Apologies if it's a trivial question.

Is it possible to permute ($P$) any arbitrary diagonal matrix $D \in M_n$ such that it yields a scaled identity matrix, i.e., $PDP^* = \alpha I_n$, where $\alpha \in F$ is some scalar?

If yes, then can we prove it? If not, then why?

Thank you so much


If I understand your problem correctly, then you are asking if there exists a matrix $P\in\operatorname{Sym}(n)$ s.t. $PDP^T=\alpha I_n,$ where $D$ is a $n\times n$ diagonal matrix. The answer is clearly no, because $PDP^T$ will have the same entries where the rows and columns are shuffled compared to $D$. I'm excluding the fact that all diagonal elements of $D$ are $\alpha,$ of course (trivial).

  • $\begingroup$ Thank you so much for the clarification. Yes agreed when $D$ is already a multiple of Identity matrix, then it is clear. $\endgroup$ – user550103 May 16 '18 at 14:26
  • $\begingroup$ I accidentally down voted and it won't let me undo $\endgroup$ – badatmath May 20 '18 at 12:14
  • $\begingroup$ Yes you can by "down voting again", i.e. using the down-arrow once again. that should neutralize the action $\endgroup$ – emma May 20 '18 at 12:39

Only if the diagonal matrix is the scaled identity matrix to begin with. Here's a proof suppose $PDP^*=\alpha I$ where $P$ is Hermitian that is $P^{-1}=P^*$.

Then multiply on the left by $P^*$ to get $DP^*=\alpha P^*$

Then multiply on the right by $P$ to get $D=\alpha I$

  • $\begingroup$ Thank you for your explanation. $\endgroup$ – user550103 May 16 '18 at 14:27

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