Exercise 8.D.6

Proof. No vector in the span of

is in $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->N$, because applying $N$ to any such vector we get a linear of combination of

which is linearly independent. The vectors above are all in $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->N$, therefore, by the Fundamental Theorem of Linear Maps (3.22), the dimension of $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->N$ is at most the dimension of $V$ minus the dimension of the span of the vectors above, that is, at most $n$. Since $N^{m_1}{v}_1,\dots ,N^{m_n}\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->N$ is linearly independent and has length $n$, it must be a basis of $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->N$. □