Unitary matrices and Spectral Theorem

Question

Due to the Spectral theorem and Shur's decomposition, if $A$ is a unitary matrix, then

$$A = QDQ^{-1} \quad (1)$$

where $D$ is diagonal and $Q$ unitary.

Now, let $A$ belongs to the center of SU(n) and $P$ to SU(n), so

$$A = PAP^{-1} \quad (2)$$

Therefore, because of Eq. (1) and Eq. (2) we have that

$$PAP^{-1} = QDQ^{-1} \quad (3)$$

My question arises from some readings that ensure that $A \equiv D$, but I don't see how to prove it. Any ideas? Maybe Eq. (1) is a unique decomposition?

Thanks in advance! ;)

Answer 1

To prove

$A \equiv D, \tag 1$

we use what we've got, viz.

$A = QDQ^{-1}, \tag 2$

by casting it in the form

$D = Q^{-1}AQ; \tag 3$

since $A$ is in the center, we may take this equation a tad further:

$D = Q^{-1}AQ = AQ^{-1}Q = AI = A, \tag 4$

as required.

I think the fact that $A$ is central in $SU(n)$ implies that $A = cI$, $c \in \Bbb C$, which would of course resolve everything; but I can't quite remember for sure. What I do know for sure is $A$ central makes similarity transformations on $A$, $A \to B^{-1}AB = AB^{-1}B = AI = A$ pretty damn simple, since $A$ is fixed by any such.