DIAGONALIZATION OF MATRIX

When is A diagonalizable? (The answer lies in examining the eigenvalues and eigenvectors of A.)

Note that

Altogether

Equivalently,

Example 1 :

Diagonalize the following matrix, if possible.

Solution :

Step 1 :

Find the eigenvalues of A.

= (2 - λ)²(1 - λ) = 0

Eigen values of A : λ = 2 and λ = 1.

Step 2 : 

Find three linearly independent eigenvectors of A.

By solving

(A - λI)x = 0,

for each value of λ, we obtain the following :

Step 3 :

Construct P from the vectors in step 2.

Step 4 :

Construct D from the corresponding eigenvalues.

Step 5 :

Check your work by verifying that AP = PD.

Example 2 :

Diagonalize the following matrix, if possible.

Solution : 

Since this matrix is triangular, the eigenvalues are λ₁ = 2 and λ₂ = 4. By solving (A - λI)x = 0 for each eigenvalue, we would find the following :

Every eigenvector of A is a multiple of v₁ or v₂ which means there are not three linearly independent eigenvectors of A and by Theorem, A is not diagonalizable.

Example 3 :

Solution :

Since A has three eigenvalues :

λ₁ = 2, λ₂ = 6, λ₃ = 1

and since eigenvectors corresponding to distinct eigenvalues are linearly independent, A has three linearly independent eigenvectors and it is therefore diagonalizable.

Example 4 :

Diagonalize the following matrix, if possible.

Solution:

Eigenvalues : -2 and 2 (each with multiplicity 2).

Solving (A -λI)x = 0 yields the following eigenspace basis sets.

{v₁, v₂, v₃, v₄} is linearly independent.

---> P = [v₁ v₂ v₃ v₄] is invertible

---> A = PDP^-1, where

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