Eigenvalues and Eigenvectors

Unit-6, Lecture-1, 2, 3
Lecture-1, 2

1 Eigenvalue and Eigenvector

A number λ ∈ F is called an eigenvalue (latent root) of a linear transfor-

mation T : V −→ V if ∃ u ∈ V , u 6= 0 such that T (u) = λu. The non

zero vector u is called eigenvector corresponding to the eigenvalue λ. As we

know that we can find a square matrix representation A for a given linear

transformation T : V −→ V , so a linear transformation means a matrix and

vice versa.

How o find Eigenvalues? Let A = [aij ] be a square matrix of order n. (

A may be singular or non-singular). If X is a eigenvector corresponding to

the eigenvalue λ then, AX = λX ⇒ (A − λI)X = 0.

If the homogeneous system of equation (A − λI)X = 0 has a non-trivial

solution, then Rank(A − λI) < n = number of variables ⇒ the coefficient

matrix (A − λI) is singular i.e.

|A − λI| = pn (λ) = (−1)n (λn − c2 λn−1 + c2 λn−2 − ... + (−1)n cn ) = 0

or λn − c1 λn−1 + . . . + (−1)n−1 cn−1 λ + (−1)n cn = 0. (This is called

characteristic equation, and its roots are called characteristic roots/

eigenvalues)

If λ1 , λ2 , . . . , λn are roots of characteristic equation, then by the relation

between characteristic roots and coefficients of polynomial equation, we have,

λ1 + λ2 + . . . + λn = c1 = a11 + a22 + ... + ann = T race(A).

2
 λ1 λ2 + λ1 λ3 + ... + λn−1 λn = c2
.. .. .. .. .. ..
. . . . . .

λ1 .λ2 . . . . λn = (−1)n cn = det(A)

Spectrum of A- It is the set of all eigenvalues of A

Spectrum Radius of A- max{|λ| : λ ∈ spectrmof A}

Eigenspace - If A is an n×n square matrix and λ is an eigenvalue of A, then

the union of the zero vector 0 and the set of all eigenvectors corresponding

to eigenvalues λ is a subspace of Rn known as the eigenspace of λ.

Characteristic space/Eigenspace corresponding to λ is given by {x ∈ V :

T (x) = λx}.

Remark 1.1. If |A| = 0 ⇒ one of the eigenvalue is 0, converse is also true.

Remark 1.2. If A is a triangular matrix then eigenvalues of A are the diagonal

entries of A.

Let V be a finite dimensional vector space over a field F and T : V → V

be a linear operator on V . Then x ∈ V , x 6= 0, is called a characteristic

vector corresponding to eigenvalue λ if T (x) = λx.

Theorem 1.1. Let V be a finite dimensional vector space over a field F and T :

V → V is a linear transformation on V . Then the following are equivalent:

(i) λ is a eigen value of T .

(ii) The operator (T − λI) is singular.

(iii) |T − λI| = 0.

Proof. If λ is a characteristic value of T , then ∃ x 6= 0 such that T x = λx.

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Therefore (T − λI)x = 0. Homogeneous system of equations have non trivial

solution only if (T − λI) is singular i.e. |T − λI| = 0.

1.1 Eigenvalue Problem

Consider a homogeneous system of equation in matrix form AX = λX,

where λ is a scalar. This system of equation always has a trival solution. We

need to find values of λ for which non trivial solution of this homogeneous

system exists. The values of λ for which nontrival solution of AX = λX

exists are eigen values and corresponding eigenvector are solutions. If X is a

solution, then αX is also a solution, where α is scalar. Hence eigenvector is

not unique (unique up to constant multiple). Thus problem of determining

the eigenvalues and corresponding eigenvectors of a square matrix A is called

eigenvalue problem.

Definition 1.1. A square matrix A = [aij ] is called

−1
1. Unitary if AT = (A) .

2. Orthogonal if AT = A−1 . i.e. AAT = I.

3. Hermitian if A = AT or A = (A)T .

4. Skew Hermitian if A = −AT .

T T
5. Positive definite if X AX > 0, for X 6= 0 and X AX = 0, iff

X = 0.

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Properties of Eigenvalues and Eigenvectors If λ is an eigenvalue of the

matrix A and X is eigenvector corresponding to λ, then

1. Matrix αA has eigenvalue αλ and the eigenvector is same. ( AX = λX

implies αAX = αλX)

2. Matrix Am has eigenvalue λm , and the corresponding eigenvector is

same.(AX = λX ⇒ Am X = λm X)

3. Matrix (A−kI) has eigenvalue λ−k, and the corresponding eigenvector

is same. (AX = λX ⇒ (A − k I)X = AX − kIX = λX − kX =

(λ − k)X)

4. Matrix A−1 has eigenvalue λ−1 ,and the corresponding eigenvector is

same. ( AX = λX implies A−1 AX = λA−1 X implies A−1 X = λ−1 X.

1
5. Matrix (A−k I)−1 has eigen value , and the corresponding eigen-
λ−k
vector is same..

6. Eigenvalues of matrix A and AT are same.

7. For a real matrix A, if α + iβ is an eigenvalue, then α − iβ is also an

eigenvalue.

8. If X is a eigenvector corresponding to the eigenvalue λ, then cX (c

is a constant) is also an eigenvector for corresponding to the same

eigenvalue. So, the eigenvector corresponding to a eigenvalue is not

unique, i.e. eigenvector is unique up to constant multiple.

5
 9. Trace of A = sum of eigen values of A.

10. Det A = Product of eigen values of A.

1
11. If λ is an eigenvalue of an orthogonal matrix Am then λ
is also an

eigenvalue.

12. Eigenvectors corresponding to the distinct eigenvalues are linearly in-

dependent.
 
 5 4 
Example 1.2. Find the eigenvalues and eigenvectors of the matrix A =  .
1 2
Solution-

Eigenvalues of A are given by A − λI = 0 ,i.e

5−λ 4
= 0 ⇒ (λ − 6)(λ − 1) = 0 ⇒ λ = 6, 1.
1 2−λ
Thus the eigenvalues of A are λ = 6, 1.

If X = (x, y)T be aeigenvector corresponding

   to the eigenvalue
 λ, then

 5−λ 4  x   0 
(A − λI)X = 0 ⇒    =  
1 2−λ y 0
Case-I. Eigenvector corresponding to λ = 6.

We
 have,    
 5−6 4  x   0 
   =  
1 2−6 y 0
    
 −1 4   x   0 
⇒   =  
1 −4 y 0
⇒ x − 4y = 0 ⇒ x = 4y = 4k (say).

6
 Thus a eigenvectors
   corresponding to λ = 6 is given by (4, 1)T and the

 4 

 

eigenspace is k   .
1 

 
(k is a scalar)
Case-II. Eigenvector corresponding to λ = 1.

We
 have,    
 5−1 4  x   0 
   =  
1 2−1 y 0
    
 4 4  x   0 
⇒   =  
1 1 y 0
⇒ x + y = 0 ⇒ x = −y = k (say).

Thus a eigenvector
  corresponding
 to λ = 1 is (1, −1)T and the eigenspace

 1 

 

is given by k   .
−1

 

(k is a scalar)

Now consider the following example.

Example 1.3. Findeigenvalues

 and corresponding
  eigenvectors-

 1 1 0   1 1 0   1 0 0 
     
 0 1 1  (ii)  0 1 0  (iii)  0 1 0 
(i)      
     
0 0 1 0 0 1 0 0 1
Solution- For each of the above problem, we obtain the characteristic equation

as (λ − 1)3 = 0 ⇒ eigenvalues are given by λ = 1, 1, 1. Eigenvalue λ = 1 is

with multiplicity 3 here.

Let us find the eigenvectors X = (x, y, z)T as below-

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     
 0 1 0  x   0 
    
(i) (A − λX) =  0 0  y  =  0 
1     
    
0 0 0 z 0
⇒ y = 0, z = 0, and x is arbitrary.
  



 1  


   
Thus the eigenvectors are x  0
 
 .

  
 
0 

 
(x is a scalar)
In this case we find
 only oneLIeigenvector.
  
 0 1 0  x   0 
    
(ii) (A − λX) =   0 0  y  =  0
0     

    
0 0 0 z 0
⇒ y = 0, z, and x are arbitrary. Taking x = 0, z = 1 and then z =

0, x = 1, we find two linearly independent eigenvector X1 = (0, 0, 1)T and

X2 = (1, 0, 0)T . Any other eigenvector will be linear combination of these

two vectors.     



  1   0 



    
Thus the eigenvectors are x 
 0  + z  0 
   .

    
 
0 1 

 
(x, z are scalars)
In this case we find
 two LI eigenvectors.
   
 0 0 0  x   0 
    
(ii) (A − λX) =   0 0 0  y  =  0 
   
    
0 0 0 z 0
⇒ y, z, and x are arbitrary. Taking x = 1, , y = 0z = 0, x = 0, , y =

1z = 0 and x = 0, y = 0, z = 1, we find three linearly independent eigenvector

X1 = (1, 0, 0)T , X2 = (0, 1, 0)T , and X3 = (0, 0, 1)T . Any other eigenvector

8
will be linear combination of 
these
 three
 vectors.
   



 1   0  0 


   

  

Thus the eigenvectors are x  0 +y 1

+z 0 
   .

      
 
0 0 1 

 
(x, y, z are scalars)
In this case we find three LI eigenvectors.

Thus the number of linearly independent eigenvectors corresponding to a

given eigenvalue λ depends on the rank of (A − λI). In the above example,

the rank of (A − λI) are 2, 1, and 0 respectively.

Remark 1.3. Eigenvectors corresponding to distinct eigenvalues are linearly

independent (LI).

Remark 1.4. If λ is an eigenvalue with multiplicity m of a square matrix A

of order n, then the number of LI eigenvectors associated with λ is given by

p = n − r,where r = Rank(A − λI), 1 ≤ p ≤ m.

Example 1.4. If λ is an eigenvalue of an invertible matrix A , then show that

|A|
is the eigen value of adj(A).
λ
Solution- AX = λX (Given)
1 |A|
⇒ adj(A)X = |A|A−1 X = |A| X = X
λ λ
|A|
⇒ is the eigenvalue of adj(A).
λ

Example 1.5. Find the

 eigenvalues, eigenvectors and eigenspaces of the ma-
 3 2 4 
 
trix A = 
 2 0 2 

 
4 2 3

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Solution-

Characteristic equation of A is

|A − λX| = 0
3−λ 2 4
⇒ 2 0−λ 2 = 0 ⇒ −λ3 + 6λ2 + 15λ + 8 = 0 ⇒ −(λ +

4 2 3−λ
1)2 (λ − 8) = 0 ⇒ λ = 8, −1, −1

Thus the eigenvalues of A are 8, -1( with multiplicity 2).

Eigenvectors-

Eigenvector X = (x, y, z)T corresponding to the eigenvalue λ is given by

(A −
λI)X = 0    
 3−λ 2 4  x   0 
    
 2 0−λ 2  y  =  0 
    
    
4 2 3−λ z 0

Case-I
 Eigenvector
 for λ
=8 

 −5 2 4  x   0 
    
 2 −8 2   y  =  0 
    
    
4 2 −5 z 0
    
 1 −4 1   x   0 
     1
⇒ −5 2 4  y  =  0
  
(using
 2
and R21 )
    
4 2 −5 z 0

10
     
 1 −4 1   x   0 
    
⇒  0 −18 9   y  =  0 (using R2 → R2 + 5R1 and R3 →
   
    
0 18 −9 z 0
R3 − 4R
1 )    
 1 −4 1   x   0 
     1
⇒  0 −2 1   y  =  0 (using R3 → R3 + R2 and 9 R2 )
   
    
0 0 0 z 0
⇒ x − 2y = 0 and −2y + z = 0 ⇒ x = z and 2y = z.

Taking y = k we get x = 2k, z = 2k

Thus the eigenvector corresponding to λ = 8 is X = (2, 1, 2)T .

Eigenspace is {k(2, 1, 2)T }k isanyscalar

Case-II. Eigenvector
    for λ= −1

 4 2 4  x   0 
    
 2 1 2  y  =  0 
    
    
4 2 4 z 0
⇒ 2x + y + 2z = 0 or y = −2(x + z), x and z are arbitrary.

Taking x = 1, z = 0 we get y − −2, taking x = 0, z = 1 we get y = −2

Thus two LI eigenvectors are X1 = (1, −2, 0)T and X2 = (0, −2, 1)T .

Eigenspace is {k1 (1, −2, 0)T + k2 (0, −2, 1)T }{k1 k2 aretwoscalars}

Characteristic Roots of Some Special Matrices

1. All the eigenvalues of a Hermitian matrix (A = AT ) are real.

Proof.- AX = λX ⇒ X ? (AX) = X ? (λX) ⇒ (X ? (AX))? = (X ? (λX))? ⇒

X ? A? X = X ? λ̄X ⇒ X ? AX = X ? λ̄X ⇒ X ? λX = λ̄X ? X ⇒ X ? X(λ −

11
 λ̄) = 0 ⇒ λ = λ̄

2. All the eigenvalues of a real symmetric matrix are real. (real symmetric

matrix can be taken as complex Hermitian matrix)

3. All nonzero eigen values of a skew Hermitian matrix A? = −A) are

pure imaginary.

Proof-Let λ is a eigenvalue of A, AX = λX. Since, A is skew hermitian,

we have (iA)? = −iA? = iA ⇒ iA is hermitian, ⇒ i(AX) = i(λX) ⇒

(iA)X = iλX ⇒ iλ is real ⇒ λ = 0 or purely imaginary.

4. The modulus of eigenvalue of an orthogonal matrix (AAT = AT A = I)

is 1.

Proof-AX = λX ⇒ (AX)T AX = (AX)T λX ⇒ X T AT AX = (λX)T λX ⇒

X T X = λ2 X T X ⇒ X T X(λ2 − 1) = 0 ⇒ λ2 = 1 ⇒ |λ| = 1

5. Modulus of eigenvalues of unitary matrix (A? A = I) is 1.

Proof- AX = λX ⇒ (AX)? = (λX)? ⇒ X ? A? = λ̄X ? ⇒ X ? A? AX =

λ̄X ? AX ⇒ X ? = λ̄λX ? X ⇒ X ? X(λλ̄ − 1) = 0Rightarrowλλ̄ = 1 or

|λ| = 1

6. Eigenvalues of a positive definite matrix are real and positive.

7. All leading minors of a positive definite matrix are positive.

12
 Lecture-3
Theorem 1.2. (Cayley-Hamilton Theorem) Every square matrix A of order

n satisfies its own characteristic equation. i.e.,

If |A − λI| = λn − c2 λn−1 + c2 λn−2 − ...(−1)n−1 cn−1 + (−1)n cn = 0, then

An − c1 An−1 + . . . + (−1)n−1 cn−1 A + (−1)n cn I = 0.

Proof. Let |A − λI| = λn − c1 λn−1 + c2 λn−2 − ... + (−1)n Cn = 0 (1)

Since cofactors of the elements of the determinant |(A − λI)| are polyno-

mial in λ of degree n − 1 or less. Therefore the elements of adjoint matrix

(transpose of cofactor matrix) are also polynomial in λ of degree n−1 or less.

Hence, adj(A − λI) can be expressed as a polynomial in λ of degree n − 1

with coefficients Bi , which are square matrix of order n having elements as

functions of the elements of A, i.e.,

adj(A − λI) = B1 λn−1 + B2 λn−2 + . . . + Bn−1 λ + Bn (2)

Also we have,

(A − λI) adj(A − λI) = |(A − λI)| I.

(A − λI)(B1 λn−1 + B2 λn−2 + . . . + Bn−1 λ + Bn )

= λn I − c1 λn−1 I + c2 λn−2 I + . . . + (−1)n−1 cn−1 A + (−1)n cn I.

Comparing the coefficients of powers of λ

−B1 = I

AB1 − B2 = −c1 I

AB2 − B3 = c2 I
..
.

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 ABn−1 − Bn = (−1)n cn−1 I

ABn = (−1)n cn I

Pre-multiplying these equations by An , An−1 , . . . , A, I and adding we have

An − c1 An−1 + . . . + (−1)n−1 cn−1 A + (−1)n cn I = 0.

Deduction-

(−1)n n−1
• Using this theorem we can obtain A−1 = − (A − c1 An−2 + ... +
cn
(−1)n−1 cn−1 I)

• We can obtain An as An = c1 An−1 − c2 An−2 . . . + (−1)n−1 cn−1 A +

(−1)n cn I
 
 1 4 
Example 1.6. Verify Cayley-Hamilton theorem for the matrix A =  
2 3
and find inverse also. Express A5 − 4A4 − 7A3 + 11A2 − A − 10I as a linear

polynomial in A.

Solution- Characteristic equation of A is given by

|A − λI| = 0 ⇒ λ2 − 4λ − 5 = 0 (1)

By Cayley-Hamilton theorem

A2 − 4A − 5I = 0 (2)

Here,        
 9 16   4 16   5 0   0 0 
A2 − 4A − 5I =  − − = =0
8 17 8 12 0 5 0 0
⇒ Cayley-Hamilton theorem is verified.

14
 −1
Multiplying equation (2) by A , we get     
1  1 4   1 0   −3 4 
 
 1
A−4I−5A−1 = 0 ⇒ A−1 =  − 4  = 
5 0 1  5
 
 2 3 2 −1


Finally, A5 − 4A4 − 7A3 + 11A2 − A − 10I = (A2 − 4A − 5I)(A3 − 2A +

3I) + A + 5I = A + 5I ( by equation (2))

Example 1.7. Find eigen values and eigen vectos of the matrix

 
 1 1 2 
 
 −1 2 1  .
A= 
 
0 1 3

Answer: |A − λI| = 6 − 11λ + 6λ2 − λ3 = (λ − 1)(λ − 2)(λ − 3) = 0.

λ = 1, 2, 3 and Eigen vectors X1 = {a(−1, −2, 1)T }, X2 = {a(−1, −1, 1)T },

X3 = {a(1, 0, 1)T } respectively.

Example 1.8. Verify Cayley-Hamilton theorem for the matrix

 
 1 2 0 
 
A=
 −1 1 2  .

 
1 2 1

Hence (i) find A3 and A−1 (ii) Verify that eigen values of A2 are squares of

eigen values of A. (iii) Find spectral radius.

15
Solution- |A−λI| = −λ3 +3λ2 −λ+3 = (λ−3)(λ2 +1) = 0. Then λ = 3, ±i.

   
 −1 4 4   −1 10 12 
   
A2 = 
 0 3 4  and A3 = A2 .A =  1 11 10
 
.

   
0 6 5 −1 16 17

 
 −3 −2 4 
1 1
T hen A−1

= [A2 − 3A + I] =  3 1 −2 .
3 3



−3 0 3

Eigen values of A2 are 9, −1, −1. Spectral radius = Max|λi | = 3

Example 1.9. If

 
 1 0 0 
 
A=
 1 0 1 .

 
0 1 0

Then show that An =A

n−2
+ A2 − I, for
 all n ≥ 3.
 1−λ 0 0 
 
Solution- |A−λI| =  1 −λ 1  = λ3 −λ2 −λ+1 = (1−λ)(λ2 −1) =
 
 
0 1 −λ
0.

A3 − A2 − A + I = 0

A3 − A2 = A − I

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 A4 − A3 = A2 − A
..
.

An−1 − An−2 = An−3 − An−4

An − An−1 = An−2 − I.

Adding these equations we have An −A2 = An−2 −I. Using these equations
n 1
recursively An = An−4 + A2 − I = An−4 + 2(A2 − I) = A2 − (n − 2)I.
2 2

 
 1 0 0 
 
A2 = 
 1 1 0 .

 
1 0 1

     
 1 0 0   1 0 0   1 0 0 
     
Hence A50 = 25 
 1 1 0  − 24  0 1 0
 
 =  25 1 0 
  
     
1 0 1 0 0 1 25 0 1

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