DIAGONALIZATION OF MATRIX

Diagonalizable :

A square matrix A is said to be diagonalizable if A is similar to a diagonal matrix. That is, if A = PDP-1 where P is invertible and D is a diagonal matrix.

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

Note that

Altogether

Equivalently,

Theorem (Diagonalization)

An n x n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.

In fact, A = PDP-1, with D a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A.

In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P.

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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