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.

  • 1
    $\begingroup$ In the above context, orthonormal usually means with respect to real matrices. Also look up the Schur decomposition (of matrices). $\endgroup$
    – copper.hat
    Apr 25, 2022 at 19:42
  • 1
    $\begingroup$ 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 $\mathbb{C}^n$. $\endgroup$
    – Muduri
    Apr 25, 2022 at 22:34

1 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.

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.


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