[Math] Why is $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ not diagonalizable

Question

I have a number of sufficient conditions as to when a matrix $A$ is diagonalizable, namely:

1. When $A$ is symmetric
2. When $A$ has distinct eigenvalues

Given $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$

$A$ has nondistinct eigenvalues $\lambda = 0$ with algebraic multiplicity $2$, is there some conditions that says this is when the matrix fails to be diagonalizable?

Answer 1

Assume it is diagonalisable the eigenvalue being $0$ with multiplicity $2$. This means that there exists an invertible matrix $P$ such that

$$\begin{bmatrix}0 & 1\\0 &0\end{bmatrix}=P\cdot\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\cdot P^{-1}=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$$

A contradiction