Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 2 / b / iv / 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 $S$ and $T$ have a common eigenbasis, which is exactly simultaneous diagonalization.