Exercise 5.A.29

Proof. Let $n=k$ and use the same notation from 3.22. If $\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T\ge 1$, then $0$ is an eigenvalue of $T$ an the corresponding eigenvectors are the non-zero vectors in $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T$. Because maximum length of a linearly independent list outside $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T$ is $k$, by 5.10 it follows that there exists at most other $k$ eigenvalues, as desired. □