Let $V$ be a finite-dimensional vector space and $T:V\to V$ a diagonalizable map. Show that there is an inner product on $V$ w.r.t. which $T$ is symmetric.
For this question, I suppose a basis for $V$ consisting of eigenvectors, corresponding to eigenvalues $\lambda_i$, and define =vw. However, when I check if it is symmetric, I find $\langle T(v), w\rangle$ does not equal to $\langle v, T(w)\rangle$. Is my hypothetical inner product wrong? And I also try to suppose a orthonormal basis for $V$ but get nothing.
I really appreciate that I can have some hints. Thanks in advance.
If $(v_1,\ldots,v_n)$ is a basis of $V$ of eigenvectors of $T$, define $\langle\cdot,\cdot\rangle$ such that$$(\forall i,j\in\{1,2,\ldots,n\}):\langle v_i,v_j\rangle=\begin{cases}1&\text{ if }i=j\\0&\text{ otherwise.}\end{cases}$$Then, if $\lambda_i$ is the eigenvalue of the eigenvector $v_i$,$$\langle Tv_i,v_j\rangle=\langle v_i,Tv_j\rangle=\begin{cases}\lambda_i&\text{ if }i=j\\0&\text{ otherwise.}\end{cases}$$Therefore, with respect to $\langle\cdot,\cdot\rangle$, $T$ is symmetric.