๐)
Solution:
Quadratic form is 2 x12 โ 2x2
2 + 6x32 + 2x1x2 โ 6x1x3 + 6x2x3
Engineering Mathematics -II
48
(ii)(๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
)
Solution:
Quadratic form is x12 + 3x2
2 + 6x32 + 2x1x2 + 4x1x3 + 2x2x3.
NATURE OF QUADRATIC FORM DETERMINED BY PRINCIPAL MINORS
Let A be a square matrix of order n say ๐ด = [(
๐11 ๐12 ๐13 โฏ ๐1๐
๐21๐22 ๐23 โฑ ๐2๐
โฆ โฏ โฆ .
)]
The principal sub determinants of A are defined as below.
๐ 1 = ๐11
๐ 2 = |๐11 ๐12
๐21 ๐22|
๐ 3 = |
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
|
. . .
. . .
. . .
๐ ๐ = |๐ด|
The quadratic form ๐ = ๐๐๐ด๐ is said to be
1. Positive definite: If ๐ 1,๐ 2,๐ 3โฆโฆโฆ.๐ ๐ > 0
2. Positive semidefinite: If ๐ 1,๐ 2,๐ 3โฆโฆโฆ.๐ ๐ โฅ 0 and atleast one ๐ ๐ = 0
3. Negative definite: If ๐ 1,๐ 3,๐ 5โฆโฆโฆ. < 0 and๐ 2,๐ 4,๐ 6โฆโฆโฆ. > 0
4. Negative semidefinite: If ๐ 1,๐ 3,๐ 5โฆโฆโฆ. < 0 and๐ 2,๐ 4,๐ 6โฆโฆโฆ. > 0 and atleast one ๐ ๐ = 0
5. Indefinite: In all other cases
Example: 1.48 Determine the nature of the Quadratic form 12๐ฑ๐๐ + ๐๐ฑ๐
๐ + ๐๐๐ฑ๐๐ + ๐๐ฑ๐๐ฑ๐
Solution:
A = (12 1 01 3 00 0 12
)
๐ 1 = ๐11=12 > 0
๐ 2 = |๐11 ๐12
๐21 ๐22| = |
12 11 3
| = 35 > 0
๐ 3 = |
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
| = |12 1 01 3 00 0 12
| = 430 > 0 , Postive definite
Example: 1.49 Determine the nature of the Quadratic form ๐ฑ๐๐ + ๐๐ฑ๐
๐
Solution:
Let A = (1 0 00 2 00 0 0
)
Engineering Mathematics -II
49
๐ 1 = ๐11 = 1 > 0
๐ 2 = |๐11 ๐12
๐21 ๐22| = 2 > 0
๐ 3 = |
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
| = |1 0 00 2 00 0 0
| = 0 , Positive semidefinite
Example: 1.50 Determine the nature of the Quadratic form ๐ฑ๐ โ ๐ฒ๐ + ๐๐ณ๐ + ๐๐ฑ๐ฒ + ๐๐ฒ๐ณ + ๐๐ณ๐ฑ
Solution:
Let A = (1 2 32 โ1 13 1 4
)
๐ 1 = ๐11 = 1 > 0
๐ 2 = |๐11 ๐12
๐21 ๐22| = โ5 < 0
๐ 3 = |
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
| = |1 2 32 โ1 13 1 4
| = 0 , Indefinite
Example: 1.51 Determine the nature of the Quadratic form ๐ฑ๐ฒ + ๐ฒ๐ณ + ๐ณ๐ฑ
Solution:
Let A = (0 1/2 1/2
1/2 0 1/21/2 1/2 0
)
๐ 1 = ๐11 = 0
๐ 2 = |๐11 ๐12
๐21 ๐22| = โ1/4 < 0
๐ 3 = |
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
| = |0 1/2 1/2
1/2 0 1/21/2 1/2 0
| =1
4> 0 , Indefinite
RANK, INDEX AND SIGNATURE OF A REAL QUADRATIC FORMS
Let ๐ = ๐๐๐ด๐ be quadratic form and the corresponding canonical form is ๐1๐ฆ12 + ๐2๐ฆ2
2 + โฏ+๐๐๐ฆ๐2.
The rank of the matrix A is number of non โzero Eigen values of A. If the rank of A is โrโ, the
canonical form of Q will contain only โrโ terms .Some terms in the canonical form may be positive or zero or
negative.
The number of positive terms in the canonical form is called the index(p) of the quadratic form.
The excess of the number of positive terms over the number of negative terms in the canonical
form .i.e,๐ โ (๐ โ ๐) = 2๐ โ ๐ is called the signature of the quadratic form and usually denoted by s. Thus
๐ = 2๐ โ ๐.
Example: 1.52 Reduce the Quadratic form๐ฑ๐๐ + ๐๐ฑ๐
๐ + ๐ฑ๐๐ โ ๐๐ฑ๐๐ฑ๐ + ๐๐ฑ๐๐ฑ๐ + ๐๐ฑ๐๐ฑ๐ to canonical form
through an orthogonal transformation .Find the nature rank, index, signature and also find the non
zero set of values which makes this Quadratic form as zero.
Solution:
Engineering Mathematics -II
50
Given A = (1 โ1 0
โ1 2 10 1 1
)
The characteristic equation is 3 โ s1
2 + s2โ s3=0
s1 = sum of the main diagonal element
= 1 + 2 + 1 = 4
s2 = sum of the minors of the main diagonalelement
= |2 11 1
| + |1 00 1
| + |1 โ1
โ1 2| = 1 + 1 + 1 = 3
s3 = |A| = |1 โ1 0
โ1 2 10 1 1
| = 0
Characteristic equation is 3 โ 42 + 3 = 0
โ = 0 ; (2 โ 4+ 3) = 0
โ = 0,1,3
To find the Eigen vectors:
Case (i) When = 0 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (1 โ1 0
โ1 2 10 1 1
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
x1 โ x2 + 0x3 = 0โฆ (1)
โx1 + 2x2 + x3 = 0โฆ (2)
0x1 + x2 + x3 = 0โฆ (3)
From (1) and (2)
x1
โ1=
x2
โ1=
x3
2โ1
x1
โ1=
x2
โ1=
x3
1
X1 = (โ1โ11
)
Case (ii) When = 3 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (1 โ 3 โ1 0โ1 2 โ 3 10 1 1 โ 3
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โ2x1 โ x2 + 0x3 = 0โฆ (4)
โx1 โ x2 + x3 = 0โฆ (5)
0x1 + x2 โ 2x3 = 0โฆ (6)
From (4) and (5)
x1
โ1=
x2
2=
x3
2โ1
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51
x1
โ1=
x2
2=
x3
1
X2 = (โ121
)
Case (iii) When = 1 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (1 โ 1 โ1 0โ1 2 โ 1 10 1 1 โ 1
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
0x1 โ x2 + 0x3 = 0โฆ (7)
โx1 + x2 + x3 = 0โฆ (8)
0x1 + x2 + 0x3 = 0โฆ (9)
From (7) and (8)
x1
โ1โ0=
x2
0โ0=
x3
0โ1
x1
โ1=
x2
0=
x3
โ1
x1
1=
x2
0=
x3
1
X3 = (101)
Hence the corresponding Eigen vectors are X1 = (11
โ1) ; X2 = (
โ121
) ; X3 = (101)
To check X1, X2& X3 are orthogonal
X1TX2 = (1 1 โ1)(
โ121
) = โ1 + 2 โ 1 = 0
X2TX3 = (โ1 2 1)(
101) = โ1 + 0 + 1 = 0
X3TX1 = (1 0 1) (
11
โ1) = 1 + 0 โ 1 = 0
Normalized Eigen vectors are
(
1
โ31
โ3โ1
โ3)
(
โ1
โ62
โ61
โ6)
(
1
โ2
01
โ2
)
Normalized modal matrix
N =
(
1
โ3
โ1
โ6
1
โ21
โ3
2
โ60
โ1
โ3
1
โ6
1
โ2)
Engineering Mathematics -II
52
NT =
(
1
โ3
1
โ3
โ1
โ3โ1
โ6
2
โ6
1
โ61
โ20
1
โ2)
Thus the diagonal matrix D = NTAN
=
(
1
โ3
โ1
โ6
1
โ21
โ3
2
โ60
โ1
โ3
1
โ6
1
โ2)
(1 โ1 0
โ1 2 10 1 1
)
(
1
โ3
1
โ3
โ1
โ3โ1
โ6
2
โ6
1
โ61
โ20
1
โ2)
D = (0 0 00 1 00 0 3
)
Canonical form = ๐๐DY where Y = (
y1
y2
y3
)
๐๐DY = (y1, y2, y3) (0 0 00 1 00 0 3
)(
y1
y2
y3
)
= 0๐ฆ12 + ๐ฆ2
2 + 3๐ฆ32
Rank = 2
Index = 2
Signature = 2 โ 0 = 2
Nature is positive semi definite.
To find non zero set of values:
Consider the transformation X = NY
(
x1
x2
x3
) =
(
1
โ3
โ1
โ6
1
โ21
โ3
2
โ60
โ1
โ3
1
โ6
1
โ2)
(
y1
y2
y3
)
x1 =y1
โ3โ
y2
โ6+
y3
โ2
x2 =y1
โ3+
2y2
โ6+ 0y3
x3 =โy1
โ3+
y2
โ6+
y3
โ2
Put y2 = 0 & y3 = 0
x1 =y1
โ3; x2 =
y1
โ3; x3 =
โy1
โ3
Put y1 = โ3
x1 = 1; x2 = 1; x3 = โ1 which makes the Quadratic equation zero.
Engineering Mathematics -II
53
Example: 1.53 Reduce the Quadratic form ๐ฑ๐๐ + ๐ฑ๐
๐ + ๐ฑ๐๐ โ ๐๐ฑ๐๐ฑ๐ to canonical form through an
orthogonal transformation .Find the nature rank,index,signature and also find the non zero set of
values which makes this Quadratic form as zero
Solution:
A = (1 โ1 0
โ1 1 00 0 1
)
The characteristic equation is 3 โ s1
2 + s2โ s3=0
s1 = sum of the main diagonal element
= 1 + 1 + 1 = 3
s2 = sum of the minors of the main diagonalelement
= |1 00 1
| + |1 00 1
| + |1 โ1
โ1 1| = 1 + 1 + 0 = 2
s3 = |A| = |1 โ1 0
โ1 1 00 0 1
| = 0
Characteristic equation is 3 โ 32 + 2 = 0
โ = 0 ; (2 โ 3+ 2) = 0
โ = 0,1,2
To find the Eigen vectors:
Case (i) When = 0 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (1 โ1 0
โ1 1 00 0 1
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
x1 โ x2 + 0x3 = 0โฆ (1)
โx1 + x2 + 0x3 = 0โฆ (2)
0x1 + 0x2 + x3 = 0โฆ (3)
From (1) and (2)
x1
1โ0=
x2
0+1=
x3
0
x1
1=
x2
1=
x3
0
X1 = (110)
Case (ii) When = 1 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (1 โ 1 โ1 0โ1 1 โ 1 00 0 1 โ 0
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
0x1 โ x2 + 0x3 = 0โฆ (4)
Engineering Mathematics -II
54
โx1 + 0x2 + 0x3 = 0โฆ(5)
0x1 + 0x2 + 0x3 = 0โฆ (6)
From (4) and (5)
x1
0=
x2
0=
x3
โ1
X2 = (00
โ1)
Case (iii) When = 2 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (1 โ 2 โ1 0โ1 1 โ 2 00 0 1 โ 2
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โx1 โ x2 + 0x3 = 0โฆ (7)
โx1 โ x2 + 0x3 = 0โฆ (8)
0x1 + 0x2 โ x3 = 0โฆ (9)
From (7) and (8)
x1
1โ0=
x2
0โ1=
x3
0โ0
x1
1=
x2
โ1=
x3
0
X3 = (1
โ10
)
Hence the corresponding Eigen vectors are X1 = (110) ; X2 = (
00
โ1) ; X3 = (
1โ10
)
To check X1, X2& X3 are orthogonal
X1TX2 = (1 1 0) (
00
โ1) = 0 + 0 + 0 = 0
X2TX3 = (0 0 โ1)(
1โ10
) = 0 + 0 + 0 = 0
X3TX1 = (1 โ1 0)(
110) = 1 โ 1 + 0 = 0
Normalized Eigen vectors are
(
1
โ21
โ2
0
)(00
โ1)(
1
โ2โ1
โ2
0
)
Normalized modal matrix
Engineering Mathematics -II
55
N = (
1
โ20
1
โ21
โ20
โ1
โ2
0 โ1 0
)
NT = (
1
โ2
1
โ20
0 0 โ11
โ2
1
โ20
)
Thus the diagonal matrix D = NTAN
= (
1
โ2
1
โ20
0 0 โ11
โ2
1
โ20
)(1 โ1 0
โ1 1 00 0 1
)(
1
โ20
1
โ21
โ20
โ1
โ2
0 โ1 0
)
D = (0 0 00 1 00 0 2
)
Canonical form = ๐๐DY where Y = (
y1
y2
y3
)
๐๐DY = (y1, y2, y3) (0 0 00 1 00 0 2
)(
y1
y2
y3
)
= 0๐ฆ12 + ๐ฆ2
2 + 2๐ฆ32
Rank = 2
Index = 2
Signature = 2 - 0 = 2
Nature is positive semi definite.
To find non zero set of values:
Consider the transformation X = NY
(
x1
x2
x3
) = (
1
โ20
1
โ21
โ20
โ1
โ2
0 โ1 0
)(
y1
y2
y3
)
x1 =y1
โ2+ 0 +
y3
โ2
x2 =y1
โ2+ 0 โ
y3
โ2x3 = 0 โ y2 โ 0
Put y2 = 0 & y3 = 0
x1 =y1
โ2; x2 =
y1
โ2; x3 = 0
Put y1 = โ2
x1 = 1; x2 = 1; x3 = 0 which makes the Quadratic equation zero.
Engineering Mathematics -II
56
Example: 1.54 Reduce the Quadratic form ๐๐ฑ๐๐ + ๐ฑ๐
๐ + ๐ฑ๐๐ + ๐๐ฑ๐๐ฑ๐ โ ๐๐ฑ๐๐ฑ๐ โ ๐๐ฑ๐๐ฑ๐ to canonical form
through an orthogonal transformation .Find the nature rank, index, signature
Solution:
A = (2 1 โ11 1 โ2
โ1 โ2 1)
The characteristic equation is 3 โ s1
2 + s2โ s3=0
s1 = sum of the main diagonal element
= 2 + 1 + 1 = 4
s2 = sum of the minors of the main diagonalelement
= |1 โ2
โ2 1| + |
2 โ1โ1 1
| + |2 11 1
| = โ3 + 1 + 1 = โ1
s3 = |A| = |2 1 โ11 1 โ2
โ1 โ2 1| = โ4
Characteristic equation is 3 โ 42 โ + 4 = 0
= โ1,1,4
To find the Eigen vectors:
Case (i) When = โ1 the Eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (2 + 1 1 โ1
1 1 + 1 โ2โ1 โ2 1 + 1
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
3x1 + x2 โ x3 = 0โฆ (1)
x1 + 2x2 โ 2x3 = 0โฆ (2)
โx1 โ 2x2 + 2x3 = 0โฆ (3)
From (1) and (2)
x1
โ2+2=
x2
โ1+6=
x3
6โ1
x1
0=
x2
5=
x3
5
X1 = (011)
Case (ii) When = 1 the Eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (2 โ 1 1 โ1
1 1 โ 1 โ2โ1 โ2 1 โ 1
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
x1 + x2 โ x3 = 0โฆ (4)
x1 + 0x2 โ 2x3 = 0โฆ (5)
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57
โx1 โ 2x2 + 0x3 = 0โฆ (6)
From (4) and (5)
x1
โ2+0=
x2
โ1+2=
x3
0โ1
x1
โ2=
x2
1=
x3
โ1
X2 = (2
โ11
)
Case (iii) When = 4 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (2 โ 4 1 โ1
1 1 โ 4 โ2โ1 โ2 1 โ 4
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โ2x1 + x2 โ x3 = 0โฆ (7)
x1 โ 3x2 โ 2x3 = 0โฆ (8)
โx1 โ 2x2 โ 3x3 = 0โฆ(9)
From (7) and (8)
x1
โ2โ3=
x2
โ1โ4=
x3
6โ1
x1
โ5=
x2
โ5=
x3
5
x1
1=
x2
1=
x3
โ1
X3 = (11
โ1)
Hence the corresponding Eigen vectors are X1 = (011) ; X2 = (
2โ11
) ; X3 = (11
โ1)
To check X1, X2& X3 are orthogonal
X1TX2 = (0 1 1) (
2โ11
) = 0 โ 1 + 1 = 0
X2TX3 = (2 โ1 1)(
11
โ1) = 2 โ 1 โ 1 = 0
X3TX1 = (1 1 โ1)(
011) = 0 + 1 โ 1 = 0
Normalized Eigen vectors are
(
01
โ21
โ2
)
(
2
โ6โ1
โ61
โ6)
(
1
โ31
โ3โ1
โ3)
Normalized modal matrix
Engineering Mathematics -II
58
N =
(
02
โ6
1
โ31
โ2
โ1
โ6
1
โ31
โ2
1
โ6
โ1
โ3)
NT =
(
01
โ2
1
โ22
โ6
โ1
โ6
1
โ61
โ3
1
โ3
โ1
โ3)
Thus the diagonal matrix D = NTAN
=
(
02
โ6
1
โ31
โ2
โ1
โ6
1
โ31
โ2
1
โ6
โ1
โ3)
(2 1 โ11 1 โ2
โ1 โ2 1)
(
01
โ2
1
โ22
โ6
โ1
โ6
1
โ61
โ3
1
โ3
โ1
โ3)
D = (โ1 0 00 1 00 0 4
)
Canonical form = ๐๐DY where Y = (
y1
y2
y3
)
๐๐DY = (y1, y2, y3) (0 0 00 1 00 0 3
)(
y1
y2
y3
)
= โ๐ฆ12 + ๐ฆ2
2 + 4๐ฆ32
Rank = 3
Index = 2
Signature = 2 โ 1 = 1
Nature is indefinite.
Example: 1.55 Reduce the Quadratic form ๐๐ + ๐๐ + ๐๐๐๐๐ โ ๐๐๐ โ ๐๐๐ to canonical form through
an orthogonal transformation .Find the nature rank, index, signature.
Solution:
A = (1 โ1 โ1
โ1 1 โ1โ1 โ1 1
)
The characteristic equation is 3 โ s1
2 + s2โ s3=0
s1 = sum of the main diagonal element
= 1 + 1 + 1 = 3
s2 = sum of the minors of the main diagonalelement
= |1 โ1
โ1 1| + |
1 โ1โ1 1
| + |1 โ1
โ1 1| = 0
s3 = |A| = |1 โ1 โ1
โ1 1 โ1โ1 โ1 1
| = โ4
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Characteristic equation is 3 โ 32 + 4 = 0
= โ1,2,2
To find the Eigen vectors:
Case (i) When = โ1 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (1 + 1 โ1 โ1โ1 1 + 1 โ1โ1 โ1 1 + 1
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
2x1 โ x2 โ x3 = 0โฆ (1)
โx1 + 2x2 โ x3 = 0โฆ (2)
โx1 โ x2 + 2x3 = 0โฆ (3)
From (1) and (2)
x1
1+2=
x2
1+2=
x3
4โ1
x1
3=
x2
3=
x3
3
X1 = (111)
Case (ii) When = 2 the eigen vector is given by(A โ I)X = 0 where X = (
๐ฅ1
๐ฅ2
๐ฅ3
)
โ (1 โ 2 โ1 โ1โ1 1 โ 2 โ1โ1 โ1 1 โ 2
)(
๐ฅ1
๐ฅ2
๐ฅ3
) = (000)
โx1 โ x2 โ x3 = 0โฆ (4)
โx1 โ x2 โ x3 = 0โฆ (5)
โx1 โ x2 โ x3 = 0โฆ (6)
Put x1 = 0 โ โx2 = x3
x2
1=
x3
โ1
X2 = (0
โ11
)
Case (iii) Let X3 = (abc) be a new vector orthogonal to both X1and X2
(๐. ๐) X1TX3 = 0 & X2
TX3 = 0
(1 1 1) (abc) = 0 & (0 โ1 1) (
abc) = 0
a + b + c = 0โฆ (7)
0a โ b + c = 0โฆ (8)
From (7) and (8)
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60
a
1+1=
b
โ1= โ
c
โ1+0
a
2=
b
โ1=
c
โ1
X3 = (2
โ1โ1
)
Hence the corresponding Eigen vectors are X1 = (111) ; X2 = (
0โ11
) ; X3 = (2
โ1โ1
)
Normalized Eigen vectors are
(
1
โ31
โ31
โ3)
(
0โ1
โ21
โ2
)
(
2
โ6โ1
โ6โ1
โ6)
Normalized modal matrix
N =
(
1
โ30
2
โ61
โ3
โ1
โ2
โ1
โ61
โ3
1
โ2
โ1
โ6)
NT =
(
1
โ3
1
โ3
1
โ3
0โ1
โ2
1
โ22
โ6
โ1
โ6
โ1
โ6)
Thus the diagonal matrix D = NTAN
=
(
1
โ3
1
โ3
1
โ3
0โ1
โ2
1
โ22
โ6
โ1
โ6
โ1
โ6)
(1 โ1 โ1
โ1 1 โ1โ1 โ1 1
)
(
1
โ30
2
โ61
โ3
โ1
โ2
โ1
โ61
โ3
1
โ2
โ1
โ6)
D = (โ1 0 00 2 00 0 2
)
Canonical form = ๐๐DY where Y = (
y1
y2
y3
)
๐๐DY = (y1, y2, y3) (0 0 00 1 00 0 3
)(
y1
y2
y3
)
= โ๐ฆ12 + 2๐ฆ2
2 + 2๐ฆ32
Rank = 3
Index = 2
Signature = 2 โ 1 = 1
Nature is indefinite.
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61
Exercise: 1.6
1. Reduce the Quadratic form2x12 + 5x2
2 + 3x32 + 4x1x2 to canonical form through an orthogonal
transformation .Find the nature, rank, index, signature.
Ans: N = (
1
โ50
2
โ52
โ50
1
โ5
0 1 0
) ๐ถ. ๐น = ๐ฆ12 + 3๐ฆ2
2 + 6๐ฆ32
; R = 3, I = 3, S = 3, N: positive definite.
2. Reduce the Quadratic form 3x12 โ 3x2
2 โ 5x32 โ 2x1x2 โ 6x2x3 โ 6x3x1 to canonical form through an
orthogonal transformation .Find the nature, rank, index, signature.
Ans: N =
(
โ3
100
1
โ14
03
โ35
2
โ141
โ10
1
โ35
3
โ14)
๐ถ. ๐น = 4๐ฆ12 โ ๐ฆ2
2 โ 8๐ฆ32 ; R = 3, I = 1,S = - 1; N: Indefinite.
3. Reduce the Quadratic form 3x12 + 2x2
2 + 3x32 โ 2x1x2 โ 2x2x3 to canonical form through an orthogonal
transformation .Find the nature, rank, index, signature
Ans: N =
(
1
โ6
1
โ2
1
โ32
โ60
โ1
โ31
โ6
โ1
โ2
1
โ3)
๐ถ. ๐น = ๐ฆ12 + 3๐ฆ2
2 + 4๐ฆ32 ; R = 3,I = 3,S = 3, N: positive definite.
4. Reduce the Quadratic form 10x12 + 2x2
2 + 5x32 โ 4x1x2 + 6x2x3 โ 10x3x1 to canonical form through an
orthogonal transformation .Find the nature, rank, index, signature.
Ans: N =
(
1
โ42
1
โ3
โ3
โ14
โ5
โ42
1
โ3
1
โ144
โ42
1
โ3
2
โ14)
๐ถ. ๐น = 3๐ฆ22 + 14๐ฆ3
2 ; R = 2, I = 2, S = 2, N: positive semidefinite.
5. Reduce the Quadratic form 5x12 + 26x2
2 + 10x32 + 6x1x2 + 4x2x3 + 14x3x1 to canonical form
through an orthogonal transformation .Find the nature, rank, index, signature.
Ans: N =
(
โ16
โ378
2
โ14
1
โ27
1
โ378
โ1
โ14
5
โ2711
โ378
3
โ14
1
โ27)
๐ถ. ๐น = 14๐ฆ22 + 27๐ฆ3
2 ; R = 2, I = 2, S = 2, N: positive semidefinite.
1.7 CAYLEY HAMILTON THEOREM
Cayley Hamilton Theorem
Statement: Every square matrix satisfies its own characteristic equation.
Uses of Cayley Hamilton Theorem:
To calculate (i) the positive integral power of A and
(ii) the inverse of a non-singular square matrix A.
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62
Problems based on Cayley โ Hamilton theorem
Example: 1.56 Show that the matrix [๐ โ๐๐ ๐
] satisfies its own characteristic equation.
Solution:
Let A = [1 โ22 1
]
The characteristic equation of the given matrix is |A โ ฮปI| = 0
ฮป2 โ S1ฮป + S1 = 0
Where S1 = sum of the main diagonal elements.
= 1 + 1 = 2
S2 = |A| = |1 โ22 1
| = 1 + 4 = 5
โด The characteristic equation is ฮป2 โ 2ฮป + 5 = 0
To prove: A2 โ 2A + 5I = O
A2 = ๐ด๐ด = (1 โ22 1
) (1 โ22 1
) = (โ3 โ44 โ3
)
A2 โ 2A + 5I = (โ3 โ44 โ3
) โ 2 (1 โ22 1
) + 5 (1 00 1
)
= (โ3 โ44 โ3
) + (โ2 4โ4 โ2
) + (5 00 5
)
= (0 00 0
) = O
Therefore, the given matrix satisfies its own characteristic equation.
Example: 1.57 Verify Cayley โ Hamilton theorem find ๐๐and ๐โ๐ when ๐ = [โ๐ โ๐ ๐โ๐ ๐ โ๐ ๐ โ๐ ๐
]
Solution:
The characteristic equation of A is |A โ ฮปI| = 0
i.e., ฮป3 โ S1ฮป2 + S2ฮป = 0 where
S1 = sum of its leading diagonal elements = 2 + 2 + 2 = 6
S2 = sum of the minors of its leading diagonal elements
= |2 โ1
โ1 2| + |
2 21 2
| + |2 โ1
โ1 2|
= (4 โ 1) + (4 โ 2) + (4 โ 1) = 3 + 2 + 3 = 8
S3 = |A| = 2(4 โ 1) + 1(โ2 + 1) + 2(1 โ 2)
= 2(3) + 1(โ1) + 2(โ1) = 6 โ 1 โ 2 = 3
โด The characteristic equation of A is ฮป3 โ S1ฮป2 + S2ฮป โ S3 = 0
๐. ๐. , ฮป3 โ 6ฮป2 + 8ฮป โ 3 = 0
By Cayley-Hamilton theorem
[Every square matrix satisfies its own characteristic equation]
(๐. ๐. ) A3 โ 6A2 + 8A โ 3I = 0 โฆ (1)
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63
Verification:
A2 = ๐ด ร ๐ด = [โ2 โ1 2โ1 2 โ1 1 โ1 2
] [โ2 โ1 2โ1 2 โ1 1 โ1 2
] = [7 โ6 9
โ5 6 โ6 5 โ5 7
]
A3 = ๐ด ร A2 = [โ2 โ1 2โ1 2 โ1 1 โ1 2
] [7 โ6 9
โ5 6 โ6 5 โ5 7
] = [29 โ28 38
โ22 23 โ28 22 โ22 29
]
โด A3 โ 6A2 + 8A โ 3I = [29 โ28 38
โ22 23 โ28 22 โ22 29
] โ 6 [7 โ6 9
โ5 6 โ6 5 โ5 7
]
+8 [โ2 โ1 2โ1 2 โ1 1 โ1 2
] โ 3 [1 0 00 1 00 0 1
]
= [29 โ28 38
โ22 23 โ28 22 โ22 29
] โ [42 โ36 54
โ30 36 โ3630 โ30 42
] + [16 โ8 16โ8 16 โ88 โ8 16
] โ [3 0 00 3 00 0 3
]
= [0 0 00 0 00 0 0
] = O
To find ๐๐:
(1) โ A3 โ 6A2 โ 8A + 3I โฆ (2)
Multiply A on both sides, we get
A4 = 6A3 โ 8A2 + 3A = 6[6A2 โ 8A + 3I ] โ 8A2 + 3A by (2)
= 36A2 โ 48A + 18I โ 8A2 + 3A
A4 = 28A2 โ 45A + 18I โฆ (3)
(1) โ A4 = 28[7 โ6 9
โ5 6 โ6 5 โ5 7
] โ 45 [2 โ1 2
โ1 2 โ1 1 โ1 2
] + 18 [1 0 00 1 00 0 1
]
= [196 โ168 252
โ140 168 โ168 140 โ140 196
] โ [90 โ45 90
โ45 90 โ4545 โ45 90
] + [18 0 00 18 00 0 18
]
= [124 โ123 162โ95 96 โ123 95 โ95 124
]
To find ๐โ๐:
(1) ร Aโ1 โ A2 โ 6A + 8I โ 3 Aโ1 = O
3 Aโ1 = A2 โ 6A + 8I
3 Aโ1 = [7 โ6 9
โ5 6 โ6 5 โ5 7
] โ 6 [2 โ1 2
โ1 2 โ1 1 โ1 2
] + 8 [1 0 00 1 00 0 1
]
= [7 โ6 9
โ5 6 โ6 5 โ5 7
] + [โ12 6 โ12 6 โ12 6โ6 6 โ12
] + [8 0 00 8 00 0 8
]
3 Aโ1 = [ 3 0 โ3 1 2 0โ1 1 3
]โ Aโ1 =1
3[ 3 0 โ3 1 2 0โ1 1 3
]
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Example: 1.58 Find ๐โ๐ if ๐ = [๐ โ๐ ๐๐ ๐ โ๐๐ ๐ โ๐
], using Cayley- Hamilton theorem.
Solution:
The characteristic equation of A is |A โ ฮปI| = 0
(๐. ๐. ) ฮป3 โ S1ฮป2 + S2ฮป โ S3 = 0 where
S1 = sum of its leading diagonal elements
= 1 + 2 + (โ1) = 2
S2 = sum of the minors of its leading diagonal elements
= |2 โ11 โ1
| + |1 42 โ1
| + |1 โ13 2
|
= (โ2 + 1) + (โ1 โ 8) + (2 + 3)
= (โ1) + (โ9) + 5 = โ5
S3 = |A| = |1 โ1 43 2 โ12 1 โ1
|
= 1(โ2 + 1) + 1(โ3 + 2) + (3 โ 4)
= 1(โ1) + 1(โ1) + 4(โ1)
= โ1 โ 1 โ 4 = โ6
โด The Characteristic equation is ฮป3 โ 2ฮป2 โ 5ฮป + 6 = 0
By Cayley Hamilton Theorem we get
[Every square matrix satisfies its own characteristic equation]
โด A3 โ 2A2 โ 5A + 6I = O โฆ (1)
To find ๐โ๐
(1) ร Aโ1 โ A2 โ 2A โ 5I + 6 Aโ1 = O
A2 โ 2A โ 5I + 6 Aโ1 = O
6 Aโ1 = โA2 + 2A + 5I
Aโ1 =1
6[โA2 + 2A + 5I] โฆ (2)
A2 = A ร A
= [1 โ1 43 2 โ12 1 โ1
] [1 โ1 43 2 โ12 1 โ1
]
= [1 โ 3 + 8 โ1 โ 2 + 4 4 + 1 โ 43 + 6 โ 2 โ3 + 4 โ 1 12 โ 2 + 12 + 3 โ 2 โ2 + 2 โ 1 8 โ 1 + 1
] = [6 1 17 0 113 โ1 8
]
โA2 + 2A + 5I = [โ6 โ1 โ1โ7 0 โ11โ3 1 โ8
] + 2 [1 โ1 43 2 โ12 1 โ1
] + 5 [1 0 00 1 00 0 1
]
= [โ6 โ1 โ1โ7 0 โ11โ3 1 โ8
] + [2 โ2 86 4 โ24 2 โ2
] + [5 0 00 5 00 0 5
]
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= [1 โ3 7
โ1 9 โ131 3 โ5
]
From (2) โ Aโ1 =1
6[
1 โ3 7โ1 9 โ131 3 โ5
]
Example: 1.59 If ๐จ = [๐ ๐๐ ๐
], then find ๐๐ง interms of A and I.
Solution:
Let ๐ด = [1 20 2
]
The characteristic equation of A is |A โ ฮปI| = 0
S1 = sum of its leading diagonal elements
= 1 + 2 = 3
S2 = |A| = |1 20 2
| = 2 โ 0 = 2
Therefore, the characteristic equation of A is
ฮป2 โ 3ฮป + 2 = 0
โ (ฮป โ 2)(ฮป โ 1) = 0
โ ฮป = 2, ฮป = 1
Hence, the Eigen values of A are 1, 2.
To find ๐๐ง
When ฮปn is divided by ฮป2 โ 3ฮป + 2
Let the quotient be Q(ฮป) and remainder be ๐ฮป + b.
ฮปn = (ฮป2 โ 3ฮป + 2)Q(ฮป) + ๐ฮป + b ... (1)
when ฮป = 1 when ฮป = 2
1n = ๐ + b 2n = 2๐ + b
2๐ + b = 2n ..... (2)
๐ + b = 2n ..... (3)
Solving (2) and (3), we get
(2) โ (3) โ ๐ = 2n โ 1n
(2) โ 2 ร (3) โ ๐ = โ2n + 2(1n)
i.e., ๐ = 2n โ 1n
๐ = 2(1n) โ 2n
Since, A2 โ 3A + 2I = O by Cayley-Hamilton Theorem
(1) โ An = ๐๐ด + ๐I
An = (2n โ 1n)A + [2(1n) โ 2n]I
Example: 1.60 Use Cayley โ Hamilton theorem to find the value of the matrix given by
(i)๐(๐) = ๐๐ โ ๐๐๐ + ๐๐๐ โ ๐๐๐ + ๐๐ โ ๐๐๐ + ๐๐๐ โ ๐๐ + ๐
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(๐ข๐ข)๐๐ โ ๐๐๐ + ๐๐๐ โ ๐๐๐ + ๐๐๐ โ ๐๐๐ + ๐๐๐ โ ๐๐ + ๐ if the matrix ๐จ = [๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
]
Solution:
The characteristic equation of A is |A โ ฮปI| = 0
ฮป3 โ S1ฮป2 + S2ฮป โ S3 = 0 where
S1 = sum of the main diagonal elements
= 2 + 1 + 2 = 5
S2 = sum of the minors of main diagonal elements
= |1 01 2
| + |2 11 2
| + |2 10 1
|
= (2 โ 0) + (4 โ 1) + (2 โ 0) = 2 + 3 + 2 = 7
= (โ1) + (โ9) + 5 = โ5
S3 = |A| = |2 1 10 1 01 1 2
|
= 2(2 โ 0) โ 1(0 โ 0) + 1(0 โ 1) = 4 โ 1 = 3
Therefore, the characteristic equation is ฮป3 โ 5ฮป2 + 7ฮป โ 3 = 0
By C โ H theorem, we get
A3 โ 5A2 + 7A โ 3I = 0โฆ (1)
Let i) ๐(A) = A8 โ 5A7 + 7A6 โ 3A5 + A4 โ 5A3 + 8A2 โ 2A + I
ii) g(A) = A8 โ 5A7 + 7A6 โ 3A5 + 8A4 โ 5A3 + 8A2 โ 2A + I
(i) A5 + A
A3 โ 5A2 + 7A โ 3I A8 โ 5A7 + 7A6 โ 3A5 + A4 โ 5A3 + 8A2 โ 2A + I
A8 โ 5A7 + 7A6 โ 3A5
(โ) A4 โ 5A3 + 8A2 โ 2A
A4 โ 5A3 + 7A2 โ 3A
(โ) A2 + A + 1 I
๐(๐ด) = (A3 โ 5A2 + 7A โ 3I )(A2 + A) + A2 + A + I
= ๐ + A2 + A + I by (1)
= A2 + A + I ... (2)
Now, A2 = [2 1 10 1 01 1 2
] [2 1 10 1 01 1 2
]
= [5 4 40 1 04 4 5
]
โด A2 + A + I = [5 4 40 1 04 4 5
] + [2 1 10 1 01 1 2
] + [1 0 00 1 00 0 1
]
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= [8 5 50 3 05 5 8
]
(ii) A5 + 8A + 35I
A3 โ 5A2 + 7A โ 3I A8 โ 5A7 + 7A6 โ 3A5 + A4 โ 5A3 + 8A2 โ 2A + 1 I
A8 โ 5A7 + 7A6 โ 3A5
(โ) 8A4 โ 5A3 + 8A2 โ 2A
8A4 โ 40A3 + 56A2 โ 24A
(โ) 35A3 โ A2 + 22A + 1 I
35A3 โ 175 A2 + 245A โ 105I
(โ) 127 A2 โ 223A + 106 I
g(๐ด) = (A3 โ 5A2 + 7A โ 3I)(A4 + 8A + 35I) + 127A2 โ 223A + 106I
= ๐ + 127A2 โ 223A + 106 I
= 127A2 โ 223A + 106 I
= 127 [5 4 40 1 04 4 5
] โ 223 [2 1 10 1 01 1 2
] + 106 [1 0 00 1 00 0 1
]
g(๐ด) = [295 285 2850 10 0
285 285 295
]
Example: 1.61 Use Cayley Hamilton theorem for the matrix ๐จ = [๐ ๐๐ ๐
] to express as a linear
polynomial in โAโ is ๐๐ โ ๐๐๐ โ ๐๐๐ + ๐๐๐๐ โ ๐ + ๐๐ ๐
Solution:
Given A = [1 42 3
]
The characteristic equation of A is |A โ ฮปI| = 0
i.e., ฮป2 โ S1ฮป + S2 = 0 where
S1 = sum of the main diagonal elements
= 1 + 3 = 4
S2 = |A| = |1 42 3
| = 3 โ 8 = โ5
โด The characteristic equation A is ฮป2 โ 4ฮป โ 5 = 0
By Cayley Hamilton Theorem we get
A2 โ 4A + 5 I = ๐ ... (1)
To find: (i) A5 โ 4A4 โ 7A3 + 11A2 โ A + 10 I
A3 โ 2A + 3I
A2 โ 4A + 5 I = ๐
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A5 โ 4A4 โ 7A3 โ 11A2 โ A โ 10I
A5 โ 4A4 โ 7A3
โ2A3 โ 11A2 โ A
โ2A3 โ 8A2 โ 10A
3A2 โ 11A โ 10I
3A2 โ 12A โ 15I
A + 5I
โด A5 โ 4A4 โ 7A3 + 11A2 โ A + 10 I = (A2 + 4A โ 5I)(A3 โ 2A + 3I) + A + 5I
= 0 + A + 5I = A + 5I by (1)
which is a linear polynomial in A.
Example: 1.62 Using Cayley Hamilton theorem find ๐โ๐ when ๐จ = [๐ ๐ ๐๐ ๐ โ๐๐ โ๐ ๐
]
Solution:
The Characteristic equation of A is |A โ ฮปI| = 0
ฮป3 โ S1ฮป2 + S2ฮป โ S3 = 0 where
S1 = sum of the main diagonal elements = 1 + 1 + 1 = 3
S2 = Sum of the minors of the main diagonal elements.
= |1 โ1
โ1 1| + |
1 31 1
| + |1 02 1
|
= (1 โ 1) + (1 โ 3) + (1 โ 0)
= 0 โ 2 + 1 = โ1
S3 = |A| = |1 0 32 1 โ11 โ1 1
|
= 1(1 โ 1) โ 0(2 + 1) + 3(โ2 โ 1)
= 0 โ 0 + 3(โ3) = โ9
โด The characteristic equation A is ฮป3 โ 3ฮป2 โ ฮป + 9 = 0
By Cayley - Hamilton Theorem every square matrix satisfies its own Characteristic equation
โด A3 โ 3A2 โ A + 9 I = ๐
Aโ1 =โ1
9[A2 โ 3A โ I] ... (1)
A2 = [1 0 32 1 โ11 โ1 1
] [1 0 32 1 โ11 โ1 1
]
= [1 + 0 + 3 0 + 0 โ 3 3 + 0 + 32 + 2 โ 1 0 + 1 + 1 6 โ 1 โ 11 โ 2 + 1 0 โ 1 โ 1 3 + 1 + 1
] = [4 โ3 63 2 40 โ2 5
]
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69
โ3๐ด = [โ3 0 โ9โ6 โ3 3โ3 3 โ3
]
(1) โ ๐ดโ1 =โ1
9 [(
4 โ3 63 2 40 โ2 5
) + (โ3 0 โ9โ6 โ3 โ3โ3 3 โ3
) โ (1 0 00 1 00 0 1
)]
=โ1
9[
0 โ3 โ3โ3 โ2 7โ3 1 1
]
=1
9[0 3 33 2 โ73 โ1 โ1
]
Example: 1.63 Verify Cayley- Hamilton for the matrix A= [๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐
]
Solution :
Given A = [1 3 74 2 31 2 1
]
The characteristic equation A is |๐ด โ ๐๐ผ| = 0
๐3 โ ๐1๐2 + ๐2ฮป๐3 = 0โฏ (1) where
S1 = Sum of the main diagonal elements
= 1 + 2 + 1 = 4
S2 = Sum of the minors of its leading diagonal elements
= |2 32 1
| + |1 71 1
| + |1 34 2
|
= (2 โ 6) + (1 โ 7) + (2 โ 12)
= โ4 โ 6 โ 10 = โ20
๐3 = |๐ด| = |1 3 74 2 31 2 1
|
= 1(2 โ 6) โ 3(4 โ 3) + 7(8 โ 2)
= โ4 โ 3(1) + 7(6)
= โ4 โ 3 + 42 = 35
โด (1) โ ๐3 โ 4๐2 โ 20๐ โ 35 = 0
By Cayley โHamilton theorem
(2) โ ๐ด3 โ 4๐ด2 โ 20๐ด โ 351 = ๐
To find ๐ด2 ๐๐๐ ๐ด3:
๐ด2 = [1 3 74 2 31 2 1
] [1 3 74 2 31 2 1
]
= [1 + 12 + 7 3 + 6 + 14 7 + 9 + 74 + 8 + 3 12 + 4 + 6 28 + 6 + 31 + 8 + 1 3 + 4 + 2 7 + 6 + 1
]
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70
= [20 23 2315 22 3710 9 14
]
๐ด3 = [20 23 2315 22 3710 9 14
] [1 3 74 2 31 2 1
]
= [20 + 92 + 23 60 + 46 + 46 140 + 69 + 2315 + 88 + 37 45 + 44 + 74 105 + 66 + 3710 + 36 + 14 30 + 18 + 28 + 70 + 27 + 14
]
= [135 152 232140 163 20860 76 111
]
๐ด3 โ 4๐ด2 โ 20๐ด โ 35 ๐ผ
= [135 152 232140 163 20860 76 111
] โ 4 [20 23 2315 22 3710 9 14
] โ 20 [1 3 74 2 31 2 1
] โ 35 [1 0 00 1 00 0 1
]
= [135 152 232140 163 20860 76 111
] + [โ80 โ92 โ92โ60 โ88 โ148โ40 โ36 โ56
] + [โ20 โ60 โ140โ80 โ40 โ60โ20 โ40 โ20
] + [โ35 0 00 โ35 00 0 โ35
]
= [0 0 00 0 00 0 0
]
โด The given matrix A satisfies its own characteristic equation.
Hence, cayley Hamilton theorem is verified.
Exercise: 1.8
1.Verify Cayley โ Hamilton Theorem and find its inverse.
(๐) [7 2 โ2
โ6 โ1 26 2 โ1
] ๐๐ง๐: ๐ดโ1 =1
3[โ3 โ2 26 5 โ2
โ6 โ2 5]
(๐) [3 1 1
โ1 5 โ11 โ1 3
] ๐๐ง๐: ๐ดโ1 =1
20[
7 โ2 โ31 4 1
โ2 2 8]
(๐ถ) [1 0 โ22 2 40 0 2
] ๐๐ง๐: ๐ดโ1
[
1 0 1
โ11
2โ2
0 01
2 ]
(๐) [1 24 3
] ๐๐ง๐: ๐ดโ1 =1
5 [โ3 24 โ1
]
(๐) [4 3 12 1 โ21 2 1
] ๐๐ง๐: ๐ดโ1 =1
11 [
5 โ1 โ7โ4 3 103 โ5 โ2
]
(๐) [1 2 โ22 5 โ43 7 โ5
] ๐๐ง๐: ๐ดโ1 = [3 โ4 2
โ2 1 0โ1 โ1 1
]
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71
(๐) [1 3 74 2 31 2 1
] ๐๐ง๐: ๐ดโ1 =1
35[โ4 11 โ5โ1 โ6 256 1 โ10
]
2. If A = [1 2 22 1 22 2 1
] , then prove that ๐ด3 โ 3๐ด2 โ 9๐ด โ 5๐ผ = 0 Hence, find ๐ด4.
๐ด๐ง๐: ๐ด4 = [209 208 208208 209 208208 208 209
]
3. Find ๐ด๐ using Cayley โHamilton theorem, taking A = [7 23 6
] also find A3.
๐๐ง๐: An = [9nโ4n
5] [
7 23 6
] + [9.4nโ4.9n
5] [
1 00 1
] and A3 = [463 266399 330
]
4. Calculate A4 for the matrix A = [2 1 20 2 30 0 5
] ๐๐ง๐: [16 32 577180 16 6090 0 625
]
5. Verify Cayley โHamilton theorem for the matrix (i) A= [3 โ1
โ1 5] (ii) A = [
1 42 3
]
6. Using cayley โHamilton theorem, compute ๐ด3 for A= [1 42 3
] ๐๐ง๐: [41 8442 83
]
7. Given that A = [1 2 10 1 โ13 โ1 1
] , Express ๐ด6 โ 5๐ด5 + 8๐ด4 โ 2๐ด3 โ 9๐ด2 + 35๐ด + 6๐ผ
as a linear polynomial in A, using cayley Hamilton theorem. ๐๐ง๐: 4A +42 ๐ผ
8. Obtain the matrix ๐ด6 โ25๐ด2 + 122๐ด ๐คโ๐๐๐ ๐ด = [0 0 22 1 0
โ1 โ1 3] ๐๐ง๐: [
โ34 0 โ20โ20 โ54 010 10 โ74
]
9. Given that A = [1 0 32 1 โ11 โ1 1
] , compute the value of
(๐ด6 โ 5๐ด5 + 8๐ด4 โ 2๐ด3 โ 9๐ด2 + 31๐ด โ 36 ๐ผ), using Cayley โ Hamilton theorem. ๐๐ง๐: [0 0 00 0 00 0 0
]
10. Find ๐ด๐ , using Cayley Hamilton theorem,when A = [5 31 3
], Hence find ๐ด4.
๐๐ง๐: ๐ด๐ = [6๐โ2๐
4] [5 3
1 3] + [
(3)2๐โ6๐
2] [
1 00 1
] , [976 960320 336
]
11. Find ๐ด๐ , using Cayley Hamilton theorem,when A = [7 32 6
] , Also find A3.
๐๐ง๐: ๐ด๐ = [9๐โ4๐
5] [
7 32 6
] + [9.4๐โ4.9๐
5] [
1 00 1
] , [463 399266 330
]