T/F: Whether a matrix is diagonalizable

[Mr Davis 97]

Homework Statement

T/F: The matrix $\begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ is diagonalizable.

Homework Equations

The Attempt at a Solution

Is there a quick way to tell whether the matrix is diagonalizable? Since it's a T/F question, that would seem to be the case. The matrix is not symmetric, so we can't conclude anything there. Also, even though it is triangular, the entries on the diagonal are not distinct, so that's a dead end too.

 

[fresh_42]

You can consider it as a matrix which is built by two blocks: $\begin{bmatrix}2&1&0\\0&2&0\\0&0&3\end{bmatrix}=\begin{bmatrix}A&0\\0&3\end{bmatrix}$. Now you only have to deal with $A$.

 

[Mr Davis 97]

You can consider it as a matrix which is built by two blocks: $\begin{bmatrix}2&1&0\\0&2&0\\0&0&3\end{bmatrix}=\begin{bmatrix}A&0\\0&3\end{bmatrix}$. Now you only have to deal with $A$.

So by brute force calculation I found that A is not diagonalizable. Why does this imply that the bigger matrix is not diagonalizable? Also, is there something characteristic of A that should tell me it's not diagonalizable, or do you have to brute force calculate it?

 

[fresh_42]

So by brute force calculation I found that A is not diagonalizable. Why does this imply that the bigger matrix is not diagonalizable?

If it was diagonalizable, then $A$ would be diagonalized as well. (Edit: And the block with $3$ doesn't contribute to $A$.)

Also, is there something characteristic of A that should tell me it's not diagonalizable, or do you have to brute force calculate it?

Yes, the $1$ beside the diagonal (on the superdiagonal) is the reason. The matrix is in the Jordan normal form. It is already almost diagonal and one can read the eigenvalues. The $1$ indicates, that there is a two-dimensional eigenspace.