Exercise 6.6.10

We use induction on the dimension $n$ of $V$. If $n=1$, $U$ and $T$ will be diagonalized simultaneously by any orthonormal basis. Suppose the statement is true for $n\le k-1$. Consider the case $n=k$. Now pick one arbitrary eigenspace $W=E_{\lambda }$ of $T$ for some eigenvalue $\lambda$. Note that $W$ is $T$-invariant naturally and $U$-invariant since

for all $w\in W$. If $W=V$, then we may apply Theorem 6.16 to the operator $U$ and get an orthonormal basis $\beta$ consisting of eigenvectors of $U$. Those vectors will also be eigenvectors of $T$. If $W$ is a proper subspace of $V$, we may apply the induction hypothesis to $T_W$ and $U_W$, which are normal by Exercise 6.4.7 and Exercise 6.4.8, and get an orthonormal basis ${\beta }_1$ for $W$ consisting of eigenvectors of $T_W$ and $U_W$. So those vectors are also eigenvectors of $T$ and $U$. On the other hand, we know that $W^{\text{⊥}}$ is also $T$- and $U$-invariant by Exercise 6.4.7. They are also normal operators by Exercise 6.4.7(d). Again, by applying the induction hypothesis we get an orthonormal basis ${\beta }_2$ for $W^{\text{⊥}}$ consisting of eigenvectors of $T$ and $U$. Since $V$ is finite dimensional, we know that $\beta ={\beta }_1\cup {\beta }_2$ is an orthonormal basis for $V$ consisting of eigenvectors of $T$ and $U$.