Unitarily diagonalizable V.S. unitary matrix

Question

We work over $\mathbb{C}$. Let $A\in M_n(\mathbb{C})$. (so our vector space is finite dimensional.) We know that

$A$ is unitary iff $AA^*=I=A^*A$ where $^*$ is conjugate transpose.

$A$ is normal iff $AA^*=A^*A$.

In particular unitary matrices are normal.

$A$ is unitarily diagonalizable iff there is a unitary matrix $P$ such that $PAP^{-1}$ is a diagonal matrix.

Question 1: Is unitarily diagonalizable the same as orthonormally diagonalizable in some books? I suppose unitary matrices are just orthonormal matrices with respect to the inner product $\langle v,w\rangle=v^*w$?

I am also quite confused between the unitary matrix and a matrix that is unitarily diagonalizable. The spectral theorem states

$A$ is normal iff $A$ is unitarily diagonalizable.

I think a corollary of spectral theorem is

$A$ is unitary iff $A$ is unitarily diagonalizable and all eigenvalues having absolute value $1$.

Combining these results it seems we get

$A$ is unitary iff $A$ is normal with all eigenvalues having absolute value $1$.

That is the distinction between unitary and unitarily diagonalizable matrices.

Question 2: Are the above statements correct? If not, could you provide an example.

Answer 1

Here is the answer to my own question.

For the first question, we say

$A$ is orthogonal if $A^tA=I=AA^t$.

$A$ is unitary if $A^H A=I=A A^H$, where $H$ means conjugate transpose.

Therefore over $\mathbb{C}$ we usually do not use 'orthonormally' diagonalizable, but unitarily diagonalizable. In this language, there exist an orthonormal (w.r.t. the standard Hermitian inner product below) eigenbasis of $T$ iff $T$ is unitarily diagonalizable in spectral theorem.

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There are two conventions of inner product, written vectors horizontally and vertically. Here we tend to use $\langle v,w\rangle=vw^H$.

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For the second question, what I wrote is correct.