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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.

Rodolfo Oviedo
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  • 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. – Ben Grossmann May 02 '22 at 23:42
  • 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. – Ben Grossmann May 02 '22 at 23:50
  • Another approach to part 2 is to use the fact that there exists a polynomial $p$ such that $A^* = p(A)$, which follows from the spectral theorem and the existence of interpolating polynomials. – Ben Grossmann May 02 '22 at 23:52
  • @BenGrossmann Part 2 follows from part 3. – Ryszard Szwarc May 03 '22 at 00:11
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    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. – Daniel Schepler May 03 '22 at 00:12
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    @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. – Daniel Schepler May 03 '22 at 00:16
  • @RyszardSzwarc Fair point – Ben Grossmann May 03 '22 at 00:23
  • @DanielSchepler You're right; I guess I missed that – Ben Grossmann May 03 '22 at 00:24
  • I have impression the Fuglede theorem takes care of the second and the third term. – Ryszard Szwarc May 03 '22 at 00:30
  • @BenGrossmann I found https://www.math.drexel.edu/~foucart/TeachingFiles/F12/M504Lect2.pdf, which proves part 3. – Rodolfo Oviedo May 03 '22 at 14:55
  • @RyszardSzwarc Please see my comment above. – Rodolfo Oviedo May 03 '22 at 14:57
  • @DanielSchepler Please see my comment above. – Rodolfo Oviedo May 03 '22 at 14:57
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    @RodolfoOviedo Thanks. Nice notes. – Ryszard Szwarc May 03 '22 at 15:01
  • @RyszardSzwarc I posted an answer below. As I am not a mathematician, I'd like to know if you consider it correct. Thanks! – Rodolfo Oviedo May 17 '22 at 22:50
  • @RodolfoOviedo You do not need the second theorem, i.e. Part 3, to prove Part 1 and Part 2. You can do it directly. For example if $AB=BA$ then $B^A^=A^B^$ which means that all $4$ operators $A,B,A^,B^$ commute with each other. Hence $$(AB)^AB==(AB)(AB)^$$ Moreover $(A+B)^(A+B)=(A+B)(A+B)^.$ The same is true for any operator of the form $p(A,A^,B,B^),$ where $p$ is a polynomial of $4$ variables. The main problem is how you get Part 3. By commutativity you know that $A=PDP^{-1}$ and $B=PEP^{-1}.$ But how do you know that you can replace $P$ with a unitary matrix ? – Ryszard Szwarc May 18 '22 at 08:57
  • @RodolfoOviedo The previous comment was too long so I could not complete it. In order to replace $P$ with $U$ the fact that $A$ and $B$ are normal should be used. – Ryszard Szwarc May 18 '22 at 09:01

1 Answers1

1

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.

Rodolfo Oviedo
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