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I know a square self-adjoint matrix $S_{vv}$ and I want to find:

$S_{rr} = HS_{vv}H^{\dagger}$

where $\dagger$ denotes conjugate transpose.

I do not know $H$ but I do know $H^{-1}$.

What is the most efficient way to solve something of this form?

I have considered rewriting it as:

$S_{rr} = H(HS_{vv})^{\dagger}$

Finding the LU decomposition of $H^{-1}$, and then doing two lots of forward and back substitutions to solve for $S_{rr}$. But it feels like there might be a better way?

Any help much appreciated.

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  • $\begingroup$ $H^{-1}$ is not singular, right? $\endgroup$ – Jeff Hammond Feb 11 '14 at 15:57
  • $\begingroup$ As a general rule, any time you are tempted to compute the explicit inverse of a matrix, there is a better way. I'm not sure what it's called when solves $R=A^{-1}V(A^{-1})^{\dagger}$ for $V$, given known $R$ and $A$, but it has at least one obvious solution. For example, I might case this as $AX^{\dagger}=V$ and solve for $X$, then solve $X=AR$ for $R$. You can use LU for this, of course. $\endgroup$ – Jeff Hammond Feb 11 '14 at 16:00
  • $\begingroup$ Thanks Jeff, yes $H^{-1}$ is not singular. $\endgroup$ – James Feb 11 '14 at 17:10
  • $\begingroup$ I agree that it can be done that way, as I indicated in my question. But I'm wondering what "the" way is to approach this problem, if such a thing exists. The thing is, like you, I don't know what to call this so it is difficult to search for a good reference. $\endgroup$ – James Feb 11 '14 at 17:14
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    $\begingroup$ If you're looking for a congruent transform, LDL is probably a better choice. $\endgroup$ – Geoff Oxberry Feb 11 '14 at 20:16

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