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If $A$ and $B$ are normal matrices and they commute $(AB=BA)$, then:

  1. $A+B $ is normal
  2. $AB $ is normal
  3. $A$ and $B$ are simultaneously diagonalizable: there is a unitary matrix $U$ such that both $U^*AU$ and $U^*BU$ are diagonal.

How can I prove the above statements?

I found the proposition above on Wikipedia, https://en.wikipedia.org/wiki/Normal_matrix#Consequences. I could have used it to prove another theorem, but I could not prove the proposition itself.

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  • $\begingroup$ Note that askers here are expected to provide context for their questions, as is explained here. It would be helpful if you could edit your question to tell us where you encountered this statement, what your thoughts are on this proof, and any other relevant thoughts that you have. $\endgroup$ May 2, 2022 at 23:42
  • $\begingroup$ Part 1 is fairly straightforward to prove using the definition of normality (i.e. that $A$ is normal iff $A^*A = AA^*$); keep in mind that $AB = BA$ implies that $A^*B^* = B^*A^*$. Part 2 is probably the trickiest; one approach is to use the following fact: if $A$ is normal and $AB = BA$, then $A^*B = BA^*$ (this result is sometimes called "Fuglede's theorem"). Part 3 is also a bit tricky; you could go for an inductive proof, or you can use the fact that any pair of commuting matrices can necessarily be simultaneously upper-triangularized with a unitary change of basis. $\endgroup$ May 2, 2022 at 23:50
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    $\begingroup$ My approach would be to prove part 3 first, and then 1 and 2 follow in a fairly straightforward manner. As for part 3, my approach would be to use the fact $A$ and $B$ commuting implies that each eigenspace of $A$ is $B$-invariant; then the eigenspaces of $A$ are mutually orthogonal and sum to the whole vector space using the spectral theorem on $A$, and then the restriction of $B$ to each of these is also unitarily diagonalizable. $\endgroup$ May 3, 2022 at 0:12
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    $\begingroup$ @BenGrossmann I don't see how to get around needing to use Fuglede's theorem or something similar for 1 as well: if you expand the definition, you need to show $A^* A + A^* B + B^* A + B^* B = A A^* + B A^* + A B^* + B B^*$, and normality of $A$ and $B$ only take care of the first and fourth terms respectively. $\endgroup$ May 3, 2022 at 0:16
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    $\begingroup$ @RodolfoOviedo Thanks. Nice notes. $\endgroup$ May 3, 2022 at 15:01

1 Answer 1

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Part 3 follows from the following theorem (proven at https://math.stackexchange.com/a/56491/434895):

Theorem. Two diagonalizable matrices $A$ and $B$ commute ($AB=BA$) if and only if they are simultaneously diagonalizable, that is, there exists an invertible matrix $P$ such that both $P^{-1}AP$ and $P^{-1}BP$ are diagonal matrices.

This theorem can be specialized to our case because, being normal, $A$ and $B$ are diagonalizable with $P=U$, where $U$ is unitary, whence $P^{-1}=U^*$:

Theorem. Two normal matrices $A$ and $B$ commute ($AB=BA$) if and only if they are simultaneously diagonalizable, that is, there exists a unitary matrix $U$ such that both $U^*AP$ and $U^*BP$ are diagonal matrices.

In what follows, we use the previous theorem to prove, first, part 2 and, second, part 1.

Let us define two diagonal matrices $D = U^* A U$ and $E = U^* B U$, whence $A = U D U^*$ and $B = U E U^*$.

$ AB = U D U^* U E U^* $
$ AB = U D E U^* $
$ (AB)^* = U D^* E^* U^* $, where $D^*$ and $E^*$ are the conjugate of $D$ and $E$

$ AB (AB)^* = U D E U^* U D^* E^* U^* = U D D^* E E^* U^* $
$ (AB)^* AB = U D^* E^* U^* U D E U^* = U D D^* E E^* U^* $

Therefore, $AB$ is normal, which proves Part 2.

$ A+B = U D U^* + U E U^* $
$ A+B = U (D U^* + E U^*) $
$ A+B = U (D+E) U^* $
$ (A+B)^* = U (D+E)^* U^* $

$ (A+B) (A+B)^* = U (D+E) U^* U (D+E)^* U^* = U (D+E)(D+E)^* U^* $
$ (A+B)^* (A+B) = U (D+E)^* U^* U (D+E) U^* = U (D+E)(D+E)^* U^* $

Therefore, $A+B$ is normal, which proves Part 1.

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