Diagonalizability: Theorems Part 1 - Example Solved Problems

DIAGONALIZABILITY - THEOREMS  PART 1

WORKED EXAMPLE PROBLEMS

Example 1

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

Solution:

 f(t) =|A ‒ λI_n| = 

 = λ² − 2λ = λ(2−2)

λ=0, λ =2.

⇒ λ=0,2 the eigenvalues are distinct.

They are diagonalizable.

Example 2

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

Solution:

Given that 

Then |A− λI| = (1−λ)²−16=1+λ²−2λ−16

= λ²−2λ−15 = (λ − 5) (λ+3)

(ie) λ = −3 and λ = 5.

 λ=−3, 5. Since the two eigenvalues are distinct, it is diagonalizable.

Example 3

Find the algebraic multiplicity of all eigenvalues of 

Solution:

⇒ (3 − t) [(3 − t) (4−t) − 0] − 1[0−0] + [0+0] = 0

⇒ (3−t)² (4−t) = 0

 f(t) = (3 − t)² (4 − t)

⇒ t = 3,3 and t=4

The algebraic multiplicity of λ=3 is 2

The algebraic multiplicity of λ=4 is 10

Example 4

Let T be a linear operator on P₂(R) defined by T(f(x)) =ƒ'(x), the matrix representation of T with respect to the standard basis β for P₂(R). Determine T is diagonalizable or not.

Solution:

Standard ordered basis for P₂(R) = { 1, x, x² }

T(1)=0

T(x)=1=1+0(x)+0(x²)

T(x²)=2x=0 (1)+2(x) + 0 (x²)

= − t [t² − 0] − 1[0] + 0[0] = − t³

The characteristic polynomial of T is [[T]_β−tI| = −t³.

Thus t has only one eigenvalue λ=0 with multiplicity m = 3.

Solving T (f(x)) = f '(x) = 0 shows that E_λ = N(T−λI)= N(T) consisting is a subspace of P₂(R), of the constant polynomials. So {1} is the basis for E_λ.

 dim (E_λ)=1.

Consequently, there is no basis for P₂(R) consisting of eigenvectors of T.

T is not diagonalizable.

Example 5

Let T be the linear operator on R³ defined by . Determine the eigenspace of T corresponding to each eigenvalue.

Solution:

Let β be the standard ordered basis for R³.

Basis β = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}

The characteristic polynomial is | [T]_β − λI |

= (4 −λ) [(3 − λ) (4 − λ)] + (λ −3)

= (λ − 3)² (λ − 5)

 Eigenvalues of T are λ₁=3 with multiplicity 2 and λ₂=5 with multiplicity 1.

For λ₁ = 5;

On expanding we get,

 −x₁+x₃ =0

 2x₁−2x₂+2x₃=0

 x₁−x₃=0

By considering the first two equations, we have

x₁ + x₃ = 0

2x₁ +2x₃ = 0

x₁+x₃ = 0

Put x₂=s, x₃=t, we get x₁=−t, x₂=s, x₃ = t.

 dim (E_λ2)=2.

Multiplicity for λ₂ is 2.

In this case the multiplicity of each eigenvalue λ_i is equal to the dimension of the corresponding eigenspace E_λi.

Note that  is linearly independent and hence basis for R³, consisting of eigenvectors T.

  T is diagonalizable.