Let $V$ be a vector space over $\mathbb{R}$. Let $\alpha: V \rightarrow V$ be a nilpotent endomorphism of $V$, i.e. $\alpha^{m}=0$ for some positive integer $m$. Prove that $\alpha$ can be represented by a strictly upper-triangular matrix (with zeros along the diagonal). [You may wish to consider the subspaces $\operatorname{ker}\left(\alpha^{j}\right)$ for $j=1, \ldots, m$.]

Show that if $\alpha$ is nilpotent, then $\alpha^{n}=0$ where $n$ is the dimension of $V$. Give an example of a $4 \times 4$ matrix $M$ such that $M^{4}=0$ but $M^{3} \neq 0$.

Let $A$ be a nilpotent matrix and $I$ the identity matrix. Prove that $I+A$ has all eigenvalues equal to 1 . Is the same true of $(I+A)(I+B)$ if $A$ and $B$ are nilpotent? Justify your answer.