Exercise 8.D.3

Proof. Let $M(N)$ denote the matrix of $N$ with respect to some Jordan basis for $N$. Then $M(N)$ is a block diagonal matrix of the form

Because $N$ is nilpotent, $0$ is the only eigenvalue of $N$ (see Exercise 7 in section 8A). Hence the diagonal entries of $M(N)$ are all $0$ and each $A_j$ has the following form

Thus every string of consecutive $1$’s corresponds to one of these blocks and its length is the same as the length of the side of the block minus $1$. It is easy to see that if $A_j$ is $n$-by-$n$, then $A_j^{n-1}\ne 0$ and $A_j^n=0$ (think of it as the matrix of an operator on a $n$ dimensional vector space with respect some basis, each basis vector is mapped to the previous one, except the first one obviously, and to send the last one to $0$ we have to apply the operator $n$ times). Exercise 9 in section 8B now implies that $M{(N)}^m\ne 0$ and $M{(N)}^{m+1}=0$. Hence the minimal polynomial of $N$ is $z^{m+1}$. □