Exercise 5.A.23

Proof. Suppose $\lambda$ is an eigenvalue of $\mathit{ST}$ and $v$ a corresponding eigenvector. Then $\mathit{STv}=\mathit{\lambda v}$. Applying $T$ to both sides, we have

If $\mathit{Tv}\ne 0$, then $\lambda$ is indeed an eigenvalue of $\mathit{TS}$.

If $\mathit{Tv}=0$, then $\lambda =0$. Because $\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T\ge 1$ (since $v$ is non-zero), it follows that $\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\mathit{TS}\ge 1$, hence there is a vector $u\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\mathit{TS}$, with $u\ne 0$, such that $\mathit{TSu}=0=\mathit{\lambda u}$. □