Diagonalizing orthonormal basis? Let $V$ be a complex vector space with an inner product $\langle\cdot,\cdot\rangle: V\times V \rightarrow \mathbb{C}$, and let $T: V \rightarrow V$ be a diagonalizable linear transformation. Using the change of basis formula, it's easy to see that there is a basis $\mathscr{B}$ such that $[T]_{\mathscr{B}}$, matrix representation of $T$ in $\mathscr{B}$ is diagonal, but it may not be orthonormal. Is there an orthonormal basis $\mathscr{B}$ such that $[T]_{\mathscr{B}}$ is diagonal? 
 A: Not necessarily. Suppose, for instance that $V=\mathbb C^2$, that $\langle\cdot,\cdot\rangle$ is the usual inner product, and that $T$ is sch that $T(1,0)=(1,0)$ and that $T(1,1)=(2,2)$. Then $T$ is not diagonalisable with respect to an orthonormal basis, since the only bases $\mathcal B$ such that $[T]_{\mathcal B}$ is diagonal are those of the form $\bigl\{\alpha(1,0),\beta(1,1)\bigr\}$, with $\alpha,\beta\neq0$.
A: If $T^* = T$ then for different eigenvalues, their corresponding eigenvectors are orthogonal, for the same eigenvalue $\lambda$ one can always find an orthonormal basis of vectors with eigenvalue $\lambda$ so in this particular case, there is an orthonormal basis of eigenvectors.
In general it is not possible. Note that there isnt a natural choice of eigenvectors but the  choice of eigenspaces is unique and natural up to permutation (spaces such that the restriction of $T$ is of the form $\lambda \, \text{Id}$). Those spaces need not be orthogonal at all.
EDIT: see @josé-carlos-santos's answer for a counter-example.
