Eigenvalues and Eigenvectors: Theorem 3 - Example Solved Problems

EIGENVALUES AND EIGENVECTORS – THEOREM 3

EXAMPLE PROBLEMS

Example 8

Find all the eigenvectors of the matrix A = 

Solution:

  ƒ(t) = |A−λI_n|

=  = λ²−2λ−3 = (λ − 3) (λ + 1)

The eigenvalues are λ₁ = 3 and λ₂ = −1.

To find eigenvector:

Case (i): Let us consider λ₁=3.

B₁ = A – λ₁I = 

−2x₁+x₂=0

4x₁−2x₂=0

Let us consider

x₁=k; x₂=2k.

Case (ii): Let us consider λ₂ = −1.

⇒ B₂ = (A−λ₂I) = 

  (A − λ₂I) X =0

⇒   2x₁+x₂=0 and 4x₁ + 2x₂ = 0.

Put x₁ = k then x₂=−2k.

is a basis of R² and consisting eigenvector of A.

 A is diagonalisable.

To find the diagonal matrix

Similarity transformation:

 

Example 9

Let V=R²; . T is a linear operator on V and β is the ordered basis. Find [T]_β and determine β is a basis consisting of eigenvectors of T.

Solution:

It is given that V=R² and T: R² → R² defined by

Here [T]_β is not a diagonal matrix.

Therefore  does not contain the eigenvectors of T.

 

Example 10

Let V=P₁(R) and T: P₁(R) → P₁(R) defined by T[a + bx]=(6a‒6b)+(12a−11b)x and β={3+ 4x, 2+3x}. Compute [T]_β and determine whether β is a basis consisting of eigenvectors of T.

Solution:

It is given that T: P₁(R) → P₁(R).

T (a + bx) = (6a − 6b) + (12a – 11b) x and β= {3+4x, 2 + 3x }

T[3+4x] = [6(3) − 6(4)] + [12(3) ‒ 11(4)]x

= (18 −24) + (36 – 44)x

= −6−8x

T[3+4x] = −2[3+4x] + 0[2x + 3x]

             ... (1)

T[2+3x] = [6(2) − 6(3)] + [12(2) ‒ 11(3)]x

= (12−18) + (24−33)x

= −6−9x

  T[2+3x] = 0(3+4x) − 3(2+3x)

From equations (1) and (2),

The corresponding matrix [T] is

 [T]_β = 

Since [T]_β is a diagonal matrix, the given basis

β ={ 3+ 4x, 2 + 3x } contains the eigenvectors of T.

 

Example 11

Let V=P₁(R) and T: P₁(R) → P₁(R) defined by T(ax + b) = (−6a + 2b)x + (− 6a+b). Find the eigenvalues of T and the basis β such that [T] is diagonal matrix.

Solution:

It is given that V=P₁(R) and T: P₁(R) → P₁(R) T(ax+b) = (−6a+2b)x + (−6a+b).

Then the matrix representation is obtained as [T] = 

⇒ | T−λI| = 0

⇒  = (−6−λ) (1− λ) + 12 = 0

 −6 +6λ −λ + λ² + 12 =0

⇒ λ² + 5λ + 6 =0

⇒ (λ+3) (λ+2)=0

The eigenvalues are λ = −2, – 3.

To find the eigenvectors, we have [T−λ]X=0, where X= 

 (−6 − 2)x₁ + 2x₂ =0 and

 − 6x₁+ (1 −λ)x₂ = 0

Case (i): When λ=−2, the above system of equations become

− 4x₁+2x₂=0 and

 6x₁+3x₂=0

 The single equation is −2x₁+x₂=0

 2x₁ = x₂

 x₁/1 = x₂/1

 X₁ = 

Case (ii) When λ=−3, the system of equations become

 −3x₁+2x₂=0 and

− 6x₁+4x₂=0

The single equation is − 3x₁ + 2x₂ = 0

 2x₂ = 3x₁

x₂/3 = x₁/2

Hence X₂ = 

From equations (1) and (2)

The basis β = 

 β= { 2x + 3, x+2} is a basis for P₁(R) which consists of T.

Now T = 

Now T =  = 4−3 = 10.

Since |T| ≠ 0, the above vectors are linearly independent.

  T is a diagonal matrix.

Hence [T] =  is a diagonal matrix.

 

Example 12

Let V=P₃(R). Let T: P₃(R) → P₃(R), defined by T[a + bx + cx² + dx³] = (−d) + [−c+d]x + [a+b−2c]x² + [−b + c −2d]x³ and β={1−x+x³, 1+x², 1, x+x²} compute [T]_β and determine whether β is a basis of eigenvectors of T.

Solution:

It is given that T: P₃(R) → P₃(R).

T[a + bx + cx² + dx³] = (− d) + [− c + d]x + [a + b − 2c]x² + [−b+c−2d]x³

and β = { 1−x+x³, 1+x², 1, x+x²)

To prove [T] is linear ac

 T[1−x+x³] = T[1−x+0x²+x³]

= T[a + bx + cx² + dx³]

By comparing these equations, we have

a=1,b=−1, c=0 and d=1

Given that

T[a + bx + cx² + dx³]

= [d]+[−c+d]x + [a+b−2c]x² + [−b+c−2d]x³.

 T[1 − x + x³] = [− 1] + [− 0 + 1]x + [1 − 1 − 2.0]x² + [1+0−2 (1)] x³

 = (− 1) + (1)x + (0)x² + (−1)x³

 T[1−x+x³] = −1+x−x³

  T[1 − x + x³] = − 1(1 − x + x³) + (0)(1 + x²) + 0(1) + 0(x + x²)

              ………..(1)

 T[1 + x²] = T [1 +0 • x + 1 • x² + 0 ⋅ x³]

Here a= 1, b = 0, c = 1 and d=0

 T [1 + x²] = [− 0] + [− 1 + 0] x + [1 + 0 − 2.1] x² + [− 0 + 1 − 2 (0)] x³

= (− 1) x + (− 1) x² + (1) x³

= −x−x² + x³

 T[1 + x²] = 1(1 − x + x³) + (− 1)(1 + x²) + 0(1) + 0(x+x²)

            ………….(2)

T[1] = T [1+0x+0x²+0x³]

Here a = 1, b=0, c=0 and d=0.

 T[1] = (− 0) + ( − 0 + 0) x + (1 +0 −2•0) x² + (− 0 + 0 − 2 • 0) x³

  = x²

  T [1] = 0 (1 − x + x³) + (1)(1 + x²) + (− 1)(1) + 0(x + x²)

              …….... (3)

T[x + x²] = T [0 +1 • x + 1 • x² + 0 • x³]

Here a = 0, b = 1, c = 1, and d=0.

= (−0)+(−1+0)x+(0+1−2) x²+(−1+1 −0) x³

T[x+x²] = − x − x²

 T[x + x²] = 0(1 − x + x³) + 0(1 + x²) + 0(1) + (−1)(x + x²)

                   …………..(4)

  From equations (1), (2), (3) and (4) we have

 T[x]β =

 This is not a diagonal matrix.

 So the basis β = {1−x + x³, 1 + x², 1, x+x²} does not contain the eigenvectors of T.

 

Example 13

For a matrix A =  ∈ M_n×n(F)

(i) Find all the eigenvalues and eigenvectors

(ii) Find a basis consisting of eigenvectors of A.

(iii) Find an invertible matrix Q and a diagonal matrix D such that Q^‒1AQ=D.

Solution:

Given matrix is A = 

 (i) To find eigenvalues

Let |A−λI| =0

 = 0

⇒ (1 − λ) (2 − λ) − 6 = 0

 λ²−3λ−4=0

  (λ+1) (λ−4)=0

⇒ λ = −1,4

 The eigenvalues are λ = −1, 4

To find the eigenvectors let us consider (A‒λI)X = 0

 (1 − λ)x₁ + 2x₂ =0

 3x₁ + (2−λ)x₂=0

Case (i): When λ=−1, then the above equations become,

 2x₁ + 2x₂ = 0

 3x₁ +3x₂ = 0

  The single equation is x₁+x₂ = 0

 x₁ = ‒x₂

x₁/‒1 = x₂/‒1

The eigenvector for λ = −1 is 

Case (ii): When λ=4, then the equations become

−3x₁+2x₂=0

3x₁−2x₂=0

We have a single equation 3x₁ − 2x₂ = 0

3x₁=2x₂

x₁/2 = x₂/3

 is a basis of R² and consisting eigenvectors of A.

A is diagonalizable.

To find the diagonal matrix, let Q = 

To find Q^‒1, let us use Cayley Hamilton theorem.

Now Q = 

 = − (− 1 − λ) (3 − λ) −2 = 0

 −3+λ−3λ+λ²−2=0

 λ²−2^λ−5=0

 Q²−2Q−5=0

Multiply by Q^‒1 on both sides, then we get

⇒ Q−2I−5Q^‒I = 0

5Q⁻¹ = Q−2I

 Q⁻¹ = 1/5 [ Q−2I ]

 

Example 14

For a matrix A = 

 (i) Find all the eigenvalues and eigenvectors.

(ii) Find a basis consisting of eigenvectors of A.

(iii) Determine an invertible matrix Q and a diagonal matrix D such that Q^‒1AQ=D.

Solution:

Given that A = 

 (i) To find the eigenvalues

 ƒ(λ) =│A−λI|

 = 

 = −λ(1−λ)²

The characteristic equation is λ(1−λ)² = 0

⇒ λ=0 or (1−λ)² = 0

⇒ λ=0 or λ = 1, 1

Hence the eigenvalues are λ= 0, 1, 1

To find the eigenvectors

Let us consider (A−λI)X = 0 where 

 (2−λ)x₁ + 0x₂ ‒1x₃ = 0

4x₁ + (1−λ)x₂ −4x₃ = 0

2x₁ +0x₂ + (−1−λ)x₃ = 0

                         ………….(1)

Case (i) when λ = 0:

Put λ=0 in the system of equations (1)

 B₁ = A − λ₁I becomes

 2x₁+0x₂−x₃ = 0

 4x₁+x₂‒4x₃ = 0

 2x₁+0x₂ − x₃ = 0

To solve the above three equations, let us use cross multiplication method,

Case (ii) when λ=1:

Put λ= 1 in the set of equations (1) (ie) B₂ = A − λ₂I becomes

x₁+0x²−x³=0

4x₁+0x₂−4x₃=0

2x₁+0x₂‒2x₃=0

⇒ x₁‒x₃=0

 x₁−x₃=0

 x₁−x₃=0

All the 3 equations are similar.

 x₁=x₃

Put x₁ = k, then x₃=k, x₂=0

The eigenvector corresponding to λ= 1 is 

Case (iii) When λ=1:

Similarly in equation x₁−x₃=0, now we put x₁ = 0.

 x₁=0, x₃=0, x₂ = any arbitrary value...

The eigenvector corresponds to λ=1 is

 N(A−λ₃I) = { , k∈ R }

The three eigenvectors are { }

(ii) The vectors β =  is the basis of A.

Now |A| = 

= 1[0−1] − 1[0−2] + 0[4−0]

= (−1)+2+0

= 1

≠ 0

Since |A| ≠ 0 the above 3 vectors are linearly independent.

 (iii) 

We know that D=Q⁻¹A Q

⇒ D = Q⁻¹A Q

 is a diagonal matrix

 

Example 15

Let T: P₂(R) → P₂(R) be defined as T[f(x)] = f(x) + (x + 1)ƒ'(x). Find eigenvalues and corresponding eigenvectors of T with respect to standard basis of P₂(R).

Solution:

We know that the standard basis of P₂ (R) is

P₂(R) = { 1, x, x² }

Given that T [f(x)] = ƒ (x) + (x + 1) ƒ′ (x)

 T(1) = 1 + (x + 1) (0) = 1 = 1.1+0x+0x²

T(x)=x+(x+1) (1) = 1 + 2x = 1.1+2x+0x²

T(x²) = x²+(x+1) (2x) = x² + 2x² + 2x

= 2x + 3x² = 0.1 + 2x + 3x²

 The corresponding matrix representation is

By expanding the determinant and solving the cubic equation we get the eigenvalues as λ= 1, 2, 3. (ie) λ₁ = 1, λ₂ = 2 & λ₃ = 3.

When λ₁ = 1; The corresponding eigenvector is

 v₁ = (1, 0, 0) = 1 ∈ P₂(R)

When λ₂=2; The eigenvector is v₂ = (1, 1, 0) = 1 + x ∈ P₂(R)

When λ₃= 3; The eigenvector is v₃ = (0, 0, 1) = x² ∈ R₂(R)

 

Example 16

For each linear operator T on V, find the eigenvalues of T and an ordered basis B for V such that [T] is a diagonal matrix.

Define: V=R³ and

T (a, b, c) = (7a−4b+10c, 4a − 3b + 8c, − 2a + b − 2c).

Solution:

Given that V=R³

T (a, b, c) = (7a−4b+ 10c, 4a − 3b+8c, − 2a + b − 2c)

Let B be the standard basis for R³

 (ie) B={(1, 0, 0), (0, 1, 0), (0, 0, 1) }

Let v₁ = (1, 0, 0), v₂ = (0, 1, 0) & v₃ = (0, 0,1)

Then T(v₁) = T(1, 0, 0) = [7 (1) − 4 (0) + 10 (0), 4 (1) − 3 (0) +8 (0), −2(1)+0−2(0)]

⇒ T (1, 0, 0) = (7, 4, − 2)

T (v₂) = T (0, 1, 0)

= [7 (0) − 4 (1) + 10 (0), 4 (0) − 3 (1) + 8 (0), − 2 (0) + (1) − 2 (0)]

T (0, 1, 0) = (−4, −3, 1)

T (v₃) = T (0, 0, 1)

= [7(0) ‒ 4(0) + 10(1), 4(0) ‒ 3(0) + 8(1), − 2(0) + (0) −2(1)]

T (0, 0, 1) = (10, 8,−2)

  The matrix representation of T is

The characteristic polynomial of T is given by

|T−λI| =  = −λ³ + 2λ² + λ − 2

= (λ+1) (λ−1) (λ−2)

 The eigenvalues are λ = −1, 1, 2

To find the eigenvectors

 [T−λI]X = 

(7 − λ) x₁ − 4x₂ + 10x₃=0

4x₁ + (−3−λ)x₂ + 8x₃ = 0

 −2x₁ + 1x₂ + (−2−λ)x₃ = 0

                    …………..(1)

Case (i) when λ=−1

The system of equations (1) becomes

 8x₁ − 4x₂ + 10x₃ = 0

 4x₁‒2x₂+8x₃=0

−2x₁+x₂−x₃=0

Solving the last two equations we get x₁ = 1, x₂=2 & x₃=0

.. The eigenvector corresponding to λ=−1 is 

Case (ii) when λ=1

The system of equations (1) becomes

5x₁‒4x₂+10x₃=0

4x₁‒5x₂+8x₃=0

−2x₁+x₂−3x₃=0

Solving the last two equations, we get x₁ = 1, x₂ = − 1, & x₃ =− 1

The eigenvector corresponding to λ=1 is 

Case (iii) when λ=2

The system of equations (1) becomes

5x₁‒4x₂+10x₃=0

4x₁‒5x₂+8x₃=0

−2x₁+x₂−4x₃=0

Solving the last two equations, we get x₁ = 2, x₂ = 0 & x₃ = −1

  The eigenvector corresponding to λ=2 is 

Therefore, the basis for the matrix T is

Now

 = 1 [1+0] − 1[− 2 − 0] + 2[− 2 + 0] = 1 + 2 −4 = −1 0

 The 3 eigenvectors are linearly independent since | A | ≠ 0.

T is a diagonal matrix

(ie) [T] =  is a diagonal matrix.

 

Example 17

For a matrix A = 

(i) Find all the eigenvalues and eigenvectors

(ii) Find a basis consisting of eigenvectors of A

(iii) Find an invertible matrix Q and a diagonal matrix D such that Q^‒1AQ=D

Solution:

Given that A = 

To find the eigenvalues

Let |A−λI|=0

⇒  = 0

On expanding the determinant we have λ³−7λ²+36=0

By solving this equation we have λ = −2, 3, 6

To find the eigenvectors; let [A‒ λI]X=0

 (1 −λ) x₁ + x₂ + 3x₃ = 0

x₁+(5− λ)x₂+x₃ = 0

3x₁+x₂+(1−λ)x₃ = 0

Case (i): When λ=−2, the eigenvector is X₁ = 

Case (ii): When λ=3, the eigenvector is X₂ = 

Case (iii): When λ=6, the eigenvector is X₃ = 

  The three eigenvectors are {,,}

 (ii) The vector β =  is the basis for A.

Let |A | =   = 1[5−1] − 1[1−3] + 3[1−15] = 4+2−42 ≠ 0.

Since |A| ≠ 0 the above 3 vectors are linearly independent.

 (iii)