= Solution
On each nonzero <eigenspace> $V_\lambda$, the restricted <linear map> $S_\lambda=S|_{V_\lambda}$ is <diagonalizable>. Indeed $m_S(S_\lambda)=0$, so the <minimal polynomial> of $S_\lambda$ divides $m_S$. A divisor of a product of distinct <linear polynomials> has the same property, proving the <diagonalizable restriction lemma>.
Choose a <basis> of $S_\lambda$-<eigenvectors> in each $V_\lambda$. Every chosen vector is simultaneously a $T$-<eigenvector>, with <eigenvalue> $\lambda$. The <direct sum> in part (ii) shows that joining these bases gives a <basis> of $V$. Therefore \b[$S$ and $T$ have a common <eigenbasis>], which is exactly <simultaneous diagonalization>.
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