Simulataneous Diagonalizability

Question

Let $S, T$ be linear operators on an $n$-dimensional vector space. Assume that $T$ has $n$ distinct eigenvalues and $S$ commutes with $T$. Prove that $S$ and $T$ are simultaneously diagonalisable. I could prove that $T$ is diagonalisable. But I dont' know what to do next.

Answer 1

$\newcommand{\diag}{\mathrm{diag}}$ By assumption, we can take $\mathscr{B} = \{\alpha_1, \ldots, \alpha_n\}$ to be an eigenbasis of $V$ such that the matrix of $S$ under $\mathscr{B}$ is a diagonal matrix $A = \diag(\lambda_1, \lambda_2, \ldots, \lambda_n)$, where $\lambda_1, \ldots, \lambda_n$ are all distinct. Denote by $B$ the matrix of $T$ under the same basis $\mathscr{B}$, then $ST = TS$ implies $AB = BA$. Write $B = (b_{ij})_{n \times n}$, then comparing the $(i, j)$ entries of $AB$ and $BA$ reveals that all the off diagonal entries of $B$ are zero, whence $B = \diag(b_{11}, b_{22}, \ldots, b_{nn})$. In summary, we have
\begin{align*} & S(\alpha_1, \ldots, \alpha_n) = (\alpha_1, \ldots, \alpha_n)\diag(\lambda_1, \ldots, \lambda_n), \\ & T(\alpha_1, \ldots, \alpha_n) = (\alpha_1, \ldots, \alpha_n)\diag(b_{11}, \ldots, b_{nn}). \end{align*}

In view of these, $S$ and $T$ are simultaneously diagonalizable.