Assume T to be the linear operator on $R^4$ whose matrix relative to the standard basis is
$$Q=\begin{bmatrix} 0 & 0 & 0 & 0 \\ a & 0 & 0 & 0 \\ 0 & b & 0 & 0\\0 & 0 & c & 0 \end{bmatrix}$$
When would T be diagonalisable?
I know that Q is not diagonalisable because the diagonal is all 0's and I thought the associated matrix is diagonalisable if and only if linear operator is diagonalisable.
The characteristic polynomial of $T$ is $x^4$, so, $0$ is the only eigenvalue of $T$. Now note that $v \in \mathbb R^4 \setminus \{0\}$ is an eigenvector of $T$ if and only if $v \in \ker T$. Thus, $T$ is diagonalizable if and only if there are four linearly independent vectors in $\ker T$, that is, $T$ is diagonalizable if and only if $\ker T = \mathbb R^4$, which means, $T$ is diagonalizable if and only if $T=0$ $(a=b=c=0)$.