# Unitarily diagonalizable V.S. unitary matrix

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.

• In the above context, orthonormal usually means with respect to real matrices. Also look up the Schur decomposition (of matrices). Apr 25, 2022 at 19:42
• What you said are mostly correct. Just one remark, $\langle v, w\rangle=vw^*$ (usually one puts this for the purpose that the second component can be antilinear) does not provide an inner product simply because you will get a matrix instead of a number. So some may set the trace of this product to be the inner product of matrices. I guess what you wanted to say is perhaps to take the inner product of the column vectors in $v$ and the line vectors in $w^*$. In that case, the two notions coincide on the level of the Hilbert space $\mathbb{C}^n$. Apr 25, 2022 at 22:34

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.

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

For the second question, what I wrote is correct.