Exercise 5.4.39

By Exercise 5.2.18, we already get the necessity. For the sufficiency, we use induction on the dimension $n$ of the spaces. If $n=1$, every operator on it is diagonalizable by the standard basis. By supposing the statement is true for $n\le k-1$, consider the case $n=k$. First, if all the operators in $C$ has only one eigenvalue, then we may pick any basis $\beta$ and know that ${[T]}_{\beta }$ is diagonal for all $T\in C$. Otherwise, there must be one operator $T$ possessing two or more eigenvalues, ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_t$. Let $W_i$ be the eigenspace of $T$ corresponding to the eigenvalue ${\lambda }_i$. We know that $V$ is the direct sum of all $W_i$’s. For the same reason in the proof of Exercise 5.4.25, we know that $W_i$ is a $U$-invariant subspace for all $U\in C$. Thus we may apply the induction hypothesis on $W_i$ and

Thus we get a basis ${\beta }_i$ for $W_i$ such that ${[U_{W_i}]}_{{\beta }_i}$ is diagonal. Let $\beta$ be the union of each ${\beta }_i$. By applying Theorem 5.25, we get the desired result.