TITLE: If a Matrix Has Only Zero as an Eigen-Value Then It Is Nilpotent
QUESTION [15 upvotes]: Prove that a matrix with only zero eigenvalues must be nilpotent.

How will I be able to prove this?

REPLY [16 votes]: We have to assume that we are considering complex matrices. Over the reals the assertion is not true, as the example
$$
\begin{bmatrix}0&0&0\\0&0&1\\0&-1&0 \end{bmatrix}
$$
shows.
Over $\mathbb C$, one can do the Schur decomposition, where $A=VTV^*$, with $V$ unitary and $T$ upper triangular. Since the diagonal of $T$ has to contain the eigenvalues of $A$, it has be zero. And it is an easy exercise that if $T$ is an $n\times n$ upper triangular with diagonal zero, then $T^n=0$. So $A^n=(VTV^*)^n=VT^nV^*=0$, and $A$ is nilpotent.