= Solution
\b[Yes, provided “commute” means pairwise commute.] Necessity follows because three <diagonal matrices> commute pairwise.
For sufficiency, first choose the common <eigenbasis> for the first two <linear maps>, say $S,T$, obtained in part (b). Their joint <eigenspaces>
$$
W_{\lambda,\mu}=\ker(S-\lambda I)\cap\ker(T-\mu I)
$$
give a <direct sum> decomposition of $V$. The third map $U$ preserves each $W_{\lambda,\mu}$, since $SUv=USv=\lambda Uv$ and $TUv=UTv=\mu Uv$. By the <diagonalizable restriction lemma>, $U|_{W_{\lambda,\mu}}$ has a <basis> of <eigenvectors> there. Every such vector already has fixed <eigenvalues> for $S$ and $T$.
Joining these bases therefore gives \b[a common <eigenbasis> for all three maps]. The same induction proves <simultaneous diagonalization> for any finite pairwise commuting family of <diagonalizable> <linear endomorphisms>.
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