Exercise 7.B.13

Proof. We will just prove (a) implies (b), since that’s the part that uses Schur’s Theorem, and we will do so by induction on $\dim <!--\; FUNCTION\; APPLICATION\; -->V$.

Suppose (a) holds. Note that if $\dim <!--\; FUNCTION\; APPLICATION\; -->V=1$, then (a) trivially implies (b), just take any non-zero vector in $V$ and divide it by its norm. Now suppose that $\dim <!--\; FUNCTION\; APPLICATION\; -->V>1$ and that (a) implies (b) for all complex inner product spaces of smaller dimension.

Choose an eigenvector $u$ of $T$ with $\left|\right|u\left|\right|=1$ (the existence of an eigenvalue is guaranteed by 5.21). Let $U=\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(u)$. Since $U$ is invariant under $T$, it follows by Exercise 3 in section 7A that $U^{\text{⊥}}$ is invariant under $T^{\text{∗}}$. By 6.50 $\dim <!--\; FUNCTION\; APPLICATION\; -->U^{\text{⊥}}=\dim <!--\; FUNCTION\; APPLICATION\; -->V-1$. Thus, by the induction hypothesis there exists an orthonormal basis of $U^{\text{⊥}}$ consisting of eigenvectors of $T^{\text{∗}}{\text{|}}_{U^{\text{⊥}}}$, which are also eigenvectors of $T$, by 7.21. Moreover, adjoining $u$ to this basis produces an orthornomal basis of $V$ consisting of eigenvectors of $T$. Therefore (b) holds. □