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Perhaps I was searching for the problem incorrectly, so I hope this isn't a repeat. If so, apologies!

Suppose you have two positive semidefinite matrices $A$ and $B$. Is there a quick or simple way to find another positive semidefinite matrix $W$ such that

$$ W A W^{\dagger} = B $$

where $W^{\dagger}$ is the conjugate transpose of $W$ and $W$ is Hermitian?

(edited to include additional assumptions)

  • Sure, we can take $W = \sqrt{B}\sqrt{A}^{-1}$ – Ben Grossmann Mar 15 '16 at 18:06
  • Given that $W^{\dagger}$ is the conjugate tranpose, that'd make $$ \sqrt{B} \sqrt{A}^{-1} A \sqrt{A}^{-1 \dagger} \sqrt{B}^{\dagger} = B $$ which doesn't satisfy the equality, right? –  Mar 15 '16 at 18:16
  • What is $W^\dagger$? Conjugate transpose? Do you include symmetry in your definition of positive semidefinite? – user251257 Mar 15 '16 at 18:18
  • I wasn't sure if it was strict enough, but I suppose yes - given that it is the conjugate transpose, these should be Hermitian matrices. –  Mar 15 '16 at 18:19
  • I was assuming that positive semidefinite implies Hermitian. Is that not true for your definition? – Ben Grossmann Mar 15 '16 at 18:20
  • Note that $\sqrt A$ is defined to be positive semidefinite and therefore Hermitian. Also, I've fallaciously assume that $A$ is invertible. – Ben Grossmann Mar 15 '16 at 18:22
  • @Omnomnomnom your $W$ isn't hermitian in general... – user251257 Mar 15 '16 at 18:23
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    Note that no such $W$ exists if $B$ has a greater rank than $A$. – Ben Grossmann Mar 15 '16 at 18:23
  • @user251257 you're right, I forgot that part – Ben Grossmann Mar 15 '16 at 18:24
  • @user251257 but how can you use those eigenbases to build a positive (semi)definite matrix? – Ben Grossmann Mar 15 '16 at 18:32
  • @user251257 Same concern as Omnomnomnom, I don't see any connection between the eigenbases of $A$ and $B$ with $W$.

    Onnomnomnom: If $A$ and $B$ have the same rank, does your solution still apply? It's not clear to me it satisfies the equality ...

     –  Mar 15 '16 at 18:36
  • Oh right. My bad. That $W$ should also be hermitian makes it much harder – user251257 Mar 15 '16 at 18:45
  • @Omnomnomnom or anyone else: It could be written as

    $$ W A - B W = 0 $$

    but then I would always find the zero matrix as the map W. Maybe I'm just spinning my wheels trying to look at it this way?

     –  Mar 15 '16 at 20:23
  • Is there a particular reason you're looking for a positive semidefinite $W$? – Ben Grossmann Mar 15 '16 at 20:26
  • @sakanojo it sort of applies; I think you can use the pseudo-inverse or do something like that. – Ben Grossmann Mar 15 '16 at 20:27
  • @sakanojo it can only be written that way if $W$ is unitary. Is that what you wanted? Because that's a lot easier (as long as $A,B$ have the same rank). – Ben Grossmann Mar 15 '16 at 20:28
  • @Omnomnomnom Maybe I was just too lax in my definition - I thought it may work for a positive semidifinite $W$, but in reality I actually need a unitary matrix $W$. We can restrict $A$ and $B$ to being unitary matrices if we need to as well ... –  Mar 15 '16 at 20:30
  • @sakanojo what kind of context is this where you can switch between positive definite and unitary?? Anyway, this will work for a unitary $W$ iff $A$ and $B$ have the same eigenvalues. This will work for some invertible $W$ iff $A$ and $B$ have the same rank. This will work for some matrix $W$ iff the rank of $A$ is at most the rank of $B$. – Ben Grossmann Mar 15 '16 at 20:41
  • @Omnomnomnom Thanks for your patience with me! I guess I'm still not seeing how the conjugate transpose of the inverse reduces to the inverse, such that your suggested solutions works above ... If my expression was correct, I don't understand how $ \sqrt{A}^{-1} A \sqrt{A}^{-1 \dagger} $ it would complete the equality. –  Mar 15 '16 at 20:56
  • Related question: Find $C$, if $A=CBC$, where $A,B,C$ are symmetric matrices. – user1551 Mar 15 '16 at 21:31
  • Note that $M^{-1 \dagger} = M^{\dagger -1} $ – Ben Grossmann Mar 15 '16 at 21:51
  • @user1551 in case you did not see my earlier request, could you please take a look at http://math.stackexchange.com/questions/1697846/simultaneous-diagonalization-of-two-bilinear-forms where the OP wants to use something he calls "Lagrange Diagonalization" to simultaneously diagonalize two quadratic forms. He did not like my answer. – Will Jagy Mar 16 '16 at 03:05

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