0
$\begingroup$

Could someone check if the solution of the problem is right?

Problem:

Let $A, B \in \mathbb{C}^{n\times n}$ be selfadjoint ,such that $[A,B] := AB − BA = 0$ Show that there is a unitary matrix $U \in \mathbb{C}^{n\times n}$ such that $U^*$A$U$ and $U^*$$B$$U$ are both diagonal.

Solution:

Let $D$1 =$U^*$A$U$ and $D$2 =$U^*$$B$$U$ .

As $A$ and $B$ are selfadjoint follows that $D^*$=$(U^*$$A$$U)^*$= $U^*$$A$$U$=$D$1, so $D$1 is hermitian, which also means that $D$1 must be diagonal ?

$\endgroup$
  • $\begingroup$ No it's wrong. you don't have the same $U$ for $A$ and $B$ in the beginning. $\endgroup$ – Yanko Jan 13 at 15:10
1
$\begingroup$

No, it is not right. You started your solution by saying “Let $D_1=U^*AU$”, without saying what $U$ is.

$\endgroup$
  • $\begingroup$ In the problem is written that U is a unitray matrix. $\endgroup$ – Kai Jan 13 at 15:12
  • 1
    $\begingroup$ @Kai The problem asks you to find such unitary $U$. $\endgroup$ – xbh Jan 13 at 15:16
  • $\begingroup$ @ xbh Ahh. I am very confused about the way of how this problem is defined. Do you think that then I should prove that if U*AU is a diagonal then U is a unitary matrix? $\endgroup$ – Kai Jan 13 at 15:25
  • $\begingroup$ @Kai No. What you are supposed to prove is that there is an unitary matrix $U$ such that both matrices $U^*AU$ and $U^*BU$ are diagonal. $\endgroup$ – José Carlos Santos Jan 13 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.