A doubt about diagonalization of hermitian and not hermitian linear operators

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In Shankar's Principle of Quantum Mechanics, Theorem 10, he says:

To every Hermitian operator $\Omega$, there exists (at least) a basis consisting of its orthonormal eigenvectors. It is diagonal in this eigenbasis and has its eigenvalues as its diagonal entries.

$$ \Omega \leftrightarrow\begin{bmatrix} \omega_{1} & 0 & 0 & \cdots & 0\\ 0 & \omega_{2} & 0 & & 0\\ 0 & 0 & \omega_{3} & & 0\\ \vdots & & & \ddots \\ 0 & 0 & 0 & & \omega_{n} \end{bmatrix} $$

After that, in the subtopic "Diagonalization of Hermitian Matrices" he says:

If $\Omega$ is a Hermitian matrix, there exists a unitary matrix $U$ (built out of the eigenvectors of $\Omega$) such that $U^{\dagger}\Omega U$ is diagonal. Thus the problem of finding a basis that diagonalizes $\Omega$ is equivalent to solving its eigenvalue problem.


My doubt appeared when I was working on the exercises. Look at it below:

Exercise $1.8.5^{*}$ Consider the matrix $$ \Omega = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos \theta\end{bmatrix}$$

(1) Show that it is unitary.

(2) Show that its eigenvalues are $e^{i\theta}$ and $e^{-i\theta}$.

(3) Find the corresponding eigenvectors; show that they are orthogonal.

(4) Verify that $U^{\dagger}\Omega U$ = (diagonal matrix), where U is the matrix of eigenvectors of $\Omega$.


On this exercise, we can easily see that $\Omega$ is not Hermitian since $\Omega^{\dagger} \neq \Omega$. Nevertheless, when I finished all the tasks of the exercise, it turns out that $U^{\dagger}\Omega U$ = (diagonal matrix) where the diagonal entries were the eigenvalues of $\Omega$, the same results we were learning before that happened when $\Omega$ was Hermitian. I want to know whether the following conclusion that I have thought is right:

If $\Omega$ is a Hermitian operator, then it is undoubtedly possible to derive a diagonal matrix, by unitary change in basis, whose diagonal entries are the $\Omega$ eigenvalues. However, if $\Omega$ is not Hermitian it may or may not be possible to find a diagonal matrix whose diagonal entries are the $\Omega$ eigenvalues.

I do not want to follow to the other pages before being sure I have understood it right. This final conclusion is correct or am I missing something?

Thank you in advance!