Find a Matrix P that diagonalizes A

[mpittma1]

Homework Statement

Find a Matrix P that diagonalizes A

Homework Equations

A = $$\begin{pmatrix}<br /> 2 & 0 & -2\\<br /> 0 & 3 & 0\\<br /> 0 & 0 & 3<br /> \end{pmatrix}$$

The Attempt at a Solution

Well right off the bat we know that this is an upper triangular matrix so the eigenvalues are the entries along the main diagonal of A.

So λ = 2, 3, 3

But if an n x n matrix A has n distinct eigenvalues, then A is diagonalizable.

In this case we only have 2 distinct eigenvalue so it shouldn't be diagonalizable...

But the answer is:
P = $$\begin{pmatrix}<br /> -2 & 0 & 1\\<br /> 0 & 1 & 0\\<br /> 1 & 0 & 0<br /> \end{pmatrix}$$


What is the proper way to start the problem to find this matrix P?

Thank You for any help in advance.

 

[Mark44]

Homework Statement

Find a Matrix P that diagonalizes A

Homework Equations

A = $$\begin{pmatrix}<br /> 2 & 0 & -2\\<br /> 0 & 3 & 0\\<br /> 0 & 0 & 3<br /> \end{pmatrix}$$

The Attempt at a Solution

Well right off the bat we know that this is an upper triangular matrix so the eigenvalues are the entries along the main diagonal of A.

So λ = 2, 3, 3

But if an n x n matrix A has n distinct eigenvalues, then A is diagonalizable.

In this case we only have 2 distinct eigenvalue so it shouldn't be diagonalizable...

But the answer is:
P = $$\begin{pmatrix}<br /> -2 & 0 & 1\\<br /> 0 & 1 & 0\\<br /> 1 & 0 & 0<br /> \end{pmatrix}$$


What is the proper way to start the problem to find this matrix P?

Thank You for any help in advance.

If you check, you'll see that the eigenspace of λ = 3 is of dimension 2, so there are two eigenvectors for this eigenvalue.

When a matrix has repeated eigenvalues, there is some terminology that distinguishes between algebraic multiplicity vs. geometric multiplicity. I think this matrix is a case of geometric multiplicity.

To find your matrix P, the important thing is to get three linearly independent eigenvectors, not whether there are three distinct eigenvalues.