Problem
Let $V$ be a finite inner product space and let $T:V \to V$ be a linear transformation. Prove that $T$ is diagonalizable if and only if the adjoint transformation $T^{*}$ is diagonalizable.
I got stuck with this problem. Since $(T^{*})^{*}=T$, it is sufficient to show the forward implication. Suppose $T$ is diagonalizable, then there exists a basis of eigenvectors $B=\{v_1,...,v_n\}$. I would like to show that $T^{*}$ also has a basis of eigenvectors.
By definition of $T^*$ We have $$\langle T(v_i),v_i \rangle=\langle v_i,T^*(v_i)\rangle$$
But $$\langle T(v_i),v_i \rangle=\alpha \langle v_i,v_i \rangle$$
I don't know what to do next, I would appreciate any hints. Thanks in advance.