Diagonalizability: Theorems Part 2 - Example Solved Problems

DIAGONALIZABILITY - THEOREMS  PART 2

WORKED EXAMPLE PROBLEMS

Example 6

Test the matrix A =  ∈ M_2×2(R), for diagonalizable.

Solution:

Given that A = 

The characteristic equation is |A−λI| = 0.

 | A−λI | =  = 0.

⇒ (1 − λ)² − 0 = 0

⇒λ=1,1

The eigenvalues of A are 1, 1. Here λ=1 with multiplicity 2.

For λ= 1, (A−λI) becomes (A−1 • I).

 [A −1•1] = 

The rank of [A−1 • I] = 1.

We know that for diagonalizability, the multiplicity of λ equals

[n-rank of (A−λI)]. Here n = 2.

 n−rank [A−I] = 2−1 = 1 ≠ 2 which is not the multiplicity of λ= 1.

Hence the given matrix A is not diagonaliable.

 

Example 7

Test the matrix A=  ∈ M_3×3(R) for diagonalizability.

Solution:

The characteristic polynomial of A is

 f(t) = |A − tI| = 

= (3−t) [(3− t) (4 − t)]

which splits the matrix A. The eigenvalues of A are 3, 3, 4.

 λ₁=4 with multiplicity 1 and λ₂=3 with multiplicity 2.

Case (i): For λ₁=4

Rank (A−4I) = 2.

n−Rank (A−4I) = 3 − 2 = 1 = multiplicity of λ₁.

 Conditions (i) and (ii) satisfied for λ₁.

Case (ii): For λ₂ = 3

Rank (A−3I) = 2.

 (ie). n−Rank (A −3I) = 3 − 2 = 1 which is not the multiplicity of λ₂.

Condition (ii) fails for λ₂.

 A is not diagonalisable.

 

Example 8

Let T be a linear operator on P₂(R) defined by T [ƒ(x)] = ƒ(1) +ƒ ′ (0) x + [ƒ′ (0) +ƒ” (0)]x². Test for diagonalisability.

Solution:

Let α be the standard ordered basis for P₂(R). And the eigenvalues of T and ordered basis for P₂(R), such that the matrix of the given transformation with respect to the new resultant basis β is a diagonal matrix.

The basis for P₂(R) is α = { 1, x, x² ́}

When f (t) = 1;

T(1)=1+0x+0x²

T (x) = 1 + (1)x + (1+0)x²

T(x²) = 1 + 0 + (0+2)x²

 

When f(t) = 1

Characteristic polynomial for β=|β−tI|

 f(t) = (1 − t) [(1 − t) (2−t)] − 1[0−0] + 1 [0 − 0]

= (1−1)² (2−t)

⇒ λ=1,1,2

Eigenvalue λ₁ = 1 with multiplicity m=2 and

The eigenvalue λ₂ = 2 with multiplicity m = 1.

Case (i): When λ₁ = 1

 rank (β−λ₁I) =

 

n−rank (β−λ₁I) = 3−1 = 2 = m (multiplicity 2)

The two conditions are satisfied.

Case (ii): When λ₂=2

 n – rank (β−λ₂I)=3−2=1=m (multiplicity 1).

The two conditions are satisfied.

T is diagonalisable.

To find the corresponding basis, We know that (A−λI)X=0.

When λ=2:

 −x₁+x₂+x₃=0

−x₂ = 0

From the above equation x₃=x₁

Now put x₁ =k;

dim E_λ=1≤Multiplicity = 1.

When λ=1;

  0+x₂+x₃=0.

  0+0+0=0

  0+x₂+x₃=0

   x₁=0; x₂ = −x₃

Now put x₃ = k.

is linearly independent.

 β = {1+x², 1, −x+x²} is basis for P₂(R) and also consisting of eigenvectors of T.

  T is diagonalizable

 f(x)=1+x² ⇒ ƒ(1)=2

 f'(x)=2x ⇒ f '(0) = 0

 f" (x)=2 ⇒  ƒ" (0)=2

T(1+x²)=2+0+(0+2)x²=2+2x²

= 2 (1 + x²) + 0 (1) + 0 (−x + x²)

T(1)=1=0 (1+x²) +1 (1) + 0 (x + x²)

T (− x + x²) = 0 + (−1)x + (−1+2) x²

= 1 (− x + x²) + 0.(1) +0 (1+x²)

[T]_β=

 

Example 9

Find Aⁿ for the given matrix A. Verify matrix  A =  is diagonalizable or not?

Solution:

The characteristic equation is

  = −t(3−t)+2 = t²−3t+2

 The eigenvalues are 1 and 2.

  λ₁ = 1, λ₂ = 2 (Here λ=t).

Since the matrix has 2 distinct eigenvalues, the matrix is diagonalizable.

Since A=QDQ^‒1 then

Aⁿ = (QDQ^‒1)ⁿ = (QDQ^‒1) (QDQ^‒1) (QDQ^‒1) ........ (QDQ^‒1) (n times)

Aⁿ=QDⁿ Q^‒1

 

Example 10

Let V=P₃(R). Define T[f(x)]=f '(x)+ƒ” (x). Test T for diagonalizability, and find the basis β and compute [T]_β.

Solution:

Given that T [f(x)] = f '(x) +ƒ'' (x)

Let us consider f(x) = c₀+c₁x + c₂x₂ + c₃x₃.          (V=P₃(R))

 f '(x)= c₁+2c₂x+3c₃x₂

 ƒ"(x)=2c₂+6c₃x

  Equation (1) becomes

 T [ƒ(x)] = [c₁+2c₂x+3c₃x²] + [2c₂+6c₃x]

The matrix representation of T in the standard basis is

 β = {1, x, x², x³)                       (V=P₃(R)).

 T [ƒ (x)] = (c₁ +2c₂) + (2c₂ + 6c₃) x + (3c₃) x².

T[f(x)] = (0c₀+c₁ +2c₂ +0c₃) + (0c₀ + 0c₁ +2c₂+6c₃)x + (0c₀ +0c₁ + 0c₂ + 3c₃)x² + (0c₀ + 0c¡ + 0c₂ + 0c₃)x3 

⇒ (− λ) [(0 − λ) (λ² − 0) ‒ 2(0 − 0) + 6(0 − 0)]

  (−1) [0(λ²−0) − 2(0−0) + 6 (0−0)] + (2)[0 (0−0) + λ(0−0) + 6(0−0)] + 0[0−0] = 0

 λ⁴ = 0

⇒ λ=0,0,0,0

Hence the eigenvalues λ=0 is with multiplicity 4.

When λ=0, we know that for diagonalizability, the multiplicity of λ equals n−rank (T−λI).

(ie) n - rank (T−0.I).

 Here n = 4;

 Rank of (T−0I) is 3.

  n − rank (T – 0I) = 4 − 3 = 1 ≠ 4

Hence [T]_β is not diagonalizable.

 

Example 11

Let V = P₂(R). Define T[f(x)] = f(0) + f(1)(x+x²). Test T for diagonalizability and find a basis β such that [T]_β is a diagonal matrix.

Solution:

Let V=P₂(R)

β={ 1, x, x² }

Given that T[f(x)] = ƒ(0) + ƒ(1) (x + x²)

Let f(x) = ax² + bx + c

Then f(0)=0+0+c=c and

 f(1) = a+b+c

T[ax² + bx + c] = c+(a+b+c)(x+x²)

= c + (a + b + c)x + (a+b+c)x²

The matrix [T]_β = 

 B = [T]_β = 

The characteristic polynomial for B is | T−λI| = 0

On expanding the above determinant we get λ³−3λ²+2λ = 0.

  The eigenvalues are λ=0, 1, 2.

Since all the eigenvalues are distinct, [T]_β is diagonalizable.

To find the eigenvectors, let (B‒λI)X = 0.

 (1−λ)x₁+0x₂+0x₃=0

1x₁+(1−λ)x₂+x₃=0

x₁+ x₂ + (1−2)x₃ = 0.

                 …………(1)

Case (i): When λ=0, the system of equations (1) becomes

x₁+0x₂+0x₃=0

x₁ + x₂+x₃=0

x₁+x₂+x₃=0

Since the last two equations are same let us consider the first two equations.

 x₁+0x₂+0x₃=0

 x₁+x₂+x₃=0

Case (ii): When λ=1, the system of equations (1) becomes

0x₁ +0x₂ +0x₃=0

x₁+0x₂+x₃=0

x₁+x₂+0x₃=0

Let us consider the last two equations

x₁+0x₂+x₃=0

x₁ + x₂ +0x₃=0

When λ= 1, the corresponding eigenvector is 

Case (iii): When λ=2, the system of equations (1) becomes

− x₁ + 0x₂ +0x₃ = 0

 x₁‒x₂+x₃=0

 x₁+x₂‒x₃=0

Let us consider the last two equations.

 x₁‒x₂+x₃=0

 x₁+x₂‒x₃=0

When λ=2 the corresponding eigenvector is 

 The basis in R³ is β = 

By converting the above basis into P₂(R), we have

 β = { 0 − 1x + 1x², −1+x+x², 0+x+x²}

Hence the required basis is

 β = {−x+x², − 1 + x + x², x + x² }

 

Example 12

Let A =  Test A for diagonalizability and if A is diagonalizable then the invertible matrix Q such that Q^‒1AQ=D.

Solution:

Let A= 

⇒   (1−λ) (1− λ) – 0 = 0

 (1− λ)² = 0

 The two eigenvalues are λ= 1, 1

Hence the eigenvalue λ = 1 is with multiplicity 2.

We know that, the condition for diagonalizable is the multiplicity of λ equals, n − Rank (A −λI). Where n is the number of eigenvalues.

Now (A−λI) becomes (A−1.I).

The rank of (A − 1∙I) = 

  n = 2 and rank of (A − I) is 1.

..n−rank(A−1) = 2−1 = 1 ≠ 2.

Hence the matrix A is not diagonalizable.

 

Example 13

 Let A =  Test A for diagonalizability and if A is diagonalizable, then find the invertible matrix Q such that Q^‒1AQ=D.

Solution:

Let A = 

Let | A−λI| =0

 = 0

⇒  (1− λ)² – 9 = 0.      (ie) 1 +λ² − 2λ − 9 = 0

⇒ λ² −2λ − 8 = 0.    (ie) (λ + 2) (λ−4)=0

The eigenvalues are λ=−2 and 4.

Both the eigenvalues are with multiplicity 1 each.

Case (i): When λ=−2, (A−λI) becomes (A + 2I).

We know that for diagonalizable, the multiplicity of λ equals n−rank (A – λI).

Rank of (A +2I) is 

The rank of (A+2I) is 1.

 n− rank of (A+2I)=2−1=1=1 (multiplicity 1)

Case (ii): When λ=4, (A−λI) becomes (A−4I).

The rank of (4−4I) = 

The rank of (A−4I) is 1.

n−rank (A−4I) = 2−1=1=1 (Multiplicity 1)

  By the condition of diagonalizability, the matrix A is diagonalizable.

To find the eigenvector:

Let us consider (A‒λI)X=0, where X= 

(1‒λ)x₁ + 3x₂ = 0 and 3x₁ + (1‒λ)x₂ = 0

When λ=−2: The system of equations (1) becomes,

 3x₁ +3x₂=0 and 3x₁ +3x₂ = 0

We have a single equation x₁ + x₂ = 0

 x₁=−x₂

Let us put x₁=−1, then x₂ = 1

 The eigenvector for λ=−2 is 

When λ=−4: The system of equations (1) becomes

 −3x₁+3x₂ = 0 and 3x₁−3x₂=0

  x₁+x₂=0

 x₁=x₂

Let us put x₁=1, then x₂ = 1.

 The eigenvector for λ= 4 is 

The basis B =  are linearly independents

The matrix Q = 

To find Q^‒1; let us apply Cayley−Hamilton theorem.

|Q−λI| = 

 (−1− λ) (1−λ)−1 = 0

⇒ − 1 + λ − λ + λ²−1=0

⇒ λ²−2=0

⇒ Q² – 2I = 0

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

Q − 2Q^‒1 =0

⇒ 2Q^‒1 = Q

Hence the matrix A is diagonalizable and the invertible matrix is Q =  and the diagonal matrix is D= .

 

Example 14

Let A= . Test A for diagonalizability and if A is diagonalizable find the invertible matrix Q such that Q^‒1AQ=D.

Solution:

Let A= 

 |A−λI| = = λ³−5λ²+3λ+9

= (λ+1) (λ−3) (λ−3)||

 The eigenvalues are λ= ‒1, 3, 3

Here λ=−1 is with multiplicity 1 and λ=3 is with multiplicity 2

Case (i): When λ=−1

Here n = 3 and rank (A + I) = 2

The multiplicity of the eigenvalue λ=−1 is 1 and is equal to 3− rank (A+I) = 3‒2 = 1

Case (ii): When λ=3

Here n = 3 and rank (A+I)=1

The multiplicity of the eigenvalue λ = 3 is 2 and is equal to

3− rank (A+I)=3−1=2.

By the condition of diagonalizability, the given matrix A is diagonalizable.

 

Example 15

Let A =  ∈ M_n×n(R). Test for diagonalizability for the given matrix A.

Solution:

Let A= 

 |A−λI| = 

= λ³− λ²+2−1

= (1 − λ) (λ² + 1)

Here the characteristic polynomial | A− λI | = (1−λ) (λ²+1)

does not split over R.

Since it is not splitable the given matrix A is not diagonalizable.

 

Example 16

Let V = P₂(R).

Define: T (ax² + bx + c) = cx² + bx + a.

Test T for diagonalizability and if T is diagonalizable, find a basis β for V such that [T] is a diagonal matrix.

Solution:

Let V=P₂(R)

Given that T (ax² + bx + c) = cx² + bx+ a

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

P₂(R) = { 1, x, x² }

T (1) = 0x²+0x + a = a = 0x²+0x + 1•a

T(x) = 0x² + bx +0 = bx = 0x² + 1•bx +0

T (x²) = cx²+0x+0=cx² + 1 cx²+0x+0

= (1 − λ) (λ² – 1)

 The eigenvalues are λ = 1 & λ²=1.               (λ ±1)

  λ=1, 1−1

Case (i) For λ= 1 (with multiplicity 2):

 (T−λI)=

The rank of (T− 1 • 1) = 1

 n− Rank (T−1.I)=3−1=2= multiplicity of λ = 1 is 2.

Hence T is diagonalizable.

To find the eigenvectors

(T−λI)X=0

⇒ 

 (0 − λ)x₁ + 0x₂ + x₃ = 0

 0x₁+ (1− λ)x₂ + 0x₃ = 0

 x₁+0x₂+(0−λ)x₃=0

                             ………..(1)

Case (i) when λ=1:

The system of equation (1) becomes

 −x₁ +0x₂ +0x₃ =0

 0x₁ + 0x₂ + 0x₃ = 0

 x₁+0x₂ ‒ x₃ = 0

⇒ x₁‒x₃=0

⇒x₁ = x₃

When x₁ = 1: x₃ = 1 & x₂=0.

The corresponding eigenvector is 

When x₁ =0; x₃=0, x₂= any arbitrary value = 1.

  The corresponding eigenvector is 

Case (ii) When λ=−1:

The system of equation (1) becomes

x₁ +0x₂ + 1x³ = 0

0x₁ + 2x₂ +0x₃ = 0

x₁+0x₂+x₃=0

⇒ x₁+x₃=0 & x₂=0

 x₁=−x₃ & x₂=0

.. When x₁ = 1; x₃=−1, x₂=0

The corresponding eigenvector is 

Hence the basis β for the matrix T is 

 The corresponding bases in P₂(R) is

  β={1+x²,x,1−x}

Hence T is diagonalizable with (1+x², x, 1−x²}

 

Example 17

Let V=R³. Define: . Check whether T is diagonalizable or not.

Solution:

 Given that V=R³ & 

Let β={(1, 0, 0), (0, 1, 0), (0, 0, 1)) be the standard basis for R³.

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

 The matrix representation of T is

The characteristic polynomial is |T−λI|=0

= λ³−2λ²+λ−2.

Since the characteristic polynomial of T does not split, T is not diagonalizable.