A linear map with an orthonormal eigenbasis

Question

Here is a proposition my professor stated in class:

Let V be a finite dimensional inner product space, with $ T: V \to V$ a linear map. Let $\gamma$ be an orthonormal basis of eigenvectors of V.

1. The standard matrix $[T]_\gamma^\gamma$ is diagonal
2. The standard matrix of the adjoint $T^*$, $([T]_\gamma^\gamma)^*$ is also diagonal. Moreover, $T \circ T^* = T^* \circ T$

However, wouldn't we have this for a non-orthonormal basis of eigenvectors of V?

Definition: let V be a finite dim. inner product space. A linear map $T: V \to V$ is said to be normal iff $T \circ T^* = T^* \circ T$

He then proceeds with an example for a linear map $T: \mathbb{C}^2 \to \mathbb{C}^2$. He finds that there are two distinct roots and states that because $T$ has an eigenbasis, then $T$ must have an orthonormal eigenbasis, since you could simply use G-S to get an orthonormal eigenbasis.

I think this is incorrect, specifically because you are not guaranteed to have eigenvectors after performing G-S on them.

My question here is: is it not the case that $T \circ T^* = T^* \circ T$ so long as you take $T$ with respect to the basis of eigenvectors, which produces a diagonal matrix?

a linear map is said to be self-adjoint $T: V \to V$ if $T = T^*$

for a real inner product space, T has an orthonormal eigenbasis iff T is self-adjoint

I have the same question here, where from my understanding, if you have an eigenbasis, then you have a diagonal matrix, and in the real case, you can always say that $T = T^*$ if you take the transformation with respect to that basis, but how does this guarantee an orthonormal eigenbasis?

Answer 1

I think the issue is that you're assuming you can simultaneously diagonalize $T$ and $T^*$ - if you can, then yes, they will commute. But if not? (i.e. if $T^*$ is not diagonal under this change of basis).

For example, take $T = \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}$ in the standard basis. Then $T^* = \begin{bmatrix} 1 & 0 \\ 1 & -1 \end{bmatrix}$, and so $TT^* = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}$ while $T^*T = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$ and if $TT^*\neq T^*T$ under the standard basis, then they're not going to be equal under some other basis.

The issue here is that $T$ and $T^*$ don't have the same eigenvectors, the eigenvalues for both matrices are $1$ and $-1$, and $T$ has corresponding eigenvectors $\begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}-1 \\ 2\end{pmatrix}$, but the eigenvectors for $T^*$ are $\begin{pmatrix}2 \\ 1\end{pmatrix}, \begin{pmatrix}0 \\ 1\end{pmatrix}$.