Finding jordan basis

Question

This matrix isn't diagonizable how would I find the third value?

Answer 1

Take a look here or on your own book:

https://en.wikipedia.org/wiki/Jordan_normal_form

In your case the Jordan canonical form for A should be:

$$\begin{bmatrix} 1&1&0\\ 0&1&0\\ 0&0&0 \end{bmatrix}$$

NOTE

When A is not diagonalizable J-form is the best we can do to approach at the diagonal form

$J$ is obtained by $A$ from a similarity transfomation $$P^{-1}AP=J$$

$P$ can be founb solving the system $$AP=PJ$$