MATRICES

Introduction:
• The matrix has a long history of application in solving linear equations. They were
known as arrays until the 1800‘s.

• The term “matrix” (Latin for “womb”, derived from mater—mother) was coined by
James Joseph Sylvester in 1850, who understood a matrix as an object giving rise to a
number of determinants today called minors, that is to say, determinants of smaller
matrices that are derived from the original one by removing columns and rows.

• An English mathematician named Cullis was the first to use modern bracket notation
for matrices in 1913 and he simultaneously demonstrated the first significant use of
the notation A = aij to represent a matrix where aij refers to the element found in
the ith row and the jth column.

Basic definition:
 Matrix:
A rectangular arrangement of elements is called as matrix

 Upper triangular Matrix:

A square matrix in which all the elements below the leading diagonal are zero
is called upper triangular matrix
2 3 4
Example: A = (0 1 2)
0 0 3
 Diagonal Matrix:
A square matrix in which all elements except the diagonal elements are zero’s
is called as diagonal matrix.
2 0 0
Example: A = (0 1 0)
0 0 3
 Lower triangular matrix:
A square matrix in which all the elements above the leading diagonal are zero
is called a lower triangular matrix
2 0 0
Example: A = (−2 1 0)
1 4 3

1
 Transpose of a matrix:
If we interchange the rows and columns of the given matrix A then the
resulting matrix is called as transpose of the given matrix. It is denoted by AT (or)A′

 Symmetric Matrix:

A matrix ‘A’ is called symmetric if 𝐴 = 𝐴𝑇

 Non-Symmetric Matrix:

A matrix ‘A’ is called non-symmetric if 𝐴 ≠ 𝐴𝑇

 Skew Symmetric Matrix:

A matrix ‘A’ is called skew symmetric if 𝐴 = − 𝐴𝑇

 Singular Matrix:

A square matrix ‘A’ is called singular if |𝐴| = 0

 Orthogonal matrix:

Matrix A is orthogonal to matrix B, if 𝐵𝑇 𝐴 = 0

 Eigen values and Eigen vectors of a real matrix:

Let A be n × n matrix. Suppose the linear transformation Y = AX transforms

X into a scalar multiple of itself.

AX = λX, X is an invariant vector

Then the unknown scalar λ is known as an eigen value of the matrix A and the
corresponding non-zero vector X as eigen vector.
That is

Let 𝐴 = 𝑎𝑖𝑗 be a square matrix. The roots of the characteristic equation | A -𝜆I| =
0 are called characteristic roots (or) Eigen values (or) latent values of the matrix ‘A’

𝑥1
𝑥2
Let 𝐴 = 𝑎𝑖𝑗 be a square matrix of order n. If there exists a non zero vector 𝑋 = 𝑥3
⋮
[𝑥𝑛 ]

Such that 𝐴𝑋 =  𝑋, then the vector X is called an Eigen vector of A

corresponding to the Eigen value 𝜆.

2
Note:

𝑥1
i) If there exists a non-zero vector 𝑋 = [𝑥 ] for a 2 × 2 matrix A, such that 𝐴𝑋 = 𝜆𝑋 ,
2
then X is called an Eigen vector corresponding to the Eigen value λ
ii) Eigen Values are unique.

iii) Eigen vectors are not unique.

 Characteristic equation:

The Characteristic equation is | 𝐴 − 𝜆𝐼| = 0

Where 𝐴 = given matrix & 𝐼 = identity matrix
That is

 If A is a square matrix of order 2 then characteristic equation is λ2 − s1 λ + s2 = 0,

where s1 = sum of main diagonal elements

s2 = Determinant value of A
 If A is a square matrix of order 3 then characteristic equation is

λ3 − s1 λ2 + s2 λ − s3 = 0,

where s1 = sum of main diagonal elements

s2 = sum of minor of the main diagonal elements

s3 = Determinant value of A
Note:

Characteristic Equation
2nd Order Matrix 3rd Order Matrix
𝜆2 − 𝑆1 𝜆 + 𝑆2 = 0 𝜆3 − 𝑆1 𝜆2 + 𝑆2 𝜆 − 𝑆3 = 0
𝑆1 = sum of main diagonal elements 𝑆1 = sum of main diagonal elements
𝑆2 = | 𝐴 | 𝑆2 = sum of minors of main diagonal elements
𝑆3 = | 𝐴 |

3
Problem:
𝟏 𝟐
1. Find the Characteristic Equation of the matrix ( )
𝟎 𝟐
solution:
1 2
Let 𝐴 = ( )
0 2
Characteristic Equation of A is 𝜆2 − 𝑆1 𝜆 + 𝑆2 = 0
𝑆1 = sum of the main diagonal elements = 1 + 2 = 3
1 2
𝑆2 =Determinant value of ‘A’= |𝐴| = | |= 2−0= 2
0 2
Characteristic Equation is 𝜆2 − 3𝜆 + 2 = 0.

Problem:
𝟐 −𝟑 𝟏
2. Find the Characteristic Equation of the matrix ( 𝟑 𝟏 𝟑)
−𝟓 𝟐 −𝟒
solution:
2 −3 1
Let 𝐴 = ( 3 1 3)
−5 2 −4
⇒ 𝑂 (𝐴 ) = 3
Characteristic Equation of A is 𝜆3 − 𝑆1 𝜆2 + 𝑆2 𝜆 − 𝑆3 = 0
𝑆1 = sum of the main diagonal elements = 2 + 1 + (−4) = −1
𝑆2 = sum of minors of main diagonal elements
1 3 2 1 2 −3
=| |+| |+| |
2 −4 −5 −4 3 1
= (−4 − 6) + (−8 + 5) + (2 + 9) = −2
𝑆3 =Determinant value of A
2 −3 1
= |𝐴 | = | 3 1 3|
−5 2 −4
= 2(−4 − 6) − (−3)(−12 + 15) + 1(6 + 5) = 0
Characteristic Equation is 𝜆3 + 𝜆2 − 2𝜆 = 0.

4
 Eigen Values (Characteristic Roots or Latent Roots):
Let 𝐴 = 𝑎𝑖𝑗 be a square matrix. The roots of the characteristic equation | A -𝜆I| = 0 are
called characteristic roots (or) Eigen values (or) latent values of the matrix ‘A’

 Eigen Vector:
𝑥1
𝑥2
Let 𝐴 = 𝑎𝑖𝑗 be a square matrix of order n. If there exists a non zero vector 𝑋 = 𝑥3
⋮
[𝑥𝑛 ]

Such that 𝐴𝑋 =  𝑋, then the vector X is called an Eigen vector of A corresponding to

the Eigen value 𝜆.

Note:

𝑥1
i) If there exists a non-zero vector X= [ 𝑥2 ] for a 3 × 3 matrix A, such that 𝐴𝑋 = 𝜆𝑋 ,
𝑥3
then X is called an Eigen vector corresponding to the Eigen value λ
ii) Eigen Values are unique.

iii) Eigen vectors are not unique.

 Conditions to use cross multiplication rule & Procedure:

i) Two equations needed.
ii) Two equations should be different.
iii) Procedure

Equations CM rule Procedure

𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 0 𝒙 𝒚 𝒛

𝑙𝑥 + 𝑚𝑦 + 𝑛𝑧 = 0 𝒚 𝑧 𝑥 𝒚

𝑏 𝑐 𝑎 𝑏
𝑚 𝑛 𝑙 𝑚
𝒙 𝒚 𝒛
⇒ = 𝑎 = 𝑎
|𝑏 𝑐 | |𝑐 | | 𝑏|
𝑚 𝑛 𝑛 𝑙 𝑙 𝑚

5
Problems based on non-symmetric matrices with non-repeated Eigen values:
Problem:
𝟑 𝟐
Find the eigen values and eigen vectors of the matrix( ).
𝟏 𝟒
Solution:
3 2
Let A = ( )
1 4
𝑂 (𝐴 ) = 2 .

Then the Characteristic equation is λ2 − s1 λ + s2 = 0,

s1 = sum of main diagonal elements = 3 + 4 = 7

s2 = Determinant value of A
3 2
=| |
1 4
= 12 − 2 = 10

The characteristic equation is λ2 − 7λ + 10 = 0

Solving this equation (λ − 2)(λ − 5) = 0

⇒λ= 2&λ=5

∴ The eigen values are λ = 2 and 5

Eigen vectors:

Consider [A − λI]X = 0
3−λ 2 ] [x1 ] [0]
⇒[ = − − − (1)
1 4 − λ x2 0
Case (i):

Substituting λ = 2 in (1)
3−2 2 x1 0
[ ] [x ] = [ ]
1 4−2 2 0
1 2 x1 0
[ ] [x ] = [ ]
1 2 2 0
x1 + 2x2 = 0 − − − −(2)

and x1 + 2x2 = 0 − − − −(3)

Equation (2) and (3) are same.

So consider x1 + 2x2 = 0

x1 = −2x2
6
 x1 x2
=
−2 1
−2
∴ The eigen vector corresponding to λ = 2 is [ ]
1
Case (ii):

Substituting λ = 5 in (1)
2 ] [x1 ] [0]
[3 − 5 =
1 4 − 5 x2 0
−2 2 x1 0
[ ][ ] = [ ]
1 −1 x2 0
−2x1 + 2x2 = 0 ⇒ x1 − x2 = 0 − − − (4)

and x1 − x2 = 0 − − − −(5)
Equation (4) and (5) are same.

consider x1 − x2 = 0

x1 = x2
x1 x2
=
1 1
1
∴ The eigen vector corresponding to λ = 5 is [ ]
1
Result:
 The eigen values are 2 and 5
−2 1
 The eigen vector corresponding to the eigen values are ( ) and ( )
1 1

𝟏 𝟎 −𝟏
1. Find the Eigen values and Eigen vectors of [𝟏 𝟐 𝟏]
𝟐 𝟐 𝟑
Solution:
1 0 −1
Let 𝐴 = [1 2 1]
2 2 3
⇒ 𝑂 (𝐴 ) = 3
Characteristic Equation of A is 𝜆3 − 𝑆1 𝜆2 + 𝑆2 𝜆 − 𝑆3 = 0
where
𝑆1 = sum of the main diagonal elements =1+2+3=6
𝑆2 = sum of minors of main diagonal elements

7
 2 1 1 −1 1 0
=| |+| |+| |
2 3 2 3 1 2
= (6 − 2) + (3 + 2) + (2 − 0) = 4 + 5 + 2 = 11
𝑆3 =Determinant value of A
1 0 −1
| |
= 𝐴 = |1 2 1 | = 1(6 − 2) − 0(3 − 2 ) + (−1)(2 − 4) = 6
2 2 3
Hence the required Characteristic Equation is 𝜆3 − 6𝜆2 + 11𝜆 − 6 = 0
Eigen values are 𝜆 = 1,2,3
Eigen vectors:
To find the Eigen vectors, solve (𝐴 −  𝐼) 𝑋 = 0.
1 0 −1 1 0 0 𝑥1 0
[(1 2 1 ) − 𝜆 (0 1 0)] [ 2 ] = [0] 𝑥
2 2 3 0 0 1 𝑥3 0
1−𝜆 0 −1 𝑥1 0
[ 1 2−𝜆 1 ] [𝑥 2 ] = [0 ] ⇒ 1
2 2 3 − 𝜆 𝑥3 0
Case (i): 𝝀 = 𝟏
0 0 −1 𝑥1 0
1 ⇒ [1 1 1 ] [ 𝑥 2 ] = [ 0]
2 2 2 𝑥3 0
0𝑥1 + 0𝑥2 − 𝑥3 = 0 ⇒ 2
𝑥1 + 𝑥2 + 𝑥3 = 0 ⇒ 3
2𝑥1 + 2𝑥2 + 2𝑥3 = 0 ⇒ 4
By CM rule,
From 2 & 3
1 𝑥 2 𝑥3 𝑥
we get 0+1 = −1+0 = 0−0
𝑥1 𝑥 𝑥3
= −12 =
1 0

1
Hence, a corresponding Eigen vector 𝑋1 = [−1]
0
Case (ii): 𝝀 = 𝟐
−1 0 −1 𝑥1 0
1 ⇒( 1 0 1 ) [𝑥 2 ] = [ 0 ]
2 2 1 𝑥3 0
−𝑥1 − 𝑥3 = 0 ⇒ 5
𝑥1 + 𝑥3 = 0 ⇒ 6
2𝑥1 + 2𝑥2 + 𝑥3 = 0 ⇒ 7
8
 By CM rule,
From 6 & 7 ,
1 𝑥2 3 𝑥 𝑥
we get 0−2 = 2−1 = 2−0
𝑥1 𝑥2 𝑥3
= =
−2 1 2

2
Hence, a corresponding Eigen vector 𝑋2 = [−1]
−2
Case (iii): 𝝀 = 𝟑
−2 0 −1 𝑥1 0
1 ⇒( 1 −1 𝑥
1 ) [ 2 ] = [0 ]
2 2 0 𝑥3 0
−2𝑥1 + 0𝑥2 − 𝑥3 = 0⇒ 8
𝑥1 − 𝑥2 + 𝑥3 = 0 ⇒ 9
2𝑥1 + 2𝑥2 + 0𝑥3 = 0 ⇒ 10
By CM rule,
From 9 & 10 ,
1 𝑥 2 3 𝑥 𝑥
we get 0−2 = 2−0 = 2+2
𝑥1 𝑥2 𝑥3
= =
−2 2 4

1
Hence, a corresponding Eigen vector𝑋3 = [−1]
−2
𝟏 −𝟏 𝟒
2. Find all the eigen values & eigen vectors of the matrix [𝟑 𝟐 −𝟏]
𝟐 𝟏 −𝟏
Solution:
1 −1 4
Let 𝐴 = [3 2 −1]
2 1 −1
⇒ 𝑂 (𝐴 ) = 3
The Characteristic Equation of A is 𝜆3 − 𝑆1 𝜆2 + 𝑆2 𝜆 − 𝑆3 = 0
where
𝑆1 = sum of the main diagonal elements = 1 + 2 − 1 = 2
𝑆2 = sum of minors of main diagonal elements
2 −1 1 4 1 −1
=| |+| |+| |
1 −1 2 −1 3 2
= (−2 + 1) + (−1 − 8) + (2 + 3) = −1 − 9 + 5 = −5
𝑆3 =Determinant value of A
9
 1 −1 4
= |𝐴| = |3 2 −1| = 1(−2 + 1) + 1(−3 + 2 ) + 4(3 − 4) = −1 − 1 − 4 = −6
2 1 −1
Hence the required Characteristic Equation is 𝜆3 − 2𝜆2 − 5𝜆 + 6 = 0
Eigen values are 𝜆 = −2, 1, 3
Eigen vectors:
To find the Eigen vectors, solve (𝐴 −  𝐼) 𝑋 = 0.
1 −1 4 1 0 0 𝑥1 0
[(3 2 −1) − 𝜆 (0 1 0)] [𝑥2 ] = [0]
2 1 −1 0 0 1 𝑥3 0
1 − λ −1 4 𝑥1 0
[ 3 2−λ 𝑥
−1 ] [ 2 ] = [0] ⇒ 1
2 1 −1 − λ 𝑥3 0
Case (i): 𝝀 = −𝟐
3 −1 4 𝑥1 0
1 ⇒ [3 4 −1] [𝑥2 ] = [0]
2 1 1 𝑥3 0
3𝑥1 − 𝑥2 + 4𝑥3 = 0 ⇒ 2
3𝑥1 + 4𝑥2 − 𝑥3 = 0 ⇒ 3
2𝑥1 + 𝑥2 + 𝑥3 = 0 ⇒ 4
By CM rule,
From 2 & 3
1 𝑥 2 𝑥
3 𝑥
we get 1−16 = 12+3 = 12+3
𝑥1 𝑥 𝑥
−15
= 152 = 153
𝑥1 𝑥2 𝑥3
= =
−1 1 1

−1
Hence, a corresponding Eigen vector 𝑋1 = [ 1 ]
1
Case (ii): 𝝀 = 𝟏
0 −1 4 𝑥1 0
1 ⇒ (3 1 −1) [𝑥2 ] = [0]
2 1 −2 𝑥3 0
0𝑥1 − 𝑥2 + 4𝑥3 = 0 ⇒ 5
3𝑥1 + 𝑥2 − 𝑥3 = 0 ⇒ 6
2𝑥1 + 𝑥2 − 2𝑥3 = 0 ⇒ 7
By CM rule,
From 6 & 7 ,

10
 1 𝑥2 3𝑥 𝑥
we get 1−4 = 12−0 = 0+3
𝑥1 𝑥 𝑥3
= 122 =
−3 3
𝑥1 𝑥2 𝑥3
= =
−1 4 1

−1
Hence, a corresponding Eigen vector 𝑋2 = [ 4 ]
1
Case (iii): 𝝀 = 𝟑
−2 −1 4 𝑥1 0
1 ⇒( 3 −1 −1 ) [ 𝑥 2 ] = [ 0]
2 1 −4 𝑥3 0
−2𝑥1 − 𝑥2 + 4𝑥3 = 0⇒ 8
3𝑥1 − 𝑥2 − 𝑥3 = 0 ⇒ 9
2𝑥1 + 𝑥2 − 4𝑥3 = 0 ⇒ 10
By CM rule,
From 9 & 10 ,
1 𝑥2 3 𝑥 𝑥
we get 4+1 = −2+12 = 3+2
𝑥1 𝑥 𝑥3
= 102 =
5 5
𝑥1 𝑥2 𝑥3
= =
1 2 1

1
Hence, a corresponding Eigen vector𝑋3 = [2]
1

Problem 2:
𝟑 𝟏 𝟒
Find the eigen values and eigen vectors of the matrix 𝐀 = [𝟎 𝟐 𝟔].
𝟎 𝟎 𝟓
Solution:
3 1 4
Let A = (0 2 6)
0 0 5
The given square matrix of order 3.

Then The characteristic equation is

λ3 − s1 λ2 + s2 λ − s3 = 0

s1 = Sum of main diagonal elements

11
= 3 + 2 + 5 = 10

s2 = Sum of the minors of the main diagonal

2 6 3 4 3 1
=| |+| |+| |
0 5 0 5 0 2
= 10 − 0 + 15 − 0 + 6 − 0 = 31

s3 = Determinant value of A
3 1 4
= |0 2 6|
0 0 5
= 3(10 − 0)

= 30
The characteristic equation is

λ3 − 10λ2 + 31λ − 30 = 0

⇒ λ = 2,3,5
To find Eigen vector

3−λ 1 4 X1
Consider [ 0 2−λ 6 ] [ X2] = 0
0 0 5 − λ X3
Case (i):

When λ = 2 , The system of equation becomes,

x1 + x2 + 4x3 = 0 − − − − − (1)

0x1 + 0x2 + 6x3 = 0 − − − −(2)

0x1 + 0x2 + 3x3 = 0 − − − −(3)
Solving (1) &(2),by using cross multiplication rule
x1 x2 x3
= =
1 4 4 1 1 1
| | | | | |
0 6 6 0 0 0
x1 x2 x3
= =
6 −6 0
x1 x2 x3
⇒ = =
1 −1 0
⇒ x1 = 1 , x2 = −1, x3 = 0
1
The first eigen vector is (−1)
0
12
Case (ii):

When λ = 3 ,The system of equation becomes ,

0x1 + x2 + 4x3 = 0 − − − − − (4)

0x1 − x2 + 6x3 = 0 − − − − − (5)

0x1 + 0x2 + 2x3 = 0 − − − − − (6)

Solving (4) &(5),by using cross multiplication rule
x1 x2 x3
= =
1 4 4 0 0 1
| | | | | |
−1 6 6 0 0 −1
x1 x2 x3
= =
6+4 0 0
x1 x2 x3
⇒ = =
10 0 0
⇒ x1 = 1 , x2 = 0, x3 = 0
1
The second eigen vector is (0)
0
Case (iii):

When λ = 5 ,The system of equation becomes ,

−2x1 + x2 + 4x3 = 0 − − − − − (7)

0x1 − 3x2 + 6x3 = 0 − − − − − (8)

0x1 + 0x2 + 0x3 = 0 − − − − − (9)

Solving (7)&(8) , by using cross multiplication rule
x1 x2 x3
= =
1 4 4 −2 −2 1
| | | | | |
−3 6 6 0 0 −3
x1 x2 x3
= =
6 + 12 0 + 12 6−0
x1 x2 x3
⇒ = =
18 12 6
⇒ x1 = 3 , x2 = 2, x3 = 1
3
The third eigen vector is (2)
1
Conclusion:

13
  The eigen values are 2,3,5
1 1 3
 The corresponding eigen vectors are (−1), (0), (2)
0 0 1

Problem 3:
𝟏 𝟏 −𝟐
Find the eigen values and eigen vectors of the matrix 𝐀 = (−𝟏 𝟐 𝟏 ).
𝟎 𝟏 −𝟏
Solution:
1 1 −2
Let A = (−1 2 1)
0 1 −1
The given square matrix of order 3 .

Then The characteristic equation is

λ3 − s1 λ2 + s2 λ − s3 = 0

s1 = Sum of main diagonal elements

= 1+2−1= 2

s2 = Sum of the minors of the main diagonal

2 1 1 −2 1 1
=| |+| |+| |
1 −1 0 −1 −1 2
= −2 − 1 − 1 − 0 + 2 + 1 = −1

s3 = Determinant value of A
1 1 −2
= |−1 2 1|
0 1 −1
= 1(−2 − 1) − 1(1 − 0) + (−2)(−1 − 0)

= 1(−3) − 1(1) − 2(−1)

= −3 − 1 + 2 = −2
The characteristic equation is

λ3 − 2λ2 − λ + 2 = 0

⇒ λ = −1, 1, 2
To find Eigen vector

14
 1−λ 1 −2 X1
Consider [ −1 2−λ 1 ] [X 2 ] = 0
0 1 −1 − λ X3
Case (i):When 𝛌 = −𝟏 ,
The system of equation becomes,

2x1 + x2 − 2x3 = 0 − − − − − (1)

−x1 + 3x2 + x3 = 0 − − − −(2)

0x1 + x2 + 0x3 = 0 − − − −(3)

Solving (1) &(2),by using cross multiplication rule
x1 x2 x3
= =
1 −2 −2 2 2 1
| | | | | |
3 1 1 −1 −1 3
x1 x2 x3
= =
7 0 7
⇒ x1 = 1 , x2 = 0, x3 = 1
1
The first eigen vector is (0)
1
Case (ii): When 𝛌 = 𝟏 ,
The system of equation becomes ,

0x1 + x2 − 2x3 = 0 − − − − − (4)

−x1 + x2 + x3 = 0 − − − − − (5)
0x1 + x2 − 2x3 = 0 − − − − − (6)
Solving (4) &(5),by using cross multiplication rule
x1 x2 x3
= =
|1 −2| |−2 0 | | 0 1|
1 1 1 −1 −1 1
x1 x2 x3
= =
3 2 1
⇒ x1 = 3 , x2 = 2, x3 = 1
3
The second eigen vector is (2)
1
Case (iii): When 𝛌 = 𝟐 ,
The system of equation becomes ,

15
−x1 + x2 − 2x3 = 0 − − − − − (7)

−x1 + 0x2 + x3 = 0 − − − − − (8)

0x1 + x2 − 3x3 = 0 − − − − − (9)

Solving (7) & (8), by using cross multiplication rule
x1 x2 x3
= =
1 −2 −2 −1| −1 1
| | | | |
0 1 1 −1 −1 0
x1 x2 x3
= =
1 3 1

⇒ x1 = 1 , x2 = 3, x3 = 1
1
The third eigen vector is 3)
(
1
Conclusion:
 The eigen values are -1,1,2
1 3 1
 The corresponding eigen vectors are (0), (2), (3)
1 1 1

Problem 4:
𝟏𝟏 −𝟒 −𝟕
(
Find the Eigen values and Eigen vectors of the following matrices 𝐀 = 𝟕 −𝟐 −𝟓)
𝟏𝟎 −𝟒 −𝟔
Solution:
11 −4 −7
Given A = ( 7 −2 −5)
10 −4 −6
The characteristic equation is λ3 − s1 λ2 + s2 λ − s3 = 0

s1 = Sum of main diagonal elements = 11 − 2 − 6 = 3

s2 = Sum of the minors of the main diagonal

2 −5| |11 −7 11 −4
= |− + |+| |
−4 −6 10 −6 7 −2
= 12 − 20 − 66 + 70 − 22 + 28 = 2
11 −4 −7
s3 = Determinant value of A = | 7 −2 −5|
10 −4 −6

16
= 11(12 − 20) + 4(−42 + 50) − 7(−28 + 20)

= 11(−8) + 4(8) − 7(−8)

= −88 + 32 + 56 = 0

The characteristic equation is λ3 − 3λ2 + 2λ = 0

⇒ λ = 0,1,2
11 − λ −4 −7 X1
To find Eigen vector [ 7 −2 − λ −5 ] [X2 ] = 0
10 −4 −6 − λ X3
Case (i):

When λ = 0 , The system of equation becomes,

11x1 − 4x2 − 7x3 = 0 − − − − − (1)

7x1 − 2x2 − 5x3 = 0 − − − −(2)

10x1 − 4x2 − 6x3 = 0 − − − −(3)

Solving (1) &(2),by using cross multiplication rule
x1 x2 x3
= =
−4 −7 −7 11 11 −4
| | | | | |
−2 −5 −5 7 7 −2
x1 x2 x3
= = ⇒ x1 = 1 , x2 = 1, x3 = 1
6 6 6

1
The first eigen vector is 1)
(
1
Case (ii):

When λ = 1 ,The system of equation becomes ,

10x1 − 4x2 − 7x3 = 0 − − − − − (4)

7x1 − 3x2 − 5x3 = 0 − − − − − (5)

10x1 − 4x2 − 7x3 = 0 − − − − − (6)

Solving (4) &(5),by using cross multiplication rule
x1 x2 x3
= =
−4 −7 −7 10 10 −4
| | | | | |
−3 −5 −5 7 7 −3
x1 x2 x
= = −23 ⇒ x1 = 1 , x2 = −1, x3 = 2
−1 1

17
 1
The second eigen vector is (−1)
2
Case (iii):

When λ = 2 ,The system of equation becomes ,

9x1 − 4x2 − 7x3 = 0 − − − − − (7)

7x1 − 4x2 − 5x3 = 0 − − − − − (8)

10x1 − 4x2 − 8x3 = 0 − − − − − (9)

Solving (7) & (8), by using cross multiplication rule
x1 x2 x3
= =
−4 −7 −7 9 9 −4
| | | | | |
−4 −5 −5 7 7 −4
x1 x x
= −42 = −83 ⇒ x1 = 2 , x2 = 1, x3 = 2
−8

2
The third eigen vector is 1)
(
2
Conclusion:
i) The eigen values are 0,1,2
1 1 2
ii) The corresponding eigen vectors are (1), (−1), (1)
1 2 2

18
 Problems based on non-symmetric matrices with repeated Eigen values:

−𝟐 𝟐 −𝟑
[
1. Find the Eigen values and Eigen vectors of 𝟐 𝟏 −𝟔]
−𝟏 −𝟐 𝟎
solution:
−2 2 −3
Let 𝐴 = [ 2 1 −6]
−1 −2 0
⇒ 𝑂 (𝐴 ) = 3
The Characteristic Equation of A is 𝜆3 − 𝑆1 𝜆2 + 𝑆2 𝜆 − 𝑆3 = 0
where
𝑆1 = sum of the main diagonal elements = - 2+1+0 = -1
𝑆2 = sum of minors of main diagonal elements
1 −6 −2 −3 −2 2
=| |+| |+| |
−2 0 −1 0 2 1
= (0 − 12) + (0 − 3) + (−2 − 4) = −12 − 3 − 6 = − 21
𝑆3 =Determinant value of A
−2 2 −3
= |𝐴 | = | 2 1 −6|
−1 −2 0
= − 2(0 − 12) – 2(0 − 6) + (−3) (−4 + 1) = 45
Hence the required Characteristic Equation is𝜆3 + 𝜆2 − 21𝜆 − 45 = 0
Eigen values are 𝜆 = 5, −3 , − 3
Eigen vectors:
To find the Eigen vectors, solve (𝐴 −  𝐼) 𝑋 = 0.
−2 2 −3 1 0 0 𝑥1 0
[( 2 1 −6) − 𝜆 (0 1 0)] [𝑥2 ] = [0]
−1 −2 0 0 0 1 𝑥3 0
−2 − 𝜆 2 −3 𝑥1 0
( 2 1−𝜆 −6 ) [𝑥2 ] = [0] ⇒ 1
−1 −2 0 − 𝜆 𝑥 3 0
Case (i): 𝜆 = 5
−7 2 −3 𝑥1 0
1 ⇒( 2 −4 𝑥
−6) [ 2 ] = [0]
−1 −2 −5 𝑥3 0
-7𝑥1 + 2𝑥2 − 3𝑥3 = 0 ⇒ 2
2𝑥1 − 4𝑥2 − 6𝑥3 = 0 ⇒ 3
−𝑥1 − 2𝑥2 − 5𝑥3 = 0 ⇒ 4

19
 By CM rule,
From 2 & 3
1 2𝑥 3 𝑥 𝑥
we get −12−12 = −6−42 = 28−4
𝑥1 2 𝑥 𝑥
= −48 = 243
−24

1
Hence, a corresponding Eigen vector 𝑋1 = [ 2 ]
−1
Case (ii): 𝜆 = −3
1 2 −3 𝑥1 0
1 ⇒( 2 4 𝑥
−6) [ 2 ] = [0]
−1 −2 3 𝑥3 0
𝑥1 + 2𝑥2 − 3𝑥3 = 0 ⇒ 5
2𝑥1 + 4𝑥2 − 6𝑥3 = 0 ⇒ 𝑥1 + 2𝑥2 − 3𝑥3 = 0 ⇒ 6
−𝑥1 − 2𝑥2 + 3𝑥3 = 0 ⇒ 𝑥1 + 2𝑥2 − 3𝑥3 = 0 ⇒ 7
Here 5 , 6 & 7 are same.
Consider 𝑥1 + 2𝑥2 − 3𝑥3 = 0 ⇒ 5
put 𝑥1 = 0,
we get 2𝑥2 = 3𝑥3
𝑥2 𝑥3
=
3 2

0
Hence, a corresponding Eigen vector 𝑋2 = [3]
2
Case (iii): 𝜆 = −3
5 ⇒ 𝑥1 + 2𝑥2 − 3𝑥3 = 0
put 𝑥2 = 0,
we get 𝑥1 = 3𝑥3
𝑥1 𝑥3
=
3 1

3
Hence, a corresponding Eigen vector 𝑋3 = [0]
1

𝟔 −𝟔 𝟓
2. Find the Eigen values and Eigen vectors of [𝟏𝟒 −𝟏𝟑 𝟏𝟎]
𝟕 −𝟔 𝟒
Solution:
The Characteristic Equation of A is 𝜆3 − 𝑆1 𝜆2 + 𝑆2 𝜆 − 𝑆3 = 0

20
where
𝑆1 = sum of the main diagonal elements = 6 − 13 + 4 = −3
𝑆2 = sum of minors of main diagonal elements= 3
𝑆3 =Determinant value of A= |𝐴| = −1
Hence the required Characteristic Equation is 𝜆3 + 3𝜆2 + 3𝜆 + 1 = 0
Eigen values are 𝜆 = −1, −1, −1
Eigen vectors:
To find the Eigen vectors, solve (𝐴 − 𝜆𝐼) 𝑋 = 0.
6 −6 5 1 0 0 𝑥1 0
[(14 −13 10) − 𝜆 (0 1 0)] [𝑥2 ] = [0]
7 −6 4 0 0 1 𝑥3 0
6−𝜆 −6 5 𝑥1 0
( 14 −13 − 𝜆 𝑥
10 ) [ 2 ] = [0]
7 −6 4 − 𝜆 𝑥3 0
0 −5 6
Hence, a corresponding Eigen vector is 𝑋1 = [5] 𝑋2 = [ 0 ] &𝑋3 = [7]
6 7 0

Problem :
𝟔 −𝟔 𝟓
Find the Eigen values and Eigen vectors of the matrix 𝐀 = [𝟏𝟒 −𝟏𝟑 𝟏𝟎]
𝟕 −𝟔 𝟒
Solution:
6 −6 5
Given A = [14 −13 10]
7 −6 4
The characteristic equation is λ3 − s1 λ2 + s2 λ − s3 = 0

s1 = Sum of main diagonal elements = 6 − 13 + 4 = −3

s2 = Sum of the minors of the main diagonal

−13 10 5| | 6 −6
=| | + |6 + |
−6 4 7 4 14 −13
= −52 + 60 + 24 − 35 − 78 + 84 = 3
6 −6 5
s3 = Determinant value of A = |14 −13 10|
7 −6 4

21
= 6(−52 + 60) + 6(56 − 70) + 5(−84 + 91) = 48 − 84 + 35

= −1
The characteristic equation is λ3 + 3λ2 + 3λ + 1 = 0

⇒ λ = −1, −1, −1
6−λ −6 5 X1
To find Eigen vector [ 14 −13 − λ 10 ] [X2 ] = 0
7 −6 4 − λ X3
Case (i):

When λ = −1 , The system of equation becomes,

7x1 − 6x2 + 5x3 = 0 − − − − − (1)

14x1 − 12x2 + 10x3 = 0 − − − −(2)

7x1 − 6x2 + 5x3 = 0 − − − −(3)

Here (1),(2)&(3) are same equations

7x1 − 6x2 + 5x3 = 0

x2 x3
Put x1 = 0 ⇒ −6x2 = −5x3 ⇒ =
5 6
0
The first eigen vector is (5)
6
x1 x3
Put x2 = 0 ⇒ 7x1 = −5x3 ⇒ =
−5 7
−5
The second eigen vector is ( 0 )
7
x1 x2
Put x3 = 0 ⇒ 7x1 = 6x2 ⇒ =
6 7
6
The third eigen vector is (7)
0
(Since the matrix is non symmetric, the corresponding eigenvectors X1 , X2 and X3 must be
linearly independent)

Conclusion:
The eigen values are -1,-1,-1
0 −5 6
The corresponding eigen vectors are (5), ( 0 ), (7)
6 7 0

22
Problem :
𝟐 𝟏 𝟎
Find the Eigen values and Eigen vectors of the matrix 𝐀 = [𝟎 𝟐 𝟏]
𝟎 𝟎 𝟐
Solution:
2 1 0
Given A = [0 2 1]
0 0 2
The characteristic equation is λ3 − s1 λ2 + s2 λ − s3 = 0

s1 = Sum of main diagonal elements

= 2+2+2= 6

s2 = Sum of the minors of the main diagonal

2 1 2 0 2 1
=| |+| |+| |
0 2 0 2 0 2
= 4 + 4 + 4 = 12

s3 = Determinant value of A
2 1 0
= |0 2 1|
0 0 2
= 2(4) = 8

The characteristic equation is λ3 − 6λ2 + 12λ − 8 = 0

⇒ λ = 2,2,2
To find Eigen vector

2−λ 1 0 X1
Consider [ 0 2−λ 1 ] [X 2 ] = 0
0 0 2 − λ X3
Case (i):

When λ = 2, The system of equation becomes,

0x1 + x2 + 0x3 = 0 − − − − − (1)

0x1 + 0x2 + 1x3 = 0 − − − −(2)

0x1 + 0x2 + 0x3 = 0 − − − −(3)

Solving (1) and (2),by using cross multiplication rule
x1 x2 x3 x1 x2 x3
1 0 = 0 0 = 0 1 ⇒ = = ⇒ x1 = 1 , x2 = 0, x3 = 0
| | | | | | 1 0 0
0 1 1 0 0 0

23
 1
The first eigen vector is (0)
0

Problems based on Symmetric matrices with non-repeated Eigen values:

𝟕 −𝟐 𝟎
1. Find the Eigen values and Eigen vectors of [−𝟐 𝟔 −𝟐]
𝟎 −𝟐 𝟓
solution:
7 −2 0
Let 𝐴 = [−2 6 −2]
0 −2 5
⇒ 𝑂(𝐴) = 3 and it is symmetric matrix
⇒ Characteristic Equation of A is 𝜆3 − 𝑆1 𝜆2 + 𝑆2 𝜆 − 𝑆3 = 0
where
𝑆1 = sum of the main diagonal elements = 7 + 6 + 5 = 18
𝑆2 = sum of minors of main diagonal elements
6 −2 7 0 7 −2
=| |+| |+| |
−2 5 0 5 −2 6
= (30 − 4) + (35 − 0) + (42 − 4) = 99
𝑆3 =Determinant value of A
7 −2 0
= |𝐴| = |−2 6 −2|
0 −2 5
= 7(30 − 4) + 2(−10 − 0 ) + 0
|𝐴| = 162
Characteristic Equation is 𝜆3 − 18𝜆2 + 99𝜆 − 162 = 0
Eigen values are 𝜆 = 3,6,9
Eigen vectors:
To find the Eigen vectors, solve (𝐴 − 𝜆𝐼) 𝑋 = 0.
7 −2 0 1 0 0 𝑥1 0
[(−2 6 −2) − 𝜆 (0 1 0)] [ 2 ] = [0] 𝑥
0 −2 5 0 0 1 𝑥3 0
7−𝜆 −2 0 𝑥1 0
( −2 6 − 𝜆 𝑥
−2 ) [ 2 ] = [0] ⇒ 1
0 −2 5 − 𝜆 𝑥3 0
Case (i): 𝝀 = 𝟑

24
 4 −2 0 𝑥1 0
1 ⇒ (−2 3 −2) [𝑥2 ] = [0]
0 −2 2 𝑥3 0
4𝑥1 − 2𝑥2 + 0𝑥3 = 0 ⇒ 2
−2𝑥1 + 3𝑥2 − 2𝑥3 = 0 ⇒ 3
0𝑥1 − 2𝑥2 + 2𝑥3 = 0 ⇒ 4
By CM rule,
From 3 & 4
we get
𝑥1 2𝑥 3 𝑥
= 0+4 = 4−0
6−4
𝑥1 𝑥2 𝑥3
= =
1 2 2

1
Hence, a corresponding Eigen vector 𝑋1 = [2]
2
Case (ii): 𝜆 = 6
1 −2 0 𝑥1 0
1 ⇒ (−2 0 −2) [𝑥2 ] = [0]
0 −2 −1 𝑥3 0
𝑥1 − 2𝑥2 + 0𝑥3 = 0 ⇒ 5
−2𝑥1 + 0𝑥2 − 2𝑥3 = 0 ⇒ 6
0𝑥1 − 2𝑥2 − 𝑥3 = 0 ⇒ 7
By CM rule,
From 6 & 7
we get
𝑥1 2𝑥 3 𝑥
= 0−2 = 4−0
0−4
𝑥1 𝑥2 𝑥
= = −23
2 1

2
Hence, a corresponding Eigen vector 𝑋2 = [ 1 ]
−2
Case (iii): 𝜆 = 9
−2 −2 0 𝑥1 0
1 ⇒ (−2 −3 𝑥
−2) [ 2 ] = [0]
0 −2 −4 𝑥3 0
−2𝑥1 − 2𝑥2 + 0𝑥3 = 0 ⇒ 8
−2𝑥1 − 3𝑥2 − 2𝑥3 = 0 ⇒ 9
0𝑥1 − 2𝑥2 − 4𝑥3 = 0 ⇒ 10

25
By CM rule,
From 9 & 10
we get
𝑥1 2𝑥 3 𝑥
= 0−8 = 4−0
12−4
𝑥1 𝑥2 𝑥3
= =
8 −8 4
𝑥1 𝑥2 𝑥3
= =
2 −2 1
2
Hence, a corresponding Eigen vector 𝑋3 = [−2]
1

Problem :
−𝟐 𝟓 𝟒
Find the Eigen values and Eigen vectors of the matrix 𝐀 = [ 𝟓 𝟕 𝟓]
𝟒 𝟓 −𝟐
Solution:
−2 5 4
Given A = [ 5 7 5]
4 5 −2
The characteristic equation is λ3 − s1 λ2 + s2 λ − s3 = 0

s1 = Sum of main diagonal elements

= −2 + 7 − 2 = 3

s2 = Sum of the minors of the main diagonal

7 5 −2 4 −2 5
=| |+| |+| |
5 −2 4 −2 5 7
= −14 − 25 + 4 − 16 − 14 − 25 = −90

𝑠3 = 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴
−2 5 4
=| 5 7 5|
4 5 −2
= −2(−14 − 25) − 5(−10 − 20) + 4(25 − 28)

= −2(−39) − 5(−30) + 4(−3) = 78 + 150 − 12 = 216

The characteristic equation is 𝜆3 − 3𝜆2 − 90𝜆 − 216 = 0

𝜆 = 3, −6,12

26
To find Eigen vector

−2 − 𝜆 5 4 𝑋1
Consider [ 5 7−𝜆 5 ] [𝑋2 ] = 0
4 5 −2 − 𝜆 𝑋3
Case (i):

When 𝜆 = 3 ,the system of equation becomes,

−5𝑥1 + 5𝑥2 + 4𝑥3 = 0 − − − − − (1)

5𝑥1 + 4𝑥2 + 5𝑥3 = 0 − − − − − (2)

4𝑥1 + 5𝑥2 − 5𝑥3 = 0 − − − − − (3)

Solving (1) &(2),by using cross multiplication rule
𝑥1 𝑥2 𝑥3
= =
|5 4| |4 −5| |−5 5|
4 5 5 5 5 4
𝑥1 𝑥2 𝑥3
= =
9 45 −45
𝑥1 𝑥2 𝑥3
⇒ = =
1 5 −5
⇒ 𝑥1 = 1 , 𝑥2 = 5, 𝑥3 = −5
Case (ii):

When 𝜆 = −6 ,the system of equation becomes,

4𝑥1 + 5𝑥2 + 4𝑥3 = 0 − − − − − (4)

5𝑥1 + 13𝑥2 + 5𝑥3 = 0 − − − − − (5)

4𝑥1 + 5𝑥2 + 4𝑥3 = 0 − − − − − (6)

Solving (4) &(5),by using cross multiplication rule
𝑥1 𝑥2 𝑥3
= =
4 4
|5 4| | | |4 5|
13 5 5 5 5 13
𝑥1 𝑥2 𝑥 𝑥 𝑥2 𝑥3
= = 273 ⇒ −11 = = ⇒ 𝑥1 = −1 , 𝑥2 = 0, 𝑥3 = 1
−27 0 0 1

−1
The second eigen vector is ( 0 )
1
Case (iii):

When 𝜆 = 12,the system of equation becomes,

−14𝑥1 + 5𝑥2 + 4𝑥3 = 0 − − − − − (7)

27
 5𝑥1 − 5𝑥2 + 5𝑥3 = 0 − − − − − (8)

4𝑥1 + 5𝑥2 − 14𝑥3 = 0 − − − − − (9)

Solving (7) &(8),by using cross multiplication rule
𝑥1 𝑥2 𝑥3
= =
4 −14
| 5 4| | | |−14 5 |
−5 5 5 5 5 −5
𝑥1 𝑥2 𝑥3 𝑥1 𝑥2 𝑥3
= = ⇒ = = ⇒ 𝑥1 = 1 , 𝑥2 = 2, 𝑥3 = 1
45 90 45 1 2 1

1
The third eigen vector is (2)
1
Conclusion:
 The eigen values are 3,6,-12
1 −1 1
 The corresponding eigen vectors are ( 5 ), ( 0 ), (2)
−5 1 1

Problems based on Symmetric matrices with repeated Eigen values

𝟎 𝟏 𝟏
1. Find the Eigen values and Eigen vectors of [𝟏 𝟎 𝟏]
𝟏 𝟏 𝟎
solution:
0 1 1
Let 𝐴 = [1 0 1]
1 1 0
⇒ 𝑂(𝐴) = 3 and it is symmetric matrix
The Characteristic Equation of A is where 𝜆3 − 𝑆1 𝜆2 + 𝑆2 𝜆 − 𝑆3 = 0
𝑆1 = sum of the main diagonal elements = 0 + 0 + 0 = 0
𝑆2 = sum of minors of main diagonal elements
0 1 0 1 0 1
=| |+| |+| |
1 0 1 0 1 0
= (0 − 1) + (0 − 1) + (0 − 1) = − 3
𝑆3 =Determinant value of A
0 1 1
= |𝐴 | = | 1 0 1| = 0(0 − 1) − 2(0 − 1 ) + 1(1 − 0 ) = 2
1 1 0
Hence the required Characteristic Equation is𝜆3 − 0𝜆2 − 3𝜆 − 2 = 0
Eigen values are 𝜆 = 2, −1 , −1

28
Eigen vectors:
To find the Eigen vectors, solve (𝐴 − 𝜆𝐼) 𝑋 = 0.
0 1 1 1 0 0 𝑥1 0
𝑥
[(1 0 1) − 𝜆 (0 1 0)] [ 2 ] = [0]
1 1 0 0 0 1 𝑥3 0
−𝜆 1 1 𝑥1 0
( 1 −𝜆 1 ) [𝑥2 ] = [0] ⇒ 1
1 1 −𝜆 𝑥3 0
Case (i): 𝜆 = 2
−2 1 1 𝑥1 0
1 ⇒( 1 −2 𝑥
1 ) [ 2 ] = [0 ]
1 1 −2 𝑥3 0
−2𝑥1 + 𝑥2 + 𝑥3 = 0 ⇒ 2
𝑥1 − 2𝑥2 + 𝑥3 = 0 ⇒ 3
𝑥1 + 𝑥2 − 2𝑥3 = 0 ⇒ 4
By CM rule,
From 2 & 3
1 𝑥
2 3 𝑥 𝑥
we get 1+2 = 1+2 = 4−1
𝑥1 𝑥2 𝑥3
= =
1 1 1

1
Hence, a corresponding Eigen vector 𝑋1 = [1]
1
Case (ii): 𝜆 = −1
1 1 1 𝑥1 0
1 ⇒ (1 1 1) [𝑥2 ]=[0]
1 1 1 𝑥3 0
𝑥1 + 𝑥2 + 𝑥3 = 0 ⇒ 5
𝑥1 + 𝑥2 + 𝑥3 = 0 ⇒ 6
𝑥1 + 𝑥2 + 𝑥3 = 0 ⇒ 7
Here 5 , 6 & 7 are same.
5 ⇒ 𝑥1 + 𝑥2 + 𝑥3 = 0
put 𝑥1 = 0 ,
𝑥2 + 𝑥3 = 0
𝑥2 = −𝑥3
𝑥2 𝑥
we get = −13
1

29
 0
Hence, a corresponding Eigen vector 𝑋2 = [ 1 ]
−1

Case (iii): 𝜆 = −1
5 ⇒ 𝑥1 + 𝑥2 + 𝑥3 = 0
Since matrix is symmetric,
𝑙
Let X3=[𝑚] ⇒ X3 is orthogonal to 𝑋1 and 𝑋2
𝑛
⇒ 𝑋3𝑇 𝑋1 = 0 & 𝑋3𝑇 𝑋2 = 0
1 0
[𝑙 𝑚 𝑛 [1] = 0 & [𝑙
] 𝑚 ]
𝑛 [ 1 ]=0
1 −1
⇒𝑙+𝑚+𝑛 =0 ⇒ 8
0𝑙 + 𝑚 − 𝑛 = 0 ⇒ 9
By CM rule,
From 8 & 9 ,
𝑙 𝑚 𝑛
we get −1−1 = 0+1 = 1−0
𝑙 𝑚 𝑛
= =
−2 1 1

2
Hence, a corresponding Eigen vector 𝑋3 = [ −1 ]
−1

30
 Properties of Eigen Values:

i). A square matrix A & its transpose AT have the same Eigen values.
ii).
Matrix Order Eigen Values
A λ1 , λ2 , λ3 , … . λ𝑛
kA kλ1 , 𝑘λ2 , kλ3 , … . 𝑘λ𝑛
𝐴𝑟 𝜆𝑟1 , 𝜆2𝑟 , 𝜆𝑟3, , … . . 𝜆𝑟𝑛,
n
1 1 1 1
𝐴−1 , , ,….
λ1 λ2 λ3 λ𝑛
𝐴 − 𝑘𝐼 λ1 − 𝑘, λ2 − 𝑘, λ3 − 𝑘, … . λ𝑛 − 𝑘
𝑘 = 𝑛𝑜𝑛 − 𝑧𝑒𝑟𝑜 𝑠𝑐𝑎𝑙𝑎𝑟
iii). Sum of the Eigen values of a matrix = Sum of the main diagonal elements of ‘A’
(or)
= Sum of the principle diagonal elements of ‘A’
(or)
= Trace of A
iv). Product of Eigen values of matrix ‘A’ = Det(A) = |A|
v). The Eigen values of a Triangular matrix are just the diagonal elements of the matrix.

Proof of Properties:
The sum of the eigen values of A is equal to the sum of the diagonal elements of A
Proof:
Let the characteristic equation is

𝜆𝑛 − 𝑠1 𝜆𝑛−1 + 𝑠2 𝜆𝑛−2 + ⋯ + (−1)𝑛 𝑠𝑛 = 0 − − − − − (1.2)

Where 𝑠1 is the sum of diagonal elements

Let 𝜆1 , 𝜆2 , 𝜆3 , … 𝜆𝑛 are the roots of A

The sum =𝜆1 + 𝜆2 + 𝜆3 + ⋯ + 𝜆𝑛

𝑐𝑜𝑒𝑓𝑓. 𝑜𝑓 𝜆𝑛−1
=−
𝑐𝑜𝑒𝑓𝑓. 𝑜𝑓 𝜆𝑛
−𝑠1
= −( )
1
= 𝑠1
∴ 𝑠𝑢𝑚 𝑜𝑓 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 = 𝑠𝑢𝑚 𝑜𝑓 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠(𝑇𝑟𝑎𝑐𝑒)

31
2. The Product of eigen values of A is equal to its determinant

Proof:
Let the characteristic equation is

𝜆𝑛 − 𝑠1 𝜆𝑛−1 + 𝑠2 𝜆𝑛−2 + ⋯ + (−1)𝑛 𝑠𝑛 = 0 − − − − − (1.2)

Where 𝑠𝑛 is determinant of A

Let 𝜆1 , 𝜆2 , 𝜆3 , … 𝜆𝑛 are the roots of A

The product=𝜆1 . 𝜆2 . 𝜆3 . … 𝜆𝑛
(−1)𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑐𝑜𝑒𝑓𝑓.
=
𝑐𝑜𝑒𝑓𝑓. 𝑜f 𝜆𝑛
(−1)𝑛 (−1)𝑛 𝑠𝑛
=
1
= (−1)2𝑛 𝑠𝑛

= 𝑠𝑛

∴ 𝑇ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 = 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑜𝑓 𝐴

3. A square matrix A and its transpose have the same eigen values
Proof:
𝑎11 𝑎12
Let 𝐴 = [𝑎 𝑎22 ]
21

𝑎11 𝑎21
𝐴𝑇 = [𝑎 𝑎22 ]
12

The characteristic equation of A is

𝑎 −𝜆 𝑎12
|𝐴 − 𝜆𝐼 | = 0 ⇒ | 11 | = 0 − − − −(1.3)
𝑎21 𝑎22−𝜆

The characteristic equation of 𝐴𝑇 is

|𝐴𝑇 − 𝜆𝐼 | = 0

𝑎 −𝜆 𝑎21
⇒ | 11 | = 0 − − − −(1.4)
𝑎12 𝑎22−𝜆
While expanding equation (1.3) and (1.4) are the same .
Hence proved

32
4. If 𝝀𝟏 , 𝝀𝟐 , 𝝀𝟑 , … , 𝝀𝒏 are eigen values of A then
𝟏 𝟏 𝟏 𝟏
i) The inverse of A has the eigen values 𝝀 , 𝝀 , 𝝀 , … , 𝝀
𝟏 𝟐 𝟑 𝒏

ii) The matrix 𝑨𝟐 has eigen values 𝝀𝟐𝒓

iii) The matrix KA has the eigen values 𝑲𝝀𝒓

Proof:

Let 𝜆𝑟 be an eigen values of A ,then

𝐴𝑋𝑟 = 𝜆𝑟 𝑋𝑟 − − − −(1.5)

Where 𝑋𝑟 is an eigen vector

Premultiply 𝐴−1 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑖𝑛 (1.5)

𝐴−1 (𝐴𝑋𝑟 ) = 𝐴−1 (𝜆𝑟 𝑋𝑟 )

⇒ 𝑋𝑟 = 𝜆𝑟 (𝐴−1 𝑋𝑟 )
1
𝐴−1 𝑋𝑟 = 𝑋
𝜆𝑟 𝑟
1
𝐻𝑒𝑛𝑐𝑒 𝐴−1 =
𝜆𝑟
1 1 1 1
∴ 𝑇ℎ𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝐴 ℎ𝑎𝑠 𝑡ℎ𝑒 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 , , ,…,
𝜆1 𝜆2 𝜆3 𝜆𝑛
ii)

Let 𝜆𝑟 be an eigen values of a 𝑛𝑡ℎ order matrix A with 𝑟 = 1 𝑡𝑜 𝑛.

⇒ 𝐴𝑋𝑟 = 𝜆𝑟 𝑋𝑟 → 1 , where 𝑋𝑟 is an eigen vector
Premultiply by A
𝐴(𝐴𝑋𝑟 ) = 𝐴(𝜆𝑟 𝑋𝑟 )
𝐴2 𝑋𝑟 = 𝜆𝑟 (𝐴𝑋𝑟 )
𝐴2 𝑋𝑟 = 𝜆𝑟 (𝜆𝑟 𝑋𝑟 ) by 1
𝐴2 𝑋𝑟 = 𝜆2𝑟 𝑋𝑟
Hence 𝜆21 , 𝜆22 , 𝜆23 … 𝜆2𝑛 𝑎𝑟𝑒 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝐴2

iii) Let 𝜆𝑟 be an eigen value of A ,then 𝐴𝑋𝑟 = 𝜆𝑟 𝑋𝑟

Where 𝑋𝑟 is an eigen vector

𝐾(𝐴𝑋𝑟 ) = 𝐾(𝜆𝑟 𝑋𝑟 )

(𝐾𝐴)𝑋𝑟 = (𝐾𝜆𝑟 )𝑋𝑟

Hence 𝐾𝜆1 , 𝐾𝜆2 , 𝐾𝜆3 , … . . 𝐾𝜆𝑟 are eigen values

Note :

33
 5. If two or more eigen values of a matrix are equal then the corresponding eigen vectors may
be LI or LD.

6. An eigen vector cannot correspond to two different eigen values.

[Link] eigen values of an orthogonal matrix are +1 or -1.
8. The eigen vector corresponding to distinct eigen values of a real symmetric matrix are
orthogonal.

Problems based on properties of eigenvalues:

𝟐 𝟎 𝟎
1. Find the Eigen values of [𝟏 𝟑 𝟎]
𝟎 𝟒 𝟒
Solution:
2 0 0
Given A=[1 3 0]
0 4 4
Clearly given matrix A is lower triangular matrix
Hence by property,
Eigenvalues are just the main diagonal elements.
Eigenvalues are 2,3,4

−1 1 1
2. Find the sum and product of the Eigen values of the matrix [ 1 −1 1]
1 1 −1
Solution:
Sum of the eigenvalues = sum of the main diagonal elements = − 1 – 1 – 1 = − 3
Product of the eigen values = Determinant value of A
−1 1 1
= |𝐴 | = | 1 −1 1|
1 2 −1
= −1 (1 − 1) − 1(−1 − 1 ) + 1(1 + 1 ) = 4

𝟖 −𝟔 𝟐
3. Find the sum and product of all eigen values of (−𝟔 𝟕 −𝟒)
𝟐 −𝟒 𝟑
Solution:

𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

= 8 + 7 + 3 = 18

34
 8 −6 2
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 = |𝐴| = | −6 7 −4|
2 −4 3
= 8(21 − 16) + 6(−18 + 8) + 2(24 − 14)

= 8(5) + 6(−10) + 2(10)

= 40 − 60 + 20 = 0

𝟏 𝟏 𝟓
(
4. Find the sum and product of all eigen values of 𝟏 𝟓 𝟏)
𝟑 𝟏 𝟏
Solution:

𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

= 1+5+1= 7
1 1 5
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 = |𝐴| = | 1 5 1|
3 1 1
= 1(5 − 1) − 1(1 − 3) + 5(1 − 15)

= 1(4) − 1(−2) + 5(−14)

= 4 + 2 − 70

= 6 − 70 = −64

𝟏 𝟏 𝟏
5. Find the sum and product of all eigen values of (𝟏 𝟐 𝟐)
𝟏 𝟐 𝟑
Solution:

𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

= 1+2+3= 6
1 1 1
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 = |𝐴| = | 1 2 2|
1 2 3
= 1( 6 − 4) − 1(3 − 2) + 1(2 − 2)

= 1( 2) − 1(1) + 1(0)

=2−1=1

35
 𝟐 𝟎 𝟏
6. Find the sum and product of all eigen values of (𝟎 𝟐 𝟎)
𝟏 𝟎 𝟐
Solution:

𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

= 2+2+2= 6
2 0 1
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 = |𝐴| = | 0 2 0|
1 0 2
= 2(4) + 1(−2) = 8 − 2 = 6

7. For a given matrix A of order 3, |𝑨| = 𝟑𝟐 and two of its eigen values of 8 & 2. Find
the
sum of the eigenvalues.
Solution:
The eigenvalues are 𝜆1 , 𝜆2 , 𝜆3 .
Product of the eigenvalues= |𝐴|
𝜆1 𝜆2 𝜆3 = 32
8 × 2 × 𝜆3 = 32
8 × 2 × 𝜆3 = 32
32
𝜆3 =
16
𝜆3 = 2
Sum of the Eigen values = 𝜆1 + 𝜆2 + 𝜆3 = 8 + 2 + 2 = 12.

8. If the sum of two eigen values and trace of a 3X3 matrix A are equal, find the value of
|𝑨|
Solution:
Let 𝜆1 , 𝜆2 , 𝜆3 be the eigen values of the given 3X3 matrix A
Sum of two eigen values = Trace of A
𝜆1 + 𝜆2 =sum of main diagonal elements=sum of the eigen values
𝜆1 + 𝜆2 = 𝜆1 + 𝜆2 +𝜆3
𝜆3 = 0
|𝐴|=Product of the eigenvalues = 𝜆1 𝜆2 𝜆3 = 0

36
 𝟐 𝟐 𝟏
9. Two eigen values of a matrix𝑨 = (𝟏 𝟑 𝟏) 𝒂𝒓𝒆 𝒆𝒒𝒖𝒂𝒍 𝒕𝒐 𝟏 𝒆𝒂𝒄𝒉 . Find the
𝟏 𝟐 𝟐
eigen values of 𝑨&𝑨−𝟏
Solution:
Given two eigen values are equal to 1

𝜆1 = 𝜆2 = 1

𝑠𝑢𝑚 𝑜𝑓 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒 = 𝑠𝑢𝑚 𝑜𝑓 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

𝜆1 + 𝜆2 + 𝜆3 = 2 + 3 + 2 = 7

1 + 1 + 𝜆3 = 7

𝜆3 = 7 − 2 = 5
The eigen values of A is 1,1,5
1
The eigen value of 𝐴−1 𝑖𝑠 1,1, (5)

𝟔 −𝟐 𝟐
10. The product of two eigen values of of the matrix [−𝟐 𝟑 −𝟏] is 16. Find the
𝟐 −𝟏 𝟑
third Eigen value.
Solution:
Let the eigen values of the matrix 𝜆1 , 𝜆2 , 𝜆3 .
Given 𝜆1 𝜆2 = 16.
Wkt , Product of the eigenvalues = |𝐴|
6 −2 2
𝜆1 𝜆2 𝜆3 = |−2 3 −1|
2 −1 3
(16) 𝜆3 = 6(9 − 1) + 2(−6 + 2) + 2(2 − 6)
16 𝜆3 = 32
32
𝜆3 =
16

𝜆3 = 2

𝟔 −𝟐 𝟐
11. The product of the two eigen values of the matrix 𝑨 = (−𝟑 𝟑 −𝟏) 𝒊𝒔 𝟏𝟒 .
𝟐 −𝟏 𝟑
𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒕𝒉𝒊𝒓𝒅 𝒆𝒊𝒈𝒆𝒏 𝒗𝒂𝒍𝒖𝒆
Solution:

37
 By property 𝜆1 𝜆2 𝜆3 = |𝐴|

⇒ |𝐴| = 28
Given that 𝜆1 𝜆2 = 14
|𝐴 | 28
∴ 𝜆3 = = =2
𝜆1 λ2 14

𝟑 𝟏𝟎 𝟓
12. If 𝟐, 𝟐, 𝟑 are the eigen values of the matrix [−𝟐 −𝟑 −𝟒]. Find the eigenvalues of
𝟑 𝟓 𝟕
AT
Solution:
3 10 5
Let 𝐴 = [−2 −3 −4]
3 5 7
Eigen values of A are 2, 2, 3
A square matrix A and its transpose AT have the same eigen values.
Hence the eigenvalues of AT are 2,2,3

𝟑 −𝟏 𝟏
13. Two of the Eigen values of of the matrix [−𝟏 𝟓 −𝟏] are 3 and 6. Find the Eigen
𝟏 −𝟏 𝟑
values of A-1.
Solution:
Let k be the third Eigen value.
Sum of the Eigen values = Sum of the main diagonal elements
3+6+k = 3+5+3
9+k=11
k=2
The Eigen values of A are 2, 3, 6.
Hence by property,
1 1 1
the Eigen values of A-1 are 2 , 3 , 6

𝟓 𝟒
14. Given that 𝑨 = ( ).Find the eigen values of 𝑨𝟐 .
𝟏 𝟐
Solution:

The characteristic equation of A is 𝜆2 − 𝑠1 𝜆 + 𝑆2 = 0

38
𝑠1 = 5 + 2 = 7
5 4|
𝑠2 = | = 10 − 4 = 6
1 2
𝜆2 − 7𝜆 + 6 = 0
The eigen values of A are 1 & 6

By property, 𝐸𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝐴2 𝑎𝑟𝑒 1 𝑎𝑛𝑑 36,

𝟒 𝟏
15. Find the eigen value of 𝟐𝑨𝟐 , 𝒊𝒇 𝑨 = ( )
𝟑 𝟐
Solution:

The characteristic equation of A is 𝜆2 − 𝑠1 𝜆 + 𝑆2 = 0

𝑠1 = 4 + 2 = 6
4 1
𝑠2 = | | =8−3 =5
3 2
𝜆2 − 6𝜆 + 5 = 0
The eigen values are 1&5

By property 𝜆21 𝑎𝑛𝑑 𝜆22 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝐴2

∴ 𝑇ℎ𝑒 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 2𝐴2 𝑎𝑟𝑒 2(1)2 &2(5)2 = 2,50

−𝟏 𝟎 𝟎
16. Given 𝑨 = [ 𝟐 −𝟑 𝟎]. Find the eigen values of 𝑨𝟐
𝟏 𝟒 𝟐
Solution:
The given matrix “A” is a lower triangular matrix.
∴ The eigenvalues of “A” are −1 , −3 , 2
By property,
Eigenvalues of 𝐴2 𝑎𝑟𝑒 (−1)2 , (−3)2 , (2)2
(ie) 1 , 9 , 4

17. If 1 & 2 are the eigen values of 2X2 matrix A, what are the eigen values of 𝑨𝟐 & 𝑨−𝟏 .
Solution:
If 𝜆1 , 𝜆2 , 𝜆3 … . . 𝜆𝑛 are the eigen values of A, then 𝜆1 𝑚 , 𝜆2 𝑚 , 𝜆3 𝑚 … . 𝜆𝑛 𝑚 are the
eigen values of 𝐴𝑚
Given:

39
1 & 2 are the eigenvalues of A.
∴ Eigenvalues of 𝐴2 are 12 𝑎𝑛𝑑 22
ie., Eigenvalues of 𝐴2 are 1 𝑎𝑛𝑑 4
and
1
Eigenvalues of 𝐴−1 are 1 and 2

𝟏 −𝟐
18. If −𝟏 is the eigen value of the matrix 𝑨 = ( ), find the eigen value of 𝑨𝟒
−𝟑 𝟐
using properties.
Solution:
Given 𝜆1 = −1
Let 𝜆2 be the second eigen value
Sum of the eigenvalues = Sum of the main diagonal elements
∴ −1 + 𝜆2 = 1 + 2
𝜆2 = 3 + 1
𝜆2 = 4
∴ The eigenvalues of 𝐴4 are (−1)4 , (4)4
𝑖𝑒., 1, 256
𝟎 𝟎 𝟐
19. Find the eigen values of 𝑨 = [𝟎 −𝟏 𝟒 ]. Also find eigen values of −𝟑𝑨 .
𝟑 𝟏 −𝟓
Solution:
Given matrix “A” is a lower triangular matrix.
∴ The eigenvalues of “A” are 2 , −1 ,3
The eigenvalues of −3𝐴 are −3(2) , −3(−1) , −3(3)
(ie) −6 , 3 , −9
𝟏 𝟏 𝟑
20. If 3 & 6 are eigen values of 𝑨 = [𝟏 𝟓 𝟏], write down all the eigen of 𝑨−𝟏
𝟑 𝟏 𝟏
Solution:
Let 𝜆 be the third eigenvalue
Sum of the eigenvalues = Sum of the main diagonal elements
∴ 3+6+𝜆 = 1+5+1 = 7
𝜆 = −2
∴ The eigenvalues of A are 3, 6, −2

40
 1 1 1
∴ The eigenvalues of 𝐴−1 are , 6 , −2
3

𝟏 𝟐
21. Prove that the eigen values of −𝟑𝑨−𝟏 𝒂𝒓𝒆 𝒕𝒉𝒆 𝒔𝒂𝒎𝒆 𝒂𝒔 𝒕𝒉𝒐𝒔𝒆 𝒐𝒇 𝑨 = ( )
𝟐 𝟏
Solution:
1 2
Let 𝐴 = ( )
2 1
The given square matrix of order 2 .
Then The characteristic equation is

𝜆2 − 𝑠1 𝜆 + 𝑠2 = 0

𝑠1 = 𝑆𝑢𝑚 𝑜𝑓 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 = 1 + 1 = 2

1 2
𝑠2 = 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴 = | | = −3
2 1
𝜆2 − 2𝜆 − 3 = 0

𝜆 = 3, −1
The eigen value of A is 3,-1 ----(1)
𝟏
E𝒊𝒈𝒆𝒏 𝒗𝒂𝒍𝒖𝒆𝒔 𝒐𝒇 𝑨−𝟏 are 𝟑 𝒂𝒏𝒅 − 𝟏
𝟏
E𝒊𝒈𝒆𝒏 𝒗𝒂𝒍𝒖𝒆𝒔 𝒐𝒇 −𝟑𝑨−𝟏 are (−𝟑) , (−𝟑) − 𝟏
𝟑

𝒊. 𝒆. − 𝟏, 𝟑 − − − −(2)

∴ 𝐹𝑟𝑜𝑚 (1)𝑎𝑛𝑑 (2) the eigen values of − 3𝐴−1 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑎s 𝑡ℎ𝑜𝑠𝑒 𝑜𝑓 𝐴

𝟐 𝟑 𝟏
22. Find the eigen value of 𝑨 = ( ) 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒆 𝒆𝒊𝒈𝒆𝒏 𝒗𝒆𝒄𝒕𝒐𝒓 ( ).
𝟎 𝟒 𝟎
Also find the second eigen value.
Solution:

By the definition we have 𝐴𝑋 = 𝜆𝑋

(𝑖. 𝑒. )(𝐴 − 𝜆𝐼 )𝑋 = 0 ⇒ (2 − 𝜆 3 1 0
)( ) = ( )
0 4−𝜆 0 0
2 − 𝜆 = 0 ⇒ 2 = 𝜆.

By property 𝜆1 + 𝜆2 = 𝑎11 + 𝑎22

⇒ 𝜆1 + 𝜆2 = 6
⇒ 𝜆2 = 4.

41
Therefore eigen values of A are 2 and 4

23. If A is an orthogonal matrix. Show that 𝑨−𝟏 𝒊𝒔 𝒂𝒍𝒔𝒐 𝒐𝒓𝒕𝒉𝒐𝒈𝒐𝒏𝒂𝒍 𝒎𝒂𝒕𝒓𝒊𝒙

Solution:
Since A is orthogonal matrix

𝐴−1 = 𝐴𝑇

⇒ 𝐴𝐴𝑇 = 𝐴𝑇 𝐴 = 𝐼

Let 𝐵 = 𝐴−1 .

𝑇𝑜 𝑝𝑟𝑜𝑣𝑒: 𝐵 𝑖𝑠 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙

ie, 𝐵𝐵𝑇 = 𝐵𝑇 𝐵 = 𝐼

𝐵𝐵𝑇 = 𝐴−1 (𝐴−1 )𝑇

= 𝐴𝑇 (𝐴𝑇 )𝑇 = 𝐴𝑇 𝐴 = 𝐼

∴ 𝐴−1 is orthogonal matrix

𝒂 𝟒
24. Find the constant a and b such that the matrix ( ) 𝒉𝒂𝒔 𝟑 𝒂𝒏𝒅 − 𝟐
𝟏 𝒃
𝒂𝒔 𝒊𝒕𝒔 𝒆𝒊𝒈𝒆𝒏 𝒗𝒂𝒍𝒖𝒆𝒔
Solution:

Given: 𝜆1 = 3 & 𝜆2 = −2
By sum of eigen values =sum of diagonal elements

3−2 =𝑎+𝑏

1 =𝑎+𝑏

𝑎+𝑏=1→ 1

𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 = |𝐴|

𝑎 4
3(−2) = | |
1 𝑏
−6 = 𝑎𝑏 − 4

0 = 𝑎𝑏 − 4 + 6

𝑎𝑏 + 2 = 0

𝑎 (1 − 𝑎 ) + 2 = 0

𝑎 − 𝑎2 + 2 = 0

42
−𝑎2 + 𝑎 + 2 = 0

⇒ 𝑎 = 2 & 𝑎 = −1

𝑤ℎ𝑒𝑛 𝑎 = 2 ⇒ 𝑏 = 1 − 𝑎 = 1 − 2 = −1

𝑤ℎe𝑛 𝑎 = −1 ⇒ 𝑏 = 1 − 𝑎 = 1 + 1 = 2,

⇒ a = 2, b = −1 & a = −1, b = 2

𝟑 𝟏 𝟒
(
25. Find the sum of the squares of the eigen values of 𝑨 = 𝟎 𝟐 𝟔)
𝟎 𝟎 𝟓
Solution:
By the property “the eigen values of a upper or lower triangular matrix are the main
diagonal elements”
Eigen values of A=3,2,5

Sum of the squares of the eigen vales of A =9+4+25=38

𝟑 𝟎 𝟎
26. Find the sum of the eigen values of the inverse of 𝑨 = (𝟖 𝟒 𝟎)
𝟔 𝟐 𝟓
Solution:
By the property “the eigen values of a upper or lower triangular matrix are the main diagonal
elements “

Eigen values of A=3,4,5

1 1 1
∴ 𝐸𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴−1 = , ,
3 4 5
1 1 1 47
𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝐴−1 = + + =
3 4 5 60

43
  CAYLEY-HAMILTON THEOREM(CHT):
 Statement:
Every square matrix satisfies its own characteristic equation.

 Uses of Cayley-Hamilton theorem:

(1) To calculate the positive integral powers of A
(2) To calculate the inverse of a square matrix A

Problem :
𝟏 −𝟐
Show that the matrix ( ) satisfies its own characteristic equation
𝟐 𝟏
Solution:
1 −2
Let 𝐴 = ( )
2 1
Characteristics equation is 𝜆2 − 𝑆1 𝜆 + 𝑆2 = 0 → 1

𝑆1 = 𝑆𝑢𝑚 𝑜𝑓 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑡𝑠 = 1 + 1 = 2

𝑆2 = |𝐴| = 1 + 4 = 5

Eq. 1 ⇒ 𝜆2 − 2𝜆 + 5 = 0

By CHT, 𝐴2 − 2𝐴 + 5𝐼 = 0

To prove: 𝐴2 − 2𝐴 + 5𝐼 = 0
1 −2 1 −2 −3 −4
𝐴2 = 𝐴. 𝐴 = ( ).( )=( )
2 1 2 1 4 −3
𝐿. 𝐻. 𝑆 = 𝐴2 − 2𝐴 + 5𝐼
−3 −4 1 −2 1 0
=( )− 2( ) +5( )
4 −3 2 1 0 1
−3 −4 2 −4 5 0 0 0
=( )−( )+( )=( )
4 −3 4 2 0 5 0 0
Therefore, the given matrix satisfies its own characteristic equation

Problem :
𝟏 𝟎
If A=( ), write 𝑨𝟐 in terms of A ad I using Cayley Hamilton theorem.
𝟎 𝟓
Solution:

44
 1 0
Let 𝐴 = ( )
0 5
Characteristics equation is 𝜆2 − 𝑆1 𝜆 + 𝑆2 = 0 → 1

𝑆1 = 𝑆𝑢𝑚 𝑜𝑓 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑡𝑠 = 1 + 5 = 6

𝑆2 = |𝐴| = 5

Eq. 1 ⇒ 𝜆2 − 6𝜆 + 5 = 0

By CHT, 𝐴2 − 6𝐴 + 5𝐼 = 0

𝐴2 = 6𝐴 − 5𝐼

1. Use Cayley-Hamilton Theorem to find (𝑨𝟒 − 𝟒𝑨𝟑 − 𝟓𝑨𝟐 + 𝑨 + 𝟐𝑰) when

𝟏 𝟐
𝑨=( )
𝟒 𝟑
Solution:
1 2
Given 𝐴 = [ ]
4 3
The characteristic equation of A is |𝐴 − 𝜆𝐼 | = 0
(ie) 𝜆2 − 𝑆1 𝜆 + 𝑆2 = 0
𝑆1 = 1 + 3 = 4
𝑆2 = |𝐴| = −5
∴ The characteristic equation of A is𝜆2 − 4𝜆 − 5 = 0 ⇒ 1
By Cayley-Hamilton Theorem we get
1 ⇒ 𝐴2 − 4𝐴 − 5𝐼 = 0 ⇒ 2
∴ 𝐴4 − 4𝐴3 − 5𝐴2 + 𝐴 + 2𝐼 = 𝐴2 (𝐴2 − 4𝐴 − 5𝐼 ) + 𝐴 + 2𝐼
= 𝐴2 (0) + 𝐴 + 2𝐼by 2
= 𝐴 + 2𝐼
1 2 1 0
=[ ]+ 2[ ]
4 3 0 1
1 2 2 0
=[ ]+[ ]
4 3 0 2
3 2
=[ ]
4 5

45
Problem :
𝟐 −𝟏 𝟐
Verify Cayley Hamilton theorem and find 𝑨𝟒 & 𝑨−𝟏 when 𝑨 = (−𝟏 𝟐 −𝟏)
𝟏 −𝟏 𝟐
Solution:
2 −1 2
𝐴 = (−1 2 −1)
1 −1 2
Characteristics equation is

𝜆3 − 𝑆1 𝜆2 + 𝑆2 𝜆 − 𝑆3 = 0 → 1

𝑆1 = 𝑆𝑢𝑚 𝑜𝑓 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑡𝑠 = 2 + 2 + 2 = 6

𝑆2 = 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑖𝑛𝑜𝑟𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙𝑒𝑙𝑒𝑚𝑒𝑡𝑠

2 −1 2 2 2 −1
=| |+| |+| |
−1 2 1 2 −1 2
= 3+2+3= 8
2 −1 2
𝑠3 = 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴 = |−1 2 −1|
1 −1 2
= 2(4 − 1) + 1(−2 + 1) + 2(1 − 2)

=3
The characteristic equation is

λ3 − 6𝜆2 + 8𝜆 − 3 = 0

By CHT, 𝐴3 − 6𝐴2 + 8𝐴 − 3𝐼 = 0 → 2

i) To prove: 𝐴3 − 6𝐴2 + 8𝐴 − 3𝐼 = 0
2 −1 2 2 −1 2
𝐴2 = 𝐴. 𝐴 = (−1 2 −1). (−1 2 −1)
1 −1 2 1 −1 2
7 −6 9
= (−5 6 −6)
5 −5 7
7 −6 9 2 −1 2
𝐴3 = 𝐴2 . 𝐴 = (−5 6 −6). (−1 2 −1)
5 −5 7 1 −1 2
29 −28 38
𝐴3 = (−22 23 −28)
22 −22 29

46
𝐿. 𝐻. 𝑆 = 𝐴3 − 6𝐴2 + 8𝐴 − 3𝐼
29 −28 38 7 −6 9 2 −1 2 1 0 0
= (−22 23 −28) − 6 (−5 6 −6) +8 (−1 2 −1) − 3 (0 1 0)
22 −22 29 5 −5 7 1 −1 2 0 0 1
0 0 0
= (0 0 0)
0 0 0
CHT is verified.

ii) 𝐴4 =?

Eqn. 2 ⇒ 𝐴3 − 6𝐴2 + 8𝐴 − 3𝐼 = 0
Multiply by A on both sides

𝐴4 − 6𝐴3 + 8𝐴2 − 3𝐴 = 0

𝐴4 = 6𝐴3 − 8𝐴2 + 3𝐴
29 −28 38 7 −6 9 2 −1 2
= (−22 23 −28) − 8 (−5 6 −6) + 3 (−1 2 −1)
22 −22 29 5 −5 7 1 −1 2
124 −123 162
𝐴4 = (−95 96 −123)
95 −95 124
iii) 𝐴−1 =?

Eqn. 1 ⇒ 𝐴3 − 6𝐴2 + 8𝐴 − 3𝐼 = 0

Multiply by 𝐴−1 on both sides

𝐴2 − 6𝐴 + 8𝐼 − 3𝐴−1 = 0

3𝐴−1 = 𝐴2 − 6𝐴 + 8𝐼
7 −6 9 2 −1 2 1 0 0
= (−5 6 −6) − 6 (−1 2 −1) + 8 (0 1 0)
5 −5 7 1 −1 2 0 0 1
3 0 −3
=( 1 2 0)
−1 1 3

1 3 0 −3
𝐴−1 = (1 2 0)
3
−1 1 3

Problem is:
Using Cayley Hamilton to find the matrix given by

47
 𝟐 𝟏 𝟏
𝐀𝟖 − 𝟓𝐀𝟕 + 𝟕𝐀𝟔 − 𝟑𝐀𝟓 + 𝐀𝟒 − 𝟓𝐀𝟑 + 𝟖𝐀𝟐 − 𝟐𝐀 + 𝐈 if the matrix 𝑨 = [𝟎 𝟏 𝟎]
𝟏 𝟏 𝟐
Solution:
2 1 1
𝐴 = [0 1 0 ]
1 1 2
⇒ 𝑂 (𝐴 ) = 3
The characteristic equation of A is λ3 − 𝑆1 λ2 + 𝑆2 𝜆 − 𝑆3 = 0

where

𝑆1 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 = 2 + 1 + 2 = 5

S2 = sum of the minors of the main diagonal elements

1 0 2 1 2 1
=| |+| |+| |
1 2 1 2 0 1

= (2 − 0) + ( 4 − 1) + (2 − 0) = 7

2 1 1
𝑆3 = |𝐴| = |0 1 0|
1 1 2

= 2( 2 − 0) − 1(0 − 0) + 1(0 − 1) = 3

The characteristic equation is λ3 − 5λ2 + 7𝜆 − 3 = 0 → 1

By Cayley-Hamilton theorem ,

1 ⇒ A3 − 5A2 + 7𝐴 − 3𝐼 = 0 → 2

Let 𝑓 (𝐴) = A8 − 5A7 + 7A6 − 3A5 + A4 − 5A3 + 8A2 − 2A + I

= 𝐴5 (A3 − 5A2 + 7𝐴 − 3𝐼) + A4 − 5A3 + 8A2 − 2A + I

= 𝐴5 (0) + A4 − 5A3 + 8A2 − 2A + I , by 𝟐

= 0 + A4 − 5A3 + 8A2 − 2A + I

= A4 − 5A3 + 7A2 + A2 − 3A + A + I

= 𝐀𝟒 − 𝟓𝐀𝟑 + 𝟕𝐀𝟐 − 𝟑𝐀 + A2 + A + I

= 𝐴(A3 − 5A2 + 7𝐴 − 3𝐼) + A2 + A + I

48
 = 𝐴(0) + A2 + A + I , by 𝟐

= 0 + 𝐴2 + 𝐴 + 𝐼

𝑓(𝐴) = 𝐴2 + 𝐴 + 𝐼 ⇒ 3

2 1 1 2 1 1 5 4 4
𝐴2 = [0 1 0 ] [0 1 0 ] = [0 1 0] ⇒ 4
1 1 2 1 1 2 4 4 5

5 4 4 2 1 1 1 0 0
3 ⇒ 𝑓 ( 𝐴 ) = [0 1 0 ] + [ 0 1 0 ] + [0 1 0]
4 4 5 1 1 2 0 0 1
8 5 5
= [0 3 0 ]
5 5 8

Problem :
𝟏 𝟎 𝟑
−𝟏
Using Cayley Hamilton theorem find 𝑨 when 𝑨 = (𝟐 𝟏 −𝟏)
𝟏 −𝟏 𝟏
Solution:
1 0 3
𝐴 = (2 1 −1)
1 −1 1
Characteristics equation is

𝜆3 − 𝑆1 𝜆2 + 𝑆2 𝜆 − 𝑆3 = 0 → 1

𝑆1 = 𝑆𝑢𝑚 𝑜𝑓 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑡𝑠 = 1 + 1 + 1 = 3

𝑆2 = 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑖𝑛𝑜𝑟𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙𝑒𝑙𝑒𝑚𝑒𝑡𝑠

1 −1 1 3 1 0
=| |+| |+| |
−1 1 1 1 2 1
= 0 − 2 + 1 = −1
1 0 3
𝑠3 = 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴 = |2 1 −1|
1 −1 1
= −9
The characteristic equation is

𝜆3 − 3𝜆2 − 𝜆 + 9 = 0

𝐴−1 =?

49
 By CHT, 𝐴3 − 3𝐴2 − 𝐴 + 9𝐼 = 0

Multiply by 𝐴−1 on both sides

𝐴2 − 3𝐴 − 𝐼 + 9𝐴−1 = 0

9𝐴−1 = −𝐴2 + 3𝐴 + 𝐼
4 −3 6 1 0 3 1 0 0
= − (3 2 4) + 3 (2 1 −1) + (0 1 0)
0 −2 5 1 −1 1 0 0 1
0 3 3
= (3 2 −7)
3 −1 −1

−1
1 0 3 3
𝐴 = (3 2 −7)
9
3 −1 −1

𝟐 −𝟏 𝟐
2. Verify Cayley –Hamilton theorem or the matrix [−𝟏 𝟐 −𝟏] and hence find A-1 &
𝟏 −𝟏 𝟐
A4
solution:
2 −1 2
Let 𝐴 = [−1 2 −1]
1 −1 2
⇒ 𝑂 (𝐴 ) = 3
The Characteristic Equation of A is 𝜆3 − 𝑆1 𝜆2 + 𝑆2 𝜆 − 𝑆3 = 0 where
𝑆1 = sum of the main diagonal elements = 2 + 2 + 2 = 6
𝑆2 =sum of minors of main diagonal elements
2 −1 2 2 2 −1
=| |+| |+| |
−1 2 1 2 −1 2
= (4 − 1) + (4 − 2) + (4 − 1) = 8
𝑆3 =Determinant value of A
2 −1 2
= |𝐴| = |−1 2 −1| = 2(4 − 1) + 1(−2 + 1) + 2(1 − 2) = 3
1 −1 2
Hence the required Characteristic Equation is 𝜆3 − 6𝜆2 + 8𝜆 − 3 = 0 ⇒ 1
By Cayley –Hamoilton theorem,
𝐴3 − 6𝐴2 + 8𝐴 − 3𝐼 = 0 ⇒ 2
Verification:

50
 TP: 𝐴3 − 6𝐴2 + 8𝐴 − 3𝐼 = 0
2 −1 2 2 −1 2 7 −6 9
2
𝐴 = 𝐴 × 𝐴 = [−1 2 −1] [−1 2 −1] = [−5 6 −6]
1 −1 2 1 −1 2 5 −5 7
2 −1 2 7 −6 9 29 −28 38
𝐴3 = 𝐴 × 𝐴2 = [−1 2 −1] [−5 6 −6] = [−22 23 −28]
1 −1 2 5 −5 7 22 −22 29
𝐴3 − 6𝐴2 + 8𝐴 − 3𝐼
29 −28 38 42 −36 56 16 −8 16 3 0 0
= [−22 23 −28] − [−30 36 −36] + [ −8 16 −8 ] − [0 3 0]
22 −22 29 30 −30 42 2 −8 16 0 0 3
0 0 0
= [0 0 0 ] = 0
0 0 0
Cayley’s Hamilton theorem is verified.
ii) 𝑨−𝟏 =?
2 ⇒ 𝐴3 − 6𝐴2 + 8𝐴 − 3𝐼 = 0
Pre multiply by 𝐴−1
𝐴2 − 6𝐴 + 8𝐼 − 3𝐴−1 = 0
3𝐴−1 = 𝐴2 − 6𝐴 + 8𝐼
7 −6 9 −12 6 −12 8 0 0
3𝐴−1 = [−5 6 −6] + [ 6 −12 6 ] + [0 8 0]
5 −5 7 −6 6 −12 0 0 8
3 0 −3
−1 [
3𝐴 = 1 2 0]
−1 1 3
3 0 −3
−1 1
𝐴 = 3[ 1 2 0]
−1 1 3
iii) 𝑨𝟒 =?
2 ⇒ 𝐴3 − 6𝐴2 + 8𝐴 − 3𝐼 = 0
Pre multiply by 𝐴
𝐴4 − 6𝐴3 + 8𝐴2 − 3𝐴 = 0
𝐴4 = 6𝐴3 − 8𝐴2 + 3𝐴 = 6[6𝐴2 − 8𝐴 + 3𝐼 ] − 8𝐴2 + 3𝐴
196 −168 252 90 −45 90 18 0 0
𝐴4 = [−140 168 −168] − [−45 90 −45] + [ 0 18 0]
140 −140 196 45 −45 90 0 0 18
124 −123 162
𝐴4 = [−95 96 −123]
95 −95 124

51
 3 0 −3
1
𝐴−1 = 3 [ 1 2 0]
−1 1 3

𝟑 −𝟏 𝟏
3. If 𝑨 = [−𝟏 𝟓 −𝟏] verify Cayley-Hamilton theorem & hence find 𝑨−𝟏 and 𝑨𝟒
𝟏 −𝟏 𝟑
Solution:

3 −1 1
Let 𝐴 = [−1 5 −1]
1 −1 3
⇒ 𝑂 (𝐴 ) = 3
The Characteristic Equation of A is 𝜆3 − 𝑆1 𝜆2 + 𝑆2 𝜆 − 𝑆3 = 0 where
𝑆1 = sum of the main diagonal elements = 3 + 5 + 3 = 11
𝑆2 =sum of minors of main diagonal elements
5 −1| |3 1 3 −1
=| + |+| |
−1 3 1 3 −1 5
= (15 − 1) + (9 − 1) + (15 − 1) = 36
𝑆3 =Determinant value of A
3 −1 1
= |𝐴| = |−1 5 −1| = 3(15 − 1) + 1(−3 + 1) + 1(1 − 5) = 36
1 −1 3
Hence the required Characteristic Equation is 𝜆3 − 11𝜆2 + 36𝜆 − 36 = 0 ⇒ 1
By Cayley –Hamilton theorem,
𝐴3 − 11𝐴2 + 36𝐴 − 36𝐼 = 0 ⇒ 2
i) Verification:
TP: 𝐴3 − 11𝐴2 + 36𝐴 − 36𝐼 = 0
3 −1 1 3 −1 1 11 −9 7
𝐴2 = 𝐴 × 𝐴 = [−1 5 −1] [−1 5 −1] = [−9 27 −9]
1 −1 3 1 −1 3 7 −9 11
3 −1 1 11 −9 7 49 −63 41
3 2
𝐴 = 𝐴 × 𝐴 = −1 5 −1] [−9 27
[ ] [
−9 = −63 153 −63]
1 −1 3 7 −9 11 41 −63 49
𝐴3 − 11𝐴2 + 36𝐴 − 36𝐼
49 −63 41 11 −9 7 3 −1 1 1 0 0
= [−63 153 −63] − 11 [−9 27 −9] + 36 [−1 5 −1] − 36 [0 1 0]
41 −63 49 7 −9 11 1 −1 3 0 0 1
0 0 0
= [0 0 0 ] = 0
0 0 0

52
 Cayley’s Hamilton theorem is verified.
ii) 𝑨−𝟏 =?
2 ⇒ 𝐴3 − 11𝐴2 + 36𝐴 − 36𝐼 = 0
Pre multiply by 𝐴−1
𝐴2 − 11𝐴 + 36𝐼 − 36𝐴−1 = 0
36𝐴−1 = 𝐴2 − 11𝐴 + 36𝐼
11 −9 7 3 −1 1 1 0 0
36𝐴−1 = [−9 27 −9] − 11 [−1 5 −1 ] + 36 [ 0 1 0]
7 −9 11 1 −1 3 0 0 1
14 2 −4
−1
36𝐴 = [ 2 8 2 ]
−4 2 14
14 2 −4
1
𝐴−1 = 36 [ 2 8 2 ]
−4 2 14
iii) 𝑨𝟒 =?
2 ⇒ 𝐴3 − 11𝐴2 + 36𝐴 − 36𝐼 = 0
Pre multiply by 𝐴
𝐴4 − 11𝐴3 + 36𝐴2 − 36𝐴 = 0
𝐴4 = 11𝐴3 − 36𝐴2 + 36𝐴
= 11[11𝐴2 − 36𝐴 + 36𝐼] − 36𝐴2 + 36𝐴
= 121𝐴2 − 36𝐴2 − 396𝐴 + 36𝐴 + 396𝐼
𝐴4 = 85𝐴2 − 360𝐴 + 396𝐼
11 −9 7 3 −1 1 1 0 0
𝐴4 = 85 [−9 27 −9] − 360 [−1 5 −1] + 396 [0 1 0]
7 −9 11 1 −1 3 0 0 1
251 −405 235
𝐴4 = [−405 891 −405]
235 −405 251

 Diagonalization:

The process of reducing a given matrix into a diagonal matrix is called diagonalisation.

53
  Diagonalization of matrices

By orthogonal transformation, Diagonal matrix 𝐷 = 𝑁 𝑇 A𝑁

Where A = given matrix

𝑥1 𝑥1 𝑥1
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )
𝑥2 𝑥2 𝑥2
N = Normalised modal matrix = 𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )
𝑥3 𝑥3 𝑥3
[𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )]

Note:
𝑥1
𝑥1 𝑙(𝑋1 )
𝑥2
If 𝑋1 = [𝑥2 ] is an Eigen vector ,then its normalized Eigen vector = 𝑙(𝑋1 )
, where
𝑥3 𝑥3
[𝑙(𝑋1 )]
𝑙(𝑋1 ) = normalized Eigen value = √𝑥12 + 𝑥22 + 𝑥32

Problems:
𝟖 −𝟔 𝟐
1. 𝑫𝒊𝒂𝒈𝒐𝒏𝒂𝒍𝒊𝒔𝒆 𝒕𝒉𝒆 𝒎𝒂𝒕𝒓𝒊𝒙 𝐀 = [−𝟔 𝟕 −𝟒]
𝟐 −𝟒 𝟑
Solution:
By orthogonal reduction,
Diagonal Matrix is 𝐷 = 𝑁 𝑇 𝐴𝑁 ⇒ 1
𝑥1 𝑥1 𝑥1
𝑙(𝑋1) 𝑙(𝑋2 ) 𝑙(𝑋3 )
𝑥2 𝑥2 𝑥2
Normalized matrix is 𝑁 = 𝑙(𝑋1) 𝑙(𝑋2 ) 𝑙(𝑋3 )
⇒ 2
𝑥3 𝑥3 𝑥3
[𝑙(𝑋1) 𝑙(𝑋2 ) 𝑙(𝑋3 )]

Normalized eigen value= 𝑙(𝑋1 ) = √𝑥12 + 𝑥22 + 𝑥32 ⇒ 3

𝑥1
𝑙(𝑋1 )
𝑥2
Normalized eigen vector= 𝑙(𝑋1 )
⇒ 4
𝑥3
(𝑙(𝑋1 ))
The characteristic equation of A is λ3 − 𝑆1 λ2 + 𝑆2 𝜆 − 𝑆3 = 0 ⇒ 5 where
𝑆1 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 = 8 + 7 + 3 = 18
S2 = sum of the minors of the main diagonal elements
7 −4 8 2 8 −6
=| |+| |+| | = (21 − 16) + (24 − 4) + (56 − 36) = 45
−4 3 2 3 −6 7

54
 8 −6 2
𝑆3 = |𝐴| = |−6 7 −4| = 8(21 − 16) + 6(−18 + 8) + 2(24 − 14) = 0
2 −4 3
5 ⇒ λ3 − 18λ2 + 45𝜆 = 0
𝜆(λ2 − 18𝜆 + 45) = 0
𝜆 = 0, 𝜆 = 3 , 𝜆 = 15
Hence the Eigen values are 𝜆 = 0 , 3 , 15
Eigen vectors :
To find the Eigen vectors , solve (𝐴 − 𝜆𝐼) = 0
8−𝜆 −6 2 𝑥1 0
[ −6 7 − 𝜆 𝑥
−4 ] [ 2 ] = [0] ⇒ 6
2 −4 3 − 𝜆 𝑥3 0
Case-(i): 𝜆 = 0
8 −6 2 𝑥1 0
6 ⇒ [−6 7 −4] [𝑥2 ] = [0]
2 −4 3 𝑥3 0
8𝑥1 − 6𝑥2 + 2𝑥3 = 0 ⇒ 7
−6𝑥1 + 7𝑥2 − 4𝑥3 = 0 ⇒ 8
2𝑥1 − 4𝑥2 + 3𝑥3 = 0 ⇒ 9
By CM rule,
From 7 and 8 ,
1 𝑥 2 3 𝑥 𝑥
we get 24−14 = −12+32 = 56−36
𝑥1 𝑥2 𝑥3
= =
1 2 2

1
Hence the corresponding Eigenvector 𝑋1 = [2]
2
3 ⇒ 𝑙(𝑋1 ) = √𝑥12 + 𝑥22 + 𝑥32 = √1 + 4 + 4 = √9 = 3
1/3
4 ⇒Normalized eigen vector= (2/3)
2/3
Case-(ii): 𝜆 = 3
5 −6 2 𝑥1 0
6 ⇒ [−6 4 −4 ] [ 𝑥 2 ] = [ 0]
2 −4 0 𝑥 3 0
5𝑥1 − 6𝑥2 + 2𝑥3 = 0 ⇒ 10
−6𝑥1 + 4𝑥2 − 4𝑥3 = 0 ⇒ 11
2𝑥1 − 4𝑥2 + 0𝑥3 = 0 ⇒ 12

55
By CM rule,
From 10 and 11 ,
1 𝑥 2 3 𝑥 𝑥
we get 0−16 = −8−0 = 24−8
𝑥1 𝑥2 𝑥
= = −23
2 1

2
Hence the corresponding Eigenvector 𝑋2 = [ 1 ]
−2
3 ⇒ 𝑙(𝑋2 ) = √𝑥12 + 𝑥22 + 𝑥32 = √4 + 1 + 4 = √9 = 3
2/3
4 ⇒Normalized eigen vector= ( 1/3 )
−2/3
Case-(iii): 𝜆 = 15
−7 −6 2 𝑥1 0
6 ⇒ [−6 −8 −4 ] [𝑥2 ] = [0]
2 −4 −12 𝑥3 0
−7𝑥1 − 6𝑥2 + 2𝑥3 = 0 ⇒ 13
−6𝑥1 − 8𝑥2 − 4𝑥3 = 0 ⇒ 14
2𝑥1 − 4𝑥2 − 12𝑥3 = 0 ⇒ 15
By CM rule,
From 13 and 14 ,
1 𝑥 2 3 𝑥 𝑥
we get 96−16 = −8−72 = 24+16
𝑥1 𝑥 𝑥3
= −22 =
2 1

2
Hence the corresponding Eigenvector 𝑋3 = [−2]
1
3 ⇒ 𝑙(𝑋3 ) = √𝑥12 + 𝑥22 + 𝑥32 = √4 + 4 + 1 = √9 = 3
2/3
4 ⇒Normalized eigen vector= (−2/3)
1/3
1 2 2
3 3 3
2 1 −2
2 ⇒ 𝑁= 3 3 3
2 −2 1
[3 3 3 ]
1 ⇒ 𝐷 = 𝑁 𝑇 𝐴𝑁

56
 1 2 2 𝑇 1 2 2
3 3 3 8 −6 2 3 3 3
2 1 −2 2 1 −2
= [−6 7 −4]
3 3 3 3 3 3
2 −2 1 2 −4 3 2 −2 1
[3 3 3 ] [3 3 3 ]
0 0 0
D= [0 3 0]
0 0 15

𝟏 𝟎
2. 𝑨 = ( ) be diagonalized? why?
𝟎 𝟏
Solution:
The characteristic equation is 𝜆2 − 𝑆1 𝜆 + 𝑆2 = 0
𝑆1 = Sum of the main diagonal elements = 1+1=2
1 0
𝑆2 = |𝐴| = | |=1-0=1
0 1
∴ 𝜆2 − 2𝜆 + 1 = 0
(𝜆 − 1)(𝜆 − 1) = 0
𝜆 = 1,1
Since the eigen values are repeated, the matrix cannot be diagonalized.

𝟑 −𝟏 𝟏
Diagonalize the matrix 𝑨 = (−𝟏 𝟓 −𝟏).
𝟏 −𝟏 𝟑
Solution:
3 −1 1
𝐴 = (−1 5 −1) ⇒ 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑎𝑡𝑟𝑖𝑥 & 𝑂(𝐴) = 3
1 −1 3
By orthogonal transformation,

Diagonal matrix 𝐷 = 𝑁 𝑇 𝐴𝑁 → 1
𝑥1 𝑥1 𝑥1
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )
𝑥2 𝑥2 𝑥2
𝑁 = 𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )
→ 2,
𝑥3 𝑥3 𝑥3
[𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )]

where 𝑙(𝑋1 ) = √𝑥12 + 𝑥22 + 𝑥32

Then The characteristic equation is

57
𝜆3 − 𝑠1 𝜆2 + 𝑠2 𝜆 − 𝑠3 = 0 → 3

𝑠1 = 𝑆𝑢𝑚 𝑜𝑓 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 = 3 + 5 + 3 = 11

𝑠2 = 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑖𝑛𝑜𝑟𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖a𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

5 −1| |3 1 3 −1
=| + |+| |
−1 3 1 3 −1 5
= 14 + 8 + 14 = 36

𝑠3 = 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴
3 −1 1
|
= −1 5 −1| = 36
1 −1 3
3 ⇒ 𝜆3 − 11𝜆2 + 36𝜆 − 36 = 0

⇒ 𝜆 = 2,3,6
To find Eigen vector

3−𝜆 −1 1 𝑋1
Consider [ −1 5−𝜆 −1 ] [𝑋2 ] = 0
1 −1 3 − 𝜆 𝑋3
(3 − 𝜆)𝑥1 − 𝑥2 +𝑥3 = 0 − − − − − (4)

−𝑥1 + (5 − 𝜆)𝑥2 − 𝑥3 = 0 − − − −(5)

𝑥1 − 𝑥2 + (3 − 𝜆)𝑥3 = 0 − − − −(6)

Case (i):When 𝝀 = 𝟐 ,
The system of equation becomes,

𝑥1 − 𝑥2 + 𝑥3 = 0 − − − − − (7)

−𝑥1 + 3𝑥2 − 𝑥3 = 0 − − − −(8)

𝑥1 − 𝑥2 + 𝑥3 = 0 − − − −(9)
Solving (7) &(8),by using cross multiplication rule
𝑥1 𝑥2 𝑥3
= =
1 − 3 −1 + 1 3 − 1
𝑥1 𝑥2 𝑥3
= =
−2 0 2
𝑥1 𝑥2 𝑥3
= =
−1 0 1
⇒ 𝑥1 = −1 , 𝑥2 = 0, 𝑥3 = 1

58
 −1
The first eigen vector is 𝑋1 = ( 0 ) ⇒ 𝑙(𝑋1 ) = √1 + 0 + 1 = √2
1
Case (ii): When 𝝀 = 𝟑 ,
The system of equation becomes ,

0𝑥1 − 𝑥2 + 𝑥3 = 0 − − − − − (10)

−𝑥1 + 2𝑥2 − 𝑥3 = 0 − − − − − (11)

𝑥1 − 𝑥2 + 0𝑥3 = 0 − − − − − (12)
Solving (10) &(11),by using cross multiplication rule
𝑥1 𝑥2 𝑥3
= =
1 − 2 −1 − 0 0 − 1
𝑥1 𝑥2 𝑥3
= =
−1 −1 −1
⇒ 𝑥1 = 1 , 𝑥2 = 1, 𝑥3 = 1
1
second eigen vector is 𝑋2 = (1) ⇒ 𝑙(𝑋2 ) = √1 + 1 + 1 = √3
1
Case (iii): When 𝝀 = 𝟔 ,
The system of equation becomes ,

−3𝑥1 − 𝑥2 + 𝑥3 = 0 − − − − − (13)

−𝑥1 − 𝑥2 − 𝑥3 = 0 − − − − − (14)

𝑥1 − 𝑥2 − 3𝑥3 = 0 − − − − − (15)
Solving (13) & (14), by using cross multiplication rule
𝑥1 𝑥2 𝑥3
= =
1 + 1 −1 − 3 3 − 1
𝑥1 𝑥2 𝑥3
= =
2 −4 2
𝑥1 𝑥 𝑥3
= −22 =
1 1

⇒ 𝑥1 = 1 , 𝑥2 = −2, 𝑥3 = 1
1
The third eigen vector is 𝑋3 = (−2) ⇒ 𝑙(𝑋3 ) = √1 + 4 + 1 = √6
1

59
 𝑥1 𝑥1 𝑥1 1 1 1
−
l(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 ) √2 √3 √6
𝑥2 𝑥2 𝑥2 1 −2
2 ⇒𝑁 = = 0
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 ) √3 √6
𝑥3 𝑥3 𝑥3 1 1 1
[𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )] [ √2 √3 √6 ]
𝑇
1 1 1 1 1 1
− −
√2 √3 √6 √2 √3 √6
1 −2 3 −1 1 1 −2
1 ⇒𝐷 = 0 (−1 5 −1) 0
√3 √6 1 −1 3 √3 √6
1 1 1 1 1 1
[ √2 √3 √6 ] [ √2 √3 √6 ]
2 0 0
𝐷 = (0 3 0)
0 0 6

Problem :
𝟔 −𝟐 𝟐
Diagonalize the matrix 𝑨 = (−𝟐 𝟑 −𝟏) by orthogonal transformation.
𝟐 −𝟏 𝟑
Solution:
6 −2 2
𝐴 = (−2 3 −1) ⇒ 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑎𝑡𝑟𝑖𝑥 & 𝑂(𝐴) = 3
2 −1 3
By orthogonal transformation,

Diagonal matrix 𝐷 = 𝑁 𝑇 𝐴𝑁 → 1
𝑥1 𝑥1 𝑥1
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )
𝑥2 𝑥2 𝑥2
𝑁 = 𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )
→ 2 , where 𝑙(𝑋1 ) = √𝑥12 + 𝑥22 + 𝑥32
𝑥3 𝑥3 𝑥3
[𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )]

Then The characteristic equation is

𝜆3 − 𝑠1 𝜆2 + 𝑠2 𝜆 − 𝑠3 = 0 → 3

𝑠1 = 𝑆𝑢𝑚 𝑜𝑓 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

= 6 + 3 + 3 = 12

60
𝑠2 = 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑖𝑛𝑜𝑟𝑠 𝑜f 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙
3 −1 6 2 6 −2
=| |+| |+| |
−1 3 2 3 −2 3
= 8 + 14 + 14 = 36

𝑠3 = 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴
6 −2 2
= |−2 3 −1| = 32
2 −1 3
The characteristic equation is

3 ⇒ 𝜆3 − 12𝜆2 + 36𝜆 − 32 = 0

⇒ 𝜆 = 8,2,2
To find Eigen vector

6−𝜆 −2 2 𝑋1
Consider [ −2 3−𝜆 −1 ] [𝑋2 ] = 0
2 −1 3 − 𝜆 𝑋3
(6 − 𝜆)𝑥1 − 2𝑥2 +2𝑥3 = 0 − − − − − (4)

−2𝑥1 + (3 − 𝜆)𝑥2 − 𝑥3 = 0 − − − −(5)

2𝑥1 − 𝑥2 + (3 − 𝜆)𝑥3 = 0 − − − −(6)

Case (i):When 𝝀 = 𝟖 ,
The system of equation becomes,

−2𝑥1 − 2𝑥2 + 2𝑥3 = 0 − − − − − (7)

−2𝑥1 − 5𝑥2 − 𝑥3 = 0 − − − −(8)

2𝑥1 − 𝑥2 − 5𝑥3 = 0 − − − −(9)

Solving (7) &(8),by using cross multiplication rule
𝑥1 𝑥2 𝑥3
= =
2 + 10 −4 − 2 10 − 4
𝑥1 𝑥2 𝑥3
= =
12 −6 6
𝑥1 𝑥2 𝑥3
= =
2 −1 1
⇒ 𝑥1 = 2 , 𝑥2 = −1, 𝑥3 = 1
2
The first eigen vector is 𝑋1 = −1) ⇒ 𝑙(𝑋1 ) = √4 + 1 + 1 = √6
(
1

61
Case (ii): When 𝝀 = 𝟐 ,
The system of equation becomes ,

4𝑥1 − 2𝑥2 + 2𝑥3 = 0 − − − − − (10)

(10)/2, 2𝑥1 − 𝑥2 + 𝑥3 = 0 − − − − − (10)

−2𝑥1 + 𝑥2 − 𝑥3 = 0 − − − − − (11)
11
, 2𝑥1 − 𝑥2 + 𝑥3 = 0 − − − − − (11)
−1
2𝑥1 − 𝑥2 + 𝑥3 = 0 − − − − − (12)

(10),(11) &(12) are same equation to 2𝑥1 − 𝑥2 + 𝑥3 = 0

Put 𝑥1 = 0, −𝑥2 + 𝑥3 = 0

𝑥2 = 𝑥3
𝑥2 𝑥3
=
1 1
⇒ 𝑥1 = 0 , 𝑥2 = 1, 𝑥3 = 1
0
second eigen vector is 𝑋2 = (1) ⇒ 𝑙(𝑋2 ) = √0 + 1 + 1 = √2
1
Case (iii): When 𝝀 = 𝟐,
4𝑥1 − 2𝑥2 + 2𝑥3 = 0 − − − − − (13)

−2𝑥1 + 𝑥2 − 𝑥3 = 0 − − − − − (14)

2𝑥1 − 𝑥2 + 𝑥3 = 0 − − − − − (15)

(10),(11) &(12) are same equation to 2𝑥1 − 𝑥2 + 𝑥3 = 0

𝑙
Let 𝑋3 = (𝑚) be the third eigen vector.
𝑛
Since given matrix is symmetric, 𝑋3 is orthogonal to 𝑋1 & 𝑋2

⇒ 𝑋1𝑇 𝑋3 = 0 & 𝑋2𝑇 𝑋3 = 0

𝑙 𝑙
(2 −1 1) (𝑚) = 0 &(0 1 1) ( 𝑚 ) = 0
𝑛 n
2𝑙 − 𝑚 + 𝑛 = 0 &

0𝑙 + 𝑚 + 𝑛 = 0
By cross multiplication rule

62
 𝑙 𝑚 𝑛
= =
−1 − 1 0 − 2 2 − 0
𝑙 𝑚 𝑛
= =
−2 −2 2
𝑙 𝑚 𝑛
= = −1
1 1

𝑙 1
𝑋3 = (𝑚) = ( 1 )
𝑛 −1
⇒ 𝑙 (𝑋3 ) = √1 + 1 + 1 = √3
𝑥1 𝑥1 𝑥1 2 0 1
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 ) √6 √3 √3
𝑥2 𝑥2 𝑥2 1 1 1
2 ⇒𝑁 = =
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 ) √6 √3 √3
𝑥3 𝑥3 𝑥3 1 1 −1
[𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )] [√6 √3 √3 ]
𝑇
2 0 1 2 0 1
√6 √3 √3 √6 √3 √3
1 1 1 6 −2 2 1 1 1
1 ⇒𝐷 = (−2 3 −1)
√6 √3 √3 2 −1 3 √6 √3 √3
1 1 −1 1 1 −1
[√6 √3 √3 ] [√6 √3 √3 ]
8 0 0
𝐷 = (0 2 0)
0 0 2

63
Quadratic Forms:
An expression in which all the terms are of degree two is called as a quadratic form.
General form

𝑄 = 𝑎11 𝑥12 + 𝑎22 𝑥22 + 𝑎33 𝑥32 + 2𝑎12 𝑥1 𝑥2 + 2𝑎13 𝑥1 𝑥3 + 2𝑎23 𝑥2 𝑥3

i.e.,

Quadratic Form = 𝑋 𝑇 𝐴𝑋 ,

𝑥1
𝑥
where 𝑋 = [ 2 ] ,
𝑥3

𝑋 𝑇 = [𝑥1 𝑥2 𝑥3 ]

𝒙 𝒚 𝒛

1 1
𝑐𝑜𝑒𝑓𝑓 𝑥 2 𝑐𝑜𝑒𝑓𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓 𝑥𝑧
𝒙 2 2
1 1
& 𝐴 = matrix of Q.F = 𝒚 𝑐𝑜𝑒𝑓𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓 𝑦 2 𝑐𝑜𝑒𝑓𝑓 𝑦𝑧
2 2
𝒛 1 1
[ 2 𝑐𝑜𝑒𝑓𝑓 𝑥𝑧 𝑐𝑜𝑒𝑓𝑓 𝑦𝑧 𝑐𝑜𝑒𝑓𝑓 𝑧 2 ]
2

Example:

2𝑥 2 − 4𝑥𝑦 + 3𝑦 2 𝑖𝑠 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑓𝑜𝑟𝑚 𝑖𝑛 𝑡𝑤𝑜 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠

Canonical Form:
A quadratic form in which all the terms are square terms is called a canonical form
Example:

𝑥12 + 3𝑥22 + 3𝑥32

Note:
 The matrix corresponding to the quadratic form
1 1
𝑐𝑜𝑒𝑓𝑓 𝑥12 𝑐𝑜𝑒𝑓𝑓 𝑥1 𝑥2 𝑐𝑜𝑒𝑓𝑓 𝑥1 𝑥3
2 2
1 1
𝑄= 𝑐𝑜𝑒𝑓𝑓 𝑥2 𝑥1 𝑐𝑜𝑒𝑓𝑓 𝑥22 𝑐𝑜𝑒𝑓𝑓 𝑥2 𝑥3
2 2
1 1
[2 𝑐𝑜𝑒𝑓𝑓 𝑥1 𝑥3 𝑐𝑜𝑒𝑓𝑓 𝑥2 𝑥3 𝑐𝑜𝑒𝑓𝑓 𝑥32 ]
2

64
Problem :

Write the matrix of the quadratic form 𝟐𝒙𝟐𝟏 − 𝟐𝒙𝟐𝟐 + 𝟒𝒙𝟐𝟑 + 𝟐𝒙𝟏 𝒙𝟐 − 𝟔𝒙𝟏 𝒙𝟑 + 𝟔𝒙𝟐 𝒙𝟑
Solution:
1 1
𝑐𝑜𝑒𝑓𝑓 𝑥12 𝑐𝑜𝑒𝑓𝑓 𝑥1 𝑥2 𝑐𝑜𝑒𝑓𝑓 𝑥1 𝑥3
2 2
1 1
𝑄= 𝑐𝑜𝑒𝑓𝑓 𝑥2 𝑥1 𝑐𝑜𝑒𝑓𝑓 𝑥22 𝑐𝑜𝑒𝑓𝑓 𝑥2 𝑥3
2 2
1 1
[2 𝑐𝑜𝑒𝑓𝑓 𝑥1 𝑥3 𝑐𝑜𝑒𝑓𝑓 𝑥2 𝑥3 𝑐𝑜𝑒𝑓𝑓 𝑥32 ]
2
1 1
2 (2) (−6)
2 2
1 1 2 1 −3
𝑄= (2) −2 (6) = [ 1 −2 3]
2 2 −3 3 4
1 1
[2 (−6) (6) 4 ]
2

Problem :

write the quadratic form corresponding to the following symmetric matrix

𝟎 −𝟏 𝟐
[−𝟏 𝟏 𝟒]
𝟐 𝟒 𝟑
Solution:
The general form is

𝑄 = 𝑎11 𝑥12 + 𝑎22 𝑥22 + 𝑎33 𝑥32 + 2𝑎12 𝑥1 𝑥2 + 2𝑎13 𝑥1 𝑥3 + 2𝑎23 𝑥2 𝑥3

𝑄 = 0𝑥12 + 𝑥22 + 3𝑥32 + 2(−1)𝑥1 𝑥2 + 2(2)𝑥1 𝑥3 + 2(4)𝑥2 𝑥3

𝑄 = 0𝑥12 + 𝑥22 + 3𝑥32 − 2𝑥1 𝑥2 + 4𝑥1 𝑥3 + 8𝑥2 𝑥3

Canonical Form By Orthogonal Transformation:

Canonical Form = 𝑌 𝑇 𝐷𝑌 ,
𝑦1
where = [𝑦2 ] ,
𝑦3

𝑌𝑇 = [𝑦1 𝑦2 𝑦3 ],

& 𝐷 = 𝑁 𝑇 𝐴𝑁

65
 Nature of Quadratic Form:

 Positive definite :
If all the Eigen values of A are positive numbers, then Q.F is Positive definite
 Negative definite:
If all the Eigen values of A are negative numbers, then Q.F is negative definite.
 Positive semi definite:
If all the Eigen values of A are positive and at least one Eigen value is zero,
then the quadratic form is said to be Positive semi-definite
 Negative semi definite:
If all the Eigen values of A are negative and at least one Eigen value is zero,
then the quadratic form is said to be negative semi-definite
 Indefinite:
If A has both positive and negative Eigen values then the quadratic form is
said to be Indefinite
Note:

[Link] Eigen Values of A Nature of Quadratic Form

1 ‘ + 𝑣𝑒’ Positive definite

2 ‘ − 𝑣𝑒’ Negative definite

3 ‘ + 𝑣𝑒’ Positive semi definite

& at least one Eigen value is zero
4 ‘ − 𝑣𝑒’ Negative semi definite
& at least one Eigen values is zero
5 ‘ + 𝑣𝑒’ & ‘ − 𝑣𝑒’ Indefinite.

In Quadaratic Form:

Index No. of ‘ + 𝑣𝑒’ eigen values

Signature Difference between no. of‘ + 𝑣𝑒’ and 𝑛𝑜. 𝑜𝑓 ‘ − 𝑣𝑒’ eigen values

66
 Rank Number. of ‘ + 𝑣𝑒’ and no. of ‘ − 𝑣𝑒’ eigen values

Problems:
1. Reduce the quadratic form 𝟔𝒙𝟐 + 𝟑𝒚𝟐 + 𝟑𝒛𝟐 − 𝟒𝒙𝒚 − 𝟐𝒚𝒛 + 𝟒𝒛𝒙 into canonical form
by an orthogonal transformation
Solution:
Canonical form :
𝐶. 𝐹 = 𝑌 𝑇 𝐷𝑌 ⇒ 1
𝑦1
where 𝑌 = (𝑦2 ) ⇒ 2
𝑦3
By orthogonal reduction,
Diagonal Matrix is 𝐷 = 𝑁 𝑇 𝐴𝑁 ⇒ 3
𝐴 = 𝑚𝑎𝑡𝑟𝑖𝑥 𝑜𝑓 𝑄. 𝐹 ⇒ 4
𝑥1 𝑥1 𝑥1
𝑙(𝑋1) 𝑙(𝑋2 ) 𝑙(𝑋3 )
𝑥2 𝑥2 𝑥2
Normalized matrix is 𝑁 = 𝑙(𝑋1) 𝑙(𝑋2 ) 𝑙(𝑋3 )
⇒ 5
𝑥3 𝑥3 𝑥3
[𝑙(𝑋1) 𝑙(𝑋2 ) 𝑙(𝑋3 )]

Normalized eigen value= 𝑙(𝑋1 ) = √𝑥12 + 𝑥22 + 𝑥32 ⇒ 6

𝑥1
𝑙(𝑋1 )
𝑥2
Normalized eigen vector= 𝑙(𝑋1 )
⇒ 7
𝑥3
(𝑙(𝑋1 ))
Q.F: 6𝑥 2 + 3𝑦 2 + 3𝑧 2 − 4𝑥𝑦 − 2𝑦𝑧 + 4𝑧𝑥 ⇒ 8
𝑨 =matrix of the Q.F
1 1
𝑐𝑜𝑒𝑓𝑓 𝑥 2 𝑐𝑜𝑒𝑓𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓 𝑥𝑧
2 2
1 1
= 𝑐𝑜𝑒𝑓𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓 𝑦 2 𝑐𝑜𝑒𝑓𝑓 𝑦𝑧
2 2
1 1
[2 𝑐𝑜𝑒𝑓𝑓 𝑥𝑧 𝑐𝑜𝑒𝑓𝑓 𝑦𝑧 𝑐𝑜𝑒𝑓𝑓 𝑧 2 ]
2
6 −2 2
𝐴 = [−2 3 −1] by 8
2 −1 3
⇒ 𝑂(𝐴) = 3 and it is symmetric matrix
Characteristic equation of A is λ3 − 𝑆1 λ2 + 𝑆2 𝜆 − 𝑆3 = 0

where

67
𝑆1 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 = 6 + 3 + 3 = 12

S2 = sum of the minors of the main diagonal elements

3 −1 6 2 6 −2
=| |+| |+| |
−1 3 2 3 −2 3

= (9 − 1) + (18 − 4) + (18 − 4) = 36

6 −2 2
𝑆3 = |𝐴| = |−2 3 −1|
2 −1 3

= 6(9 − 1) + 2(−6 + 2) + 2(2 − 6) = 32

Characteristic equation is λ3 − 12λ2 + 36𝜆 − 32 = 0

Hence the Eigen values are 𝜆 = 8,2,2

Eigen vectors :

To find the Eigen vectors, solve (𝐴 − 𝜆𝐼)𝑋 = 0

6−𝜆 −2 2 𝑥1 0
[ −2 3−𝜆 −1 ] [ 𝑥 2 = 0] ⇒ 1
] [
2 −1 3 − 𝜆 𝑥3 0

Case (i): 𝜆 = 8

−2 −2 2 𝑥1 0
1 ⇒ [−2 −5 −1] [𝑥2 ] = [0]
2 −1 −5 𝑥3 0

−2𝑥1 − 2𝑥2 + 2𝑥3 = 0 ⇒ 2

−2𝑥1 − 5𝑥2 − 𝑥3 = ⇒ 3

2𝑥1 − 𝑥2 − 5𝑥3 = 0 ⇒ 4

By CM rule,

From 2 and 3 ,

we get

𝑥1 𝑥2 𝑥3
= =
2 + 10 −4 − 2 10 − 4

68
𝑥1 𝑥2 𝑥3
= =
2 −1 1

2
Hence the corresponding Eigenvector 𝑋1 = [−1]
1

Case (ii): 𝜆 = 2

4 −2 2 𝑥1 0
1 ⇒ [−2 1 𝑥
−1] [ 2 ] = [0]
2 −1 1 𝑥3 0

4𝑥1 − 2𝑥2 + 2𝑥3 = 0

÷ 2, 2𝑥1 − 𝑥2 + 𝑥3 = 0 ⇒ 5

−2𝑥1 + 𝑥2 − 𝑥3 = 0

÷ −1, 2𝑥1 − 𝑥2 + 𝑥3 = 0 ⇒ 6

2𝑥1 − 𝑥2 + 𝑥3 = 0 ⇒ 7

5 , 6 and 7 are same.

we get, 7 ⇒

2𝑥1 − 𝑥2 + 𝑥3 = 0 ⇒ 8

𝐼𝑓 𝑥1 = 0,

8 ⇒ −𝑥2 + 𝑥3 = 0

𝑥2 𝑥3
=
1 1

0
Hence the corresponding Eigenvector 𝑋2 = [1]
1

Case (iii): 𝜆 = 2

8 ⇒ 2𝑥1 − 𝑥2 + 𝑥3 = 0

𝑙
Let 𝑋3 = [𝑚]
𝑛

Here 𝑋3 is orthogonal to 𝑋1 & 𝑋2

69
⇒ 𝑋3𝑇 𝑋1 = 0 & 𝑋3𝑇 𝑋2 = 0

𝑿𝑻𝟑 𝑿𝟏 = 𝟎 𝑿𝑻𝟑 𝑿𝟐 = 𝟎
0 2
⇒ (𝑙 𝑚 )
𝑛 1] = 0
[ ⇒ (𝑙 𝑚 )
𝑛 −1] = 0
[
1 1
0𝑙 + 𝑚 + 𝑛 = 0 ⇒ 9 2𝑙 − 𝑚 + 𝑛 = 0 ⇒ 10

By CM rule,

From 9 and 10

𝑙 𝑚 𝑛
= =
−1 − 1 0 − 2 2 − 0

𝑙 𝑚 𝑛
= =
1 1 −1

1
Hence the corresponding Eigenvector 𝑋3 = [ 1 ]
−1
2 1
0
√6 √3
−1 1 1
Normalized modal matrix is 𝑁 = √6 √2 √3
1 1 −1
[ √6 √2 √3 ]

2 −1 1
√6 √6 √6
1 1
𝑁𝑇 = 0
√2 √2
1 1 −1
[√3 √3 √3 ]

𝐷 = 𝑁 𝑇 𝐴𝑁

2 −1 1 2 1
0
√6 √6 √6 6 −2 2 √6 √3
1 1 −1 1 1
= 0 [−2 3 −1]
√2 √2 √6 √2 √3
1 1 −1 2 −1 3 1 1 −1
[ √3 √3 √3 ] [ √6 √2 √3 ]

8 0 0
𝐷 = [0 2 0]
0 0 2
70
Canonical form :

𝐶𝐹 = 𝑌 𝑇 𝐷𝑌

8 0 0 𝑦1
= (𝑦1 𝑦2 𝑦3 ) [0 2 0] (𝑦2 )
0 0 2 𝑦3

= 8y12 + 2y22 + 2y32 is the required canonical form.

2. Reduce the quadratic form 𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐 − 𝟐𝒙𝒚 − 𝟐𝒚𝒛 − 𝟐𝒛𝒙 to the canonical form

through an orthogonal transformation. Write down the transformation.

Solution:
Canonical form :

𝐶. 𝐹 = 𝑌 𝑇 𝐷𝑌 ⇒ 1

𝑦1
where 𝑌 = (𝑦2 ) ⇒ 2
𝑦3

By orthogonal reduction,

Diagonal Matrix is 𝐷 = 𝑁 𝑇 𝐴𝑁 ⇒ 3

𝐴 = 𝑚𝑎𝑡𝑟𝑖𝑥 𝑜𝑓 𝑄. 𝐹 ⇒ 4

𝑥1 𝑥1 𝑥1
𝑙(𝑋1) 𝑙(𝑋2 ) 𝑙(𝑋3 )
𝑥2 𝑥2 𝑥2
Normalized matrix is 𝑁 = 𝑙(𝑋1) 𝑙(𝑋2 ) 𝑙(𝑋3 )
⇒ 5
𝑥3 𝑥3 𝑥3
[𝑙(𝑋1) 𝑙(𝑋2 ) 𝑙(𝑋3 )]

Normalized eigen value= 𝑙(𝑋1 ) = √𝑥12 + 𝑥22 + 𝑥32 ⇒ 6

𝑥1
𝑙(𝑋1 )
𝑥2
Normalized eigen vector= 𝑙(𝑋1 )
⇒ 7
𝑥3
(𝑙(𝑋1 ))

Q.F: 𝑥 2 + 𝑦 2 + 𝑧 2 − 2𝑥𝑦 − 2𝑦𝑧 − 2𝑧𝑥 ⇒ 8

𝑨 =matrix of the Q.F

71
 1 1
𝑐𝑜𝑒𝑓𝑓 𝑥 2 𝑐𝑜𝑒𝑓𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓 𝑥𝑧
2 2
1 1
= 𝑐𝑜𝑒𝑓𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓 𝑦 2 𝑐𝑜𝑒𝑓𝑓 𝑦𝑧
2 2
1 1
[2 𝑐𝑜𝑒𝑓𝑓 𝑥𝑧 𝑐𝑜𝑒𝑓𝑓 𝑦𝑧 𝑐𝑜𝑒𝑓𝑓 𝑧 2 ]
2

1 −1 −1
𝐴 = [−1 1 −1] by 8
−1 −1 1

⇒ 𝑂(𝐴) = 3 and it is symmetric matrix

The characteristic equation of A is λ3 − 𝑆1 λ2 + 𝑆2 𝜆 − 𝑆3 = 0 ⇒ 9

where

𝑆1 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 = 1 + 1 + 1 = 3

S2 = sum of the minors of the main diagonal elements

1 −1 1 −1 1 −1
=| |+| |+| | = (1 − 1) + (1 − 1) + (1 − 1) = 0
−1 1 −1 1 −1 1

1 −1 −1
𝑆3 = |𝐴| = |−1 1 −1| = 1(1 − 1) + 1(−1 − 1) − 1(1 + 1) = −4
−1 −1 1

9 ⇒ λ3 − 3λ2 + 0𝜆 + 4 = 0

Eigen values are 𝜆 = −1, 2, 2

Eigen vectors :

To find the Eigen vectors , solve (𝐴 − 𝜆𝐼)𝑋 = 0

1−𝜆 −1 −1 𝑥1 0
[ −1 1−𝜆 𝑥
−1 ] [ 2 ] = [0] ⇒ 10
−1 −1 1 − 𝜆 𝑥3 0

Case (i): 𝜆 = −1

2 −1 −1 𝑥1 0
10 ⇒ [−1 2 −1] [𝑥2 ] = [0]
−1 −1 2 𝑥3 0

2𝑥1 − 𝑥2 − 𝑥3 = 0 ⇒ 11

−𝑥1 + 2𝑥2 − 𝑥3 = ⇒ 12

72
 −𝑥1 − 𝑥2 + 2𝑥3 = 0 ⇒ 13

By CM rule,

From 11 and 12 ,

we get

𝑥1 𝑥2 𝑥3
= =
1+2 1+2 4−1
𝑥1 𝑥2 𝑥3
= =
3 3 3
𝑥1 𝑥2 𝑥3
= =
1 1 1

1
Hence the corresponding Eigenvector 𝑋1 = [1]
1

𝑙(𝑋1 ) = √𝑥12 + 𝑥22 + 𝑥32 = √1 + 1 + 1 = √3

1
√3
1
7 ⇒Normalized eigen vector= √3
1
( √3 )

Case (ii): 𝜆 = 2

−1 −1 −1 𝑥1 0
9 ⇒ [−1 −1 −1] [𝑥2 ] = [0]
−1 −1 −1 𝑥3 0

−𝑥1 − 𝑥2 − 𝑥3 = 0 ⇒ 14

−𝑥1 − 𝑥2 − 𝑥3 = 0 ⇒ 15

−𝑥1 − 𝑥2 − 𝑥3 = 0 ⇒ 16

14 , 15 and 16 are same

we get,

16 ⇒ −𝑥1 − 𝑥2 − 𝑥3 = 0

⇒ 𝑥1 + 𝑥2 + 𝑥3 = 0 ⇒ 17

73
𝐼𝑓 𝑥1 = 0,

17 ⇒ 𝑥2 + 𝑥3 = 0

𝑥2 = −𝑥3

𝑥2 𝑥3
=
−1 1

0
Hence the corresponding Eigenvector 𝑋2 = [−1]
1

𝑙(𝑋2 ) = √𝑥12 + 𝑥22 + 𝑥32 = √0 + 1 + 1 = √2

0
−1
7 ⇒Normalized eigen vector= (√2 )
1
√2

Case (iii): 𝜆 = 2

17 ⇒ 𝑥1 + 𝑥2 + 𝑥3 = 0

𝑙
Let 𝑋3 = [𝑚]
𝑛

Here 𝑋3 is orthogonal to 𝑋1 & 𝑋2

⇒ 𝑋3𝑇 𝑋1 = 0 & 𝑋3𝑇 𝑋2 = 0

𝑿𝑻𝟑 𝑿𝟏 = 𝟎 𝑿𝑻𝟑 𝑿𝟐 = 𝟎
1 0
⇒ (𝑙 𝑚 𝑛 ) [1 ] = 0 ⇒ (𝑙 𝑚 𝑛) [−1] = 0
1 1
𝑙 + 𝑚 + 𝑛 = 0 ⇒ 18 0𝑙 − 𝑚 + 𝑛 = 0 ⇒ 19

By CM rule,

From 18 and 19 ,

we get

74
 𝑙 𝑚 𝑛
= =
1 + 1 0 − 1 −1 − 0

𝑙 𝑚 𝑛
= =
2 −1 −1

2
Hence the corresponding Eigenvector 𝑋3 = [−1]
−1

𝑙(𝑋3 ) = √𝑥12 + 𝑥22 + 𝑥32 = √4 + 1 + 1 = √6

2
√6
−1
7 ⇒Normalized eigen vector= √6
−1
( √6 )
1 2
0
√3 √6
1 −1 −1
5 ⇒ 𝑁= √3 √2 √6
1 1 −1
[ √3 √2 √6 ]

3 ⇒ 𝐷 = 𝑁 𝑇 𝐴𝑁

1 2 𝑇 1 2
0 0
√3 √6 1 −1 −1 √3 √6
1 −1 −1 1 −1 −1
= [−1 1 −1]
√3 √2 √6 √3 √2 √6
1 1 −1 −1 −1 1 1 1 −1
[ √3 √2 √6 ] [ √3 √2 √6 ]

−1 0 0
𝐷=[ 0 2 0]
0 0 2

Canonical form :

1 ⇒ 𝐶𝐹 = 𝑌 𝑇 𝐷𝑌

−1 0 0 𝑦1
= (𝑦1 𝑦2 𝑦3 ) [ 0 2 0] (𝑦2 )
0 0 2 𝑦3

= −y12 + 2y22 + 2y32 is the required canonical form.

75
3. Reduce the quadratic form 𝟖𝒙𝟐𝟏 + 𝟕𝒙𝟐𝟐 + 𝟑𝒙𝟐𝟑 − 𝟏𝟐𝒙𝟏 𝒙𝟐 − 𝟖𝒙𝟐 𝒙𝟑 + 𝟒𝒙𝟑 𝒙𝟏 into

canonical form by means of an orthogonal transformation.

Solution:
Canonical form :

𝐶. 𝐹 = 𝑌 𝑇 𝐷𝑌 ⇒ 1

𝑦1
where 𝑌 = 𝑦2 ) ⇒ 2
(
𝑦3

By orthogonal reduction,

Diagonal Matrix is 𝐷 = 𝑁 𝑇 𝐴𝑁 ⇒ 3

𝐴 = 𝑚𝑎𝑡𝑟𝑖𝑥 𝑜𝑓 𝑄. 𝐹 ⇒ 4

𝑥1 𝑥1 𝑥1
𝑙(𝑋1) 𝑙(𝑋2 ) 𝑙(𝑋3 )
𝑥2 𝑥2 𝑥2
Normalized matrix is 𝑁 = 𝑙(𝑋1) 𝑙(𝑋2 ) 𝑙(𝑋3 )
⇒ 5
𝑥3 𝑥3 𝑥3
[𝑙(𝑋1) 𝑙(𝑋2 ) 𝑙(𝑋3 )]

Normalized eigen value= 𝑙(𝑋1 ) = √𝑥12 + 𝑥22 + 𝑥32 ⇒ 6

𝑥1
𝑙(𝑋1 )
𝑥2
Normalized eigen vector= 𝑙(𝑋1 )
⇒ 7
𝑥3
(𝑙(𝑋1 ))

Q.F : 8𝑥12 + 7𝑥22 + 3𝑥32 − 12𝑥1 𝑥2 − 8𝑥2 𝑥3 + 4𝑥3 𝑥1

Let 𝑥1 = 𝑥, 𝑥2 = 𝑦 & 𝑥3 = 𝑧
Q.F : 8𝑥 2 + 7𝑦 2 + 3𝑧 2 − 12𝑥𝑦 − 8𝑦𝑧 + 4𝑥𝑧 ⇒ 8
𝑨 =matrix of the Q.F

1 1
𝑐𝑜𝑒𝑓𝑓 𝑥 2 𝑐𝑜𝑒𝑓𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓 𝑥𝑧
2 2
1 1
= 𝑐𝑜𝑒𝑓𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓 𝑦 2 𝑐𝑜𝑒𝑓𝑓 𝑦𝑧
2 2
1 1
[2 𝑐𝑜𝑒𝑓𝑓 𝑥𝑧 𝑐𝑜𝑒𝑓𝑓 𝑦𝑧 𝑐𝑜𝑒𝑓𝑓 𝑧 2 ]
2

8 −6 2
𝐴 = [−6 7 −4] by 8
2 −4 3

76
𝑂(𝐴) = 3 and it is symmetric matrix

The characteristic equation of A is λ3 − 𝑆1 λ2 + 𝑆2 𝜆 − 𝑆3 = 0 ⇒ 9

where

𝑆1 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 = 8 + 7 + 3 = 18

S2 = sum of the minors of the main diagonal elements

7 −4 8 2 8 −6
=| |+| |+| | = (21 − 16) + (24 − 4) + (56 − 36) = 45
−4 3 2 3 −6 7

8 −6 2
| |
𝑆3 = 𝐴 = |−6 7 −4| = 8(21 − 16) + 6(−18 + 8) + 2(24 − 14) = 0
2 −4 3

9 ⇒ λ3 − 18λ2 + 45𝜆 = 0

Eigen values are 𝜆 = 0, 3, 15

Eigen vectors :

To find the Eigen vectors, solve (𝐴 − 𝜆𝐼)𝑋 = 0

8−𝜆 −6 2 𝑥1 0
[ −6 7−𝜆 −4 ] [ 𝑥 2 ] = [ 0] ⇒ 10
2 −4 3 − 𝜆 𝑥3 0

Case (i): 𝜆 = 0

8 −6 2 𝑥1 0
10 ⇒ [−6 7 −4] [𝑥2 ] = [0]
2 −4 3 𝑥3 0

8𝑥1 − 6𝑥2 + 2𝑥3 = 0

4𝑥1 − 3𝑥2 + 𝑥3 = 0 ⇒ 11

−6𝑥1 + 7𝑥2 − 4𝑥3 = ⇒ 12

2𝑥1 − 4𝑥2 + 3𝑥3 = 0 ⇒ 13

By CM rule,

From 11 and 12 ,

we get

77
 𝑥1 𝑥2 𝑥3
= =
12 − 7 −6 + 16 28 − 18
𝑥1 𝑥2 𝑥3
= =
5 10 10
𝑥1 𝑥2 𝑥3
= =
1 2 2

1
Hence the corresponding Eigenvector 𝑋1 = [2]
2

𝑙(𝑋1 ) = √𝑥12 + 𝑥22 + 𝑥32 = √1 + 4 + 4 = √9 = 3

1/3
7 ⇒Normalized eigen vector= (2/3)
2/3

Case (ii): 𝜆 = 3

5 −6 2 𝑥1 0
9 ⇒ [−6 4 −4] [𝑥2 ] = [0]
2 −4 0 𝑥3 0

5𝑥1 − 6𝑥2 + 2𝑥3 = 0 ⇒ 14

−6𝑥1 + 4𝑥2 − 4𝑥3 = 0

−3𝑥1 + 2𝑥2 − 2𝑥3 = 0 ⇒ 15

2𝑥1 − 4𝑥2 + 0𝑥3 = 0

𝑥1 − 2𝑥2 + 0𝑥3 = 0 ⇒ 16

By CM rule,

From 15 and 16 ,

we get

𝑥1 𝑥2 𝑥3
= =
0 − 4 −2 − 0 6 − 2
𝑥1 𝑥2 𝑥3
= =
−4 −2 4
𝑥1 𝑥2 𝑥3
= =
2 1 −2

78
 2
Hence the corresponding Eigenvector 𝑋2 = [ 1 ]
−2

𝑙(𝑋2 ) = √𝑥12 + 𝑥22 + 𝑥32 = √4 + 1 + 4 = √9 = 3

2/3
7 ⇒Normalized eigen vector= ( 1/3 )
−2/3

Case (iii): 𝜆 = 15

−7 −6 2 𝑥1 0
9 ⇒ [−6 −8 −4 ] [𝑥2 ] = [0]
2 −4 −12 𝑥3 0

−7𝑥1 − 6𝑥2 + 2𝑥3 = 0 ⇒ 14

−6𝑥1 − 8𝑥2 − 4𝑥3 = 0

3𝑥1 + 4𝑥2 + 2𝑥3 = 0 ⇒ 15

2𝑥1 − 4𝑥2 − 12𝑥3 = 0

𝑥1 − 2𝑥2 − 6𝑥3 = 0 ⇒ 16

By CM rule,

From 15 and 16 ,

we get

𝑥1 𝑥2 𝑥3
= =
−24 + 4 2 + 18 −6 − 4
𝑥1 𝑥2 𝑥3
= =
−20 20 −10
𝑥1 𝑥2 𝑥3
= =
2 −2 1

2
Hence the corresponding Eigenvector 𝑋3 = [−2]
1

𝑙(𝑋3 ) = √𝑥12 + 𝑥22 + 𝑥32 = √4 + 4 + 1 = √9 = 3

79
 2/3
7 ⇒Normalized eigen vector= (−2/3)
1/3

1/3 2/3 2/3

5 ⇒ 𝑁 = [2/3 1/3 −2/3]
2/3 −2/3 1/3

3 ⇒ 𝐷 = 𝑁 𝑇 𝐴𝑁

1/3 2/3 2/3 𝑇 8 −6 2 1/3 2/3 2/3

= [2/3 1/3 −2/3] [−6 7 −4] [2/3 1/3 −2/3]
2/3 −2/3 1/3 2 −4 3 2/3 −2/3 1/3

0 0 0
[
𝐷= 0 3 0]
0 0 15

Canonical form :

1 ⇒ 𝐶𝐹 = 𝑌 𝑇 𝐷𝑌

0 0 0 𝑦1
( )
= 𝑦1 𝑦2 𝑦3 [0 3 0 ] (𝑦2 )
0 0 15 𝑦3

= 0y12 + 3y22 + 15y32 is the required canonical form.

4. Find the index and signature of the Q.F 𝒙𝟐𝟏 + 𝟐𝒙𝟐𝟐 − 𝟑𝒙𝟐𝟑
Solution:
Let 𝑓 (𝑥1 , 𝑥2 , 𝑥3 ) = 𝑥12 + 2𝑥22 − 3𝑥32
it is already in the canonical form.
Index = Number of positive terms in the C.F = 2
Signature = Number of positive terms- Number of negative terms = 2-1=1.

5. Determine the nature of the following quadratic form 𝒇(𝒙𝟏 , 𝒙𝟐 , 𝒙𝟑 ) = 𝒙𝟐𝟏 + 𝟐𝒙𝟐𝟐
Solution:
Let 𝑓 (𝑥1 , 𝑥2 , 𝑥3 ) = 𝑥12 + 2𝑥22
it is already in the canonical form.
The C.F contains two positive and one zero term.
Hence QF is positive semi-definite

80
 −1 0 0
6. Give the nature of a quadratic form whose matrix is [ 0 −1 0]
0 0 −2
Solution:
The Eigen values of the given matrix are -1 , - 1 , - 2
All the Eigen values are negative numbers.
Hence the nature of the Q.F is negative definite.
𝟎 𝟓 −𝟏
7. Write down the quadratic form corresponding to the matrix 𝑨 = [ 𝟓 𝟏 𝟔]
−𝟏 𝟔 𝟐
Solution:
1 1
𝑐𝑜𝑒𝑓𝑓 𝑥 2 𝑐𝑜𝑒𝑓𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓 𝑥𝑧
0 5 −1 2 2
1 1
𝐴=[ 5 1 6 ]= 𝑐𝑜𝑒𝑓𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓 𝑦 2 𝑐𝑜𝑒𝑓𝑓 𝑦𝑧
2 2
−1 6 2 1 1
[ 2 𝑐𝑜𝑒𝑓𝑓 𝑥𝑧 𝑐𝑜𝑒𝑓𝑓 𝑦𝑧 𝑐𝑜𝑒𝑓𝑓 𝑧 2 ]
2
1
𝑐𝑜𝑒𝑓𝑓 𝑥𝑦 = 5
2

⇒ 𝑐𝑜𝑒𝑓𝑓 𝑥𝑦 = 10
1
𝑐𝑜𝑒𝑓𝑓 𝑥𝑧 = −1
2

⇒ 𝑐𝑜𝑒𝑓𝑓 𝑥𝑧 = −2
1
𝑐𝑜𝑒𝑓𝑓 𝑦𝑧 = 6
2

⇒ 𝑐𝑜𝑒𝑓𝑓 𝑦𝑧 = 12
The quadratic form 𝑦 2 + 2𝑧 2 + 10 𝑥𝑦 − 2 𝑥𝑧 + 12𝑦𝑧

Problem :

Reduce the Quadratic Form 𝒙𝟐 + 𝟑𝒚𝟐 + 𝟑𝒛𝟐 − 𝟐𝒚𝒛 to its canonical form by
orthogonal reduction. Also find index, signature and nature of Quadratic Form.
Solution:

Quadratic Form is 𝑥 2 + 3𝑦 2 + 3𝑧 2 − 2𝑦𝑧 → 1

The matrix of the Q.F is
1 1
𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑥 2 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑥𝑧
2 2
1 1
𝐴= 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑦 2 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑦𝑧
2 2
1 1
( 2 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑥𝑧 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑦𝑧 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑧 2 )
2

81
 1 0 0
⇒ 𝐴 = (0 3 −1) ⇒ 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑎𝑡𝑟𝑖𝑥 & 𝑂(𝐴) = 3
0 −1 3
By orthogonal transformation,

Canonical Form = 𝑌 𝑇 𝐷𝑌 → 2
𝑦1
where 𝑌 = [𝑦2 ], 𝑌𝑇 = [𝑦1 𝑦2 𝑦3 ]
𝑦3

Diagonal matrix 𝐷 = 𝑁 𝑇 𝐴𝑁 → 3
𝑥1 𝑥1 𝑥1
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )
𝑥2 𝑥2 𝑥2
𝑁 = 𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )
→ 4 , where 𝑙(𝑋1 ) = √𝑥12 + 𝑥22 + 𝑥32
𝑥3 𝑥3 𝑥3
[𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )]

Then The characteristic equation is

𝜆3 − 𝑠1 𝜆2 + 𝑠2 𝜆 − 𝑠3 = 0 → 5

𝑠1 = 𝑆𝑢𝑚 𝑜𝑓 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

= 1+3+3= 7

𝑠2 = 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑖𝑛𝑜𝑟𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙

3 −1 1 0 1 0
=| |+| |+| |
−1 3 0 3 0 3
= 8 + 3 + 3 =14

𝑠3 = 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴
1 0 0
= |0 3 −1| = 8
0 −1 3
3 ⇒ 𝜆3 − 7𝜆2 + 14𝜆 − 8 = 0

⇒ 𝜆 = 1,2,4
To find Eigen vector
1−𝜆 0 0 𝑥1
Consider [ 0 3−𝜆 −1 ] [𝑥2 ] = 0
0 −1 3 − 𝜆 𝑥3
(1 − 𝜆)𝑥1 + 0𝑥2 +0𝑥3 = 0 − − − − − (4)

0𝑥1 + (3 − 𝜆)𝑥2 − 𝑥3 = 0 − − − −(5)

0𝑥1 − 𝑥2 + (3 − 𝜆)𝑥3 = 0 − − − −(6)

82
Case (i):When 𝝀 = 𝟏 ,
The system of equation becomes,

0𝑥1 + 0𝑥2 + 0𝑥3 = 0 − − − − − (7)

0𝑥1 + 2𝑥2 − 𝑥3 = 0 − − − −(8)

0𝑥1 − 𝑥2 + 2𝑥3 = 0 − − − −(9)

Solving (8) &(9),by using cross multiplication rule
𝑥1 𝑥2 𝑥3
= =
4−1 0−0 0−0
𝑥1 𝑥2 𝑥3
= =
3 0 0
𝑥1 𝑥2 𝑥3
= =
1 0 0
⇒ 𝑥1 = 1 , 𝑥2 = 0, 𝑥3 = 0
1
The first eigen vector is 𝑋1 = (0) ⇒ 𝑙(𝑋1 ) = √1 + 0 + 0 = √1 = 1
0
Case (ii): When 𝝀 = 𝟐 ,
The system of equation becomes ,

−𝑥1 + 0𝑥2 + 0𝑥3 = 0 − − − − − (10)

0𝑥1 + 𝑥2 − 𝑥3 = 0 − − − − − (11)

0𝑥1 − 𝑥2 + 𝑥3 = 0 − − − − − (12)
Solving (10) &(11),by using cross multiplication rule
𝑥1 𝑥2 𝑥3
= =
0 − 0 0 − 1 −1 − 0
𝑥1 𝑥2 𝑥3
= =
0 −1 −1
⇒ 𝑥1 = 0 , 𝑥2 = 1, 𝑥3 = 1
0
second eigen vector is 𝑋2 = (1) ⇒ 𝑙(𝑋2 ) = √0 + 1 + 1 = √2
1
Case (iii): When 𝝀 = 𝟒 ,
The system of equation becomes ,

−3𝑥1 + 0𝑥2 + 0𝑥3 = 0 − − − − − (13)

0𝑥1 − 𝑥2 − 𝑥3 = 0 − − − − − (14)

83
0𝑥1 − 𝑥2 − 𝑥3 = 0 − − − − − (15)
Solving (13) & (14), by using cross multiplication rule
𝑥1 𝑥2 𝑥3
= =
0−0 0−3 3−0
𝑥1 𝑥2 𝑥3
= =
0 −3 3
𝑥1 𝑥2 𝑥3
= =
0 −1 1

⇒ 𝑥1 = 0 , 𝑥2 = −1, 𝑥3 = 1
0
The third eigen vector is 𝑋3 = (−1) ⇒ 𝑙(𝑋3 ) = √0 + 1 + 1 = √2
1
𝑥1 𝑥1 𝑥1
1 0 0
𝑙(X1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 ) 1 −1
𝑥2 𝑥2 𝑥2 0
4 ⇒𝑁 = = √2 √2
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 ) 1 1
𝑥3 𝑥3 𝑥3 0
[ √2 √2 ]
[𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )]
𝑇
1 0 0 1 0 0
1 −1 1 0 0 1 −1
0 0
3 ⇒𝐷 = √2 √2 (0 3 −1) √2 √2
1 1 0 −1 3 1 1
0 0
[ √2 √2 ] [ √2 √2 ]
1 0 0
𝐷 = (0 2 0)
0 0 4
1 0 0 𝑦1
2 ⇒Canonical Form = [𝑦1 𝑦2 𝑦3 ] (0 2 0) [𝑦2 ]
0 0 4 𝑦3
Canonical Form = 𝑦12 + 2𝑦22 + 4𝑦32

Since = 1,2,4 ,
Nature of Quadratic Form = Positive definite

Index=No. of ‘ + 𝑣𝑒’ eigen values=3

Signature=Difference between no. of ‘ + 𝑣𝑒’

and ‘ − 𝑣𝑒’ eigen values =3-0=3

Rank=Number. of ‘ + 𝑣𝑒’ and ‘ − 𝑣𝑒’ eigen values

=3+0=3

84
Problem :

Reduce the Quadratic Form 𝟐𝒙𝟏 𝒙𝟐 + 𝟐𝒙𝟏 𝒙𝟑 − 𝟐𝒙_𝟐𝒙𝟑 to its canonical form by
orthogonal reduction. Also find index, signature and nature of Quadratic Form.
Solution:

Quadratic Form is 2𝑥1 𝑥2 + 2𝑥1 𝑥3 − 2𝑥_2𝑥3 → 1

The matrix of the Q.F is
1 1
𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑥 2 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑥𝑧
2 2
1 1
𝐴= 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑥𝑦 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑦 2 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑦𝑧
2 2
1 1
( 2 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑥𝑧 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑦𝑧 𝑐𝑜𝑒𝑓𝑓𝑡. 𝑜𝑓 𝑧 2 )
2
0 1 1
⇒ 𝐴 = (1 0 −1) ⇒ 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑎𝑡𝑟𝑖𝑥 & 𝑂(𝐴) = 3
1 −1 0
By orthogonal transformation,

Canonical Form = 𝑌 𝑇 𝐷𝑌 → 2
𝑦1
where 𝑌 = [𝑦2 ], 𝑌𝑇 = [𝑦1 𝑦2 𝑦3 ]
𝑦3

Diagonal matrix 𝐷 = 𝑁 𝑇 𝐴𝑁 → 3
𝑥1 𝑥1 𝑥1
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )
𝑥2 𝑥2 𝑥2
𝑁 = 𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )
→ 4 , where 𝑙(𝑋1 ) = √𝑥12 + 𝑥22 + 𝑥32
𝑥3 𝑥3 𝑥3
[𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )]

Then The characteristic equation is

𝜆3 − 𝑠1 𝜆2 + 𝑠2 𝜆 − 𝑠3 = 0 → 3

𝑠1 = 𝑆𝑢𝑚 𝑜𝑓 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

= 0+0+0= 0

𝑠2 = 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑖𝑛𝑜𝑟𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙

0 −1 0 1 0 1
=| |+| |+| |
−1 0 1 0 1 0
= −1 − 1 − 1 = −3
𝑠3 = 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴

85
 0 1 1
= |1 0 −1| = −2
1 −1 0
3 ⇒ 𝜆3 − 0𝜆2 − 3𝜆 − 2 = 0

⇒ 𝜆 = −2,1,1
To find Eigen vector

0−𝜆 1 1 𝑋1
Consider [ 1 0−𝜆 −1 ] [𝑋2 ] = 0
1 −1 0 − 𝜆 𝑋3
−𝜆𝑥1 + 𝑥2 +𝑥3 = 0 − − − − − (4)

𝑥1 −𝜆𝑥2 − 𝑥3 = 0 − − − −(5)

𝑥1 − 𝑥2 − 𝜆𝑥3 = 0 − − − −(6)

Case (i):When 𝝀 = −𝟐 ,
The system of equation becomes,

2𝑥1 + 𝑥2 + 𝑥3 = 0 − − − − − (7)

𝑥1 + 2𝑥2 − 𝑥3 = 0 − − − −(8)

𝑥1 − 𝑥2 + 2𝑥3 = 0 − − − −(9)
Solving (7) &(8),by using cross multiplication rule
𝑥1 𝑥2 𝑥3
= =
−1 − 2 1 + 2 4 − 1
𝑥1 𝑥2 𝑥3
= =
−3 3 3
𝑥1 𝑥2 𝑥3
= =
−1 1 1
⇒ 𝑥1 = −1 , 𝑥2 = 1, 𝑥3 = 1
−1
The first eigen vector is 𝑋1 = ( 1 ) ⇒ 𝑙(𝑋1 ) = √1 + 1 + 1 = √3
1
Case (ii): When 𝝀 = 𝟏 ,
The system of equation becomes ,

−𝑥1 + 𝑥2 + 𝑥3 = 0 − − − − − (10)

𝑥1 − 𝑥2 − 𝑥3 = 0 − − − − − (11)
𝑥1 − 𝑥2 − 𝑥3 = 0 − − − − − (12)

86
(10),(11) &(12) are same equation to 𝑥1 − 𝑥2 − 𝑥3 = 0

Put 𝑥1 = 0, −𝑥2 − 𝑥3 = 0

−𝑥2 = 𝑥3
𝑥2 𝑥3
=
1 −1
⇒ 𝑥1 = 0 , 𝑥2 = 1, 𝑥3 = 1
0
second eigen vector is 𝑋2 = ( 1 ) ⇒ 𝑙 (𝑋2 ) = √0 + 1 + 1 = √2
−1
Case (iii): When 𝝀 = 𝟏,

−𝑥1 + 𝑥2 + 𝑥3 = 0 − − − − − (13)

𝑥1 − 𝑥2 − 𝑥3 = 0 − − − − − (14)

𝑥1 − 𝑥2 − 𝑥3 = 0 − − − − − (15)

(10),(11) &(12) are same equation to 𝑥1 − 𝑥2 − 𝑥3 = 0

𝑙
Let 𝑋3 = 𝑚) be the third eigen vector.
(
𝑛
Since given matrix is symmetric, 𝑋3 is orthogonal to 𝑋1 & 𝑋2

⇒ 𝑋1𝑇 𝑋3 = 0 &𝑋2𝑇 𝑋3 = 0
𝑙 𝑙
(−1 1 1) (𝑚) = 0 &(0 −1 1) ( 𝑚 ) = 0
𝑛 𝑛
−𝑙 + 𝑚 + 𝑛 = 0 &

0𝑙 − 𝑚 + 𝑛 = 0
By cross multiplication rule
𝑙 𝑚 𝑛
= =
1+1 0+1 1−0
𝑙 𝑚 𝑛
= =
2 1 1
𝑙 2
𝑋3 = (𝑚) = (1)
𝑛 1
⇒ 𝑙 (𝑋3 ) = √4 + 1 + 1 = √6

87
 𝑥1 𝑥1 𝑥1 −1 0 2
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 ) √3 √2 √6
𝑥2 𝑥2 𝑥2 1 −1 1
4 ⇒𝑁 = =
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 ) √3 √2 √6
𝑥3 𝑥3 𝑥3 1 1 1
[𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )] [ √3 √2 √6]
𝑇
−1 0 2 −1 0 2
√3 √2 √6 √3 √2 √6
1 −1 1 0 1 1 1 −1 1
3 ⇒𝐷 = (1 0 −1)
√3 √2 √6 1 −1 0 √3 √2 √6
1 1 1 1 1 1
[ √3 √2 √6] [ √3 √2 √6]
−2 0 0
𝐷=( 0 1 0)
0 0 1
−2 0 0 𝑦1
2 ⇒Canonical Form = [𝑦1 𝑦2 𝑦3 ] ( 0 1 0) [𝑦2 ]
0 0 1 𝑦3
Canonical Form = −2𝑦12 + 𝑦22 + 𝑦32

Since = −2,1,1 ,
Nature of Quadratic Form =Indefinite

Index=No. of ‘ + 𝑣𝑒’ eigen values=2

Signature=Difference between no. of ‘ + 𝑣𝑒’

and ‘ − 𝑣𝑒’ eigen values

=2-1=1

Rank=Number. of ‘ + 𝑣𝑒’ and ‘ − 𝑣𝑒’ eigen values

=2+1=3

Applications:
Problem :
Find out what type of conic section the following quadratic form represents
𝑄 = 17𝑥12 − 30𝑥1 𝑥2 + 17𝑥22 = 128 .
Answer:
We have 𝑄 = 𝑋 𝑇 𝐴 𝑋

88
 17 −15] 𝑥1
Here 𝐴 = [ & 𝑋 = [𝑥 ]
−15 17 2

17 − 𝜆 −15 |
Characteristic equation is | =0
−15 17 − 𝜆
(17 − 𝜆)2 − 152 = 0

𝜆2 − 34𝜆 + 289 − 225 = 0

𝜆2 − 34𝜆 + 64 = 0
⇒ 𝜆1 = 2 & 𝜆2 = 32
Hence Canonical form is 2𝑦12 + 32𝑦22
⇒ 𝑄 = 2𝑦12 + 32𝑦22 = 128
2𝑦12 + 32𝑦22 = 128
𝑦12 𝑦22
+ =1
64 4
𝑦12 𝑦22
+ =1
82 22
𝑦12 𝑦2
Quadratic form represents the ellipse + 222 = 1
82

Problem :
8 −6 2
Find the latent roots and latent vectors of the matrix 𝐴 = (−6 7 −4)
2 −4 3
Answer:
𝑆1 = 18, 𝑆2 = 45 & 𝑆3 = 0
Characteristic equation is
𝜆3 − 18𝜆2 + 45𝜆 − 0 = 0
⇒ 𝜆1 = 0, 𝜆2 = 3 & 𝜆3 = 15
latent roots are 0, 3, & 15
1 2 2
latent vectors are 𝑋1 = [2], 𝑋2 = [ 1 ] & X 3 = [−2]
2 −2 1

89
Problem :

An elastic membrane in the 𝑥12 + 𝑥22 = 1 is stretched so that a point 𝑃: (𝑥1 , 𝑥2 ) goes over
𝑦1 5 3 𝑥1
into the point 𝑄: (𝑦1 , 𝑦2 ) given by 𝑦 = [𝑦 ] = 𝐴𝑋 = [ ] [ ] in components
2 3 5 𝑥2
𝑦1 = 5𝑥1 + 3𝑥2 , 𝑦2 = 3𝑥1 + 5𝑥2 . Find the Eigen values & Eigen vectors.

Answer:
𝑦1 5 3 𝑥1
Given 𝑦 = [𝑦 ] = 𝐴𝑋 = [ ][ ]
2 3 5 𝑥2
⇒ 𝑦 = 𝐴𝑋 → 1

& 𝑦1 = 5𝑥1 + 3𝑥2 , 𝑦2 = 3𝑥1 + 5𝑥2 → 2

Consider 𝐴𝑋 = 𝜆𝑋 → 3

2 ⇒ 5𝑥1 + 3𝑥2 = 𝑦1 = 𝐴𝑋 = 𝜆𝑋 = 𝜆𝑥1 by 1 & 3

⇒ (5 − 𝜆)𝑥1 + 3𝑥2 = 0 → 4
&
2 ⇒ 3𝑥1 + 5𝑥2 = 𝑦2 = 𝐴𝑋 = 𝜆𝑋 = 𝜆x2 by 1 & 3

⇒ 3𝑥1 + (5 − 𝜆)𝑥2 = 0 → 5
i) Eigen values=?
5−𝜆 3 |
Characteristic equation is | =0
3 5−𝜆
(5 − 𝜆 )2 − 9 = 0

𝜆2 − 10𝜆 + 25 − 9 = 0
𝜆2 − 10𝜆 + 16 = 0
⇒ 𝜆1 = 8 & 𝜆2 = 2
ii) Eigen vectors=?
If 𝝀𝟏 = 𝟖,

4 ⇒ −3𝑥1 + 3𝑥2 = 0

⇒ −𝑥1 + 𝑥2 = 0 → 6

5 ⇒ 3𝑥1 − 3𝑥2 = 0

⇒ 𝑥1 − 𝑥2 = 0 → 7

6 & 7 are same equation.

6 ⇒ 𝑥1 = 𝑥2

90
 1
First eigen vector = 𝑋1 = [ ]
1
If 𝝀𝟏 = 𝟐,

4 ⇒ 3𝑥1 + 3𝑥2 = 0

⇒ 𝑥1 + 𝑥2 = 0 → 8

5 ⇒ 3x1 + 3x2 = 0

⇒ x1 + x2 = 0 → 9

8 & 9 are same equation.

8 ⇒ x1 = −x2
x1 x2
=
−1 1
−1
Second Eigen vector = X2 = [ ]
1

Problem :

The eigenvectors of a 3 × 3 real symmetric matrix A corresponding to the eigenvalues 2,3,6

are [1 0 −1]𝑇 , [1 1 1]𝑇 , [−1 2 −1]𝑇 respectively, find the matrix
Answer:
Eigen values are 2,3,6
2 0 0
⇒ 𝐷 = [0 3 0]
0 0 6
1 1 −1
Eigen vectors are 𝑋1 = [ 0 ], 𝑋2 = [1], 𝑋1 = [ 2 ]
−1 1 −1
⇒ 𝑙 (𝑋1 ) = √1 + 0 + 1 = √2

⇒ 𝑙 (𝑋2 ) = √1 + 1 + 1 = √3

⇒ 𝑙 (𝑋3 ) = √1 + 4 + 4 = √6
𝑥1 𝑥1 𝑥1 1 1 −1
−
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 ) √2 √3 √6
𝑥2 𝑥2 𝑥2 1 2
⇒𝑁 = = 0
𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 ) √3 √6
𝑥3 𝑥3 𝑥3 1 1 −1
[𝑙(𝑋1 ) 𝑙(𝑋2 ) 𝑙(𝑋3 )] [ √2 √3 √6 ]
W.K.T 𝐷 = 𝑁 𝑇 𝐴𝑁

91
Premultiply by N & post multiply by 𝑁 𝑇
𝑁𝐷𝑁 𝑇 = 𝐴
⇒ 𝐴 = 𝑁𝐷𝑁 𝑇
𝑇
1 1 −1 1 1 −1
− −
√2 √3 √6 √2 √3 √6
1 2 2 0 0 1 2
𝐴= 0 [0 3 0] 0
√3 √6 0 0 6 √3 √6
1 1 −1 1 1 −1
[ √2 √3 √6 ] [ √2 √3 √6 ]
3 −1 1
𝐴 = [−1 5 −1]
1 −1 3

92