Exercise 8.A.12

Proof. Suppose $v_1,\dots ,v_n$ is such basis. Then $N{v}_1=0$, because the the first column of the matrix has $0$ in all its entries. The definition of matrix of linear map shows that $N{v}_2\in \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1)$. But this implies that $N^2{v}_2=0$. Similarly, $N{v}_3\in \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,v_2)$, so $N^3{v}_3=0$. Continuing like this, we see that $N^j{v}_j=0$, for each $j=1,\dots ,n$. Therefore $N^n=0$ and so $N$ is nilpotent. □