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I am looking for a way for calculating

$$ (VXV^H)^{-1}$$ where $X$ is an $N$-by-$N$ diagonal matrix and $V$ is $M$-by-$N$ matrix an (pseudo) orthonormal matrix satisfying $VV^H = I_M$. Here, $I_M$ denotes the identity matrix $M$-by-$M$ and $M<N$. Secondly $(.)^H$ denotes the conjugate transpose (Hermitian) of a matrix.

My question is if I can represent the above expression in a more compact fashion. The problem for me is that $V$ is non-square. If $M=N$, I could have written the above expression in the following form: $$ (VXV^H)^{-1} = V X^{-1} V^H$$ Which is very easy to compute since $X$ is a diagonal matrix.

Let me clarify my problem a little bit more. I have a program that calculates $$ (VXV^H)^{-1}$$ for each iteration and $X$ changes for each iteration but $V$ is constant. Therefore, I can store the value of $V^{-1}$ for further calculations if needed. Can you help me for simplifying it? If you don't have any possible way, do you know any iterative algorithm for me to calculate the approximate of it? The matrices will be quite large :)

Thank you!

Okan Erturk
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Let $W$ be a matrix generated such that $$ Q = \pmatrix{V\\ W} $$ is square with orthonormal rows, so that $QQ^H = Q^HQ = I_N$. Note that $$ QXQ^H = \pmatrix{ VXV^H & VXW^H\\ WXV^H & WXW^H}, \\ QX^{-1}Q^H = \left[QXQ^H\right]^{-1} = \pmatrix{ VXV^H & VXW^H\\ WXV^H & WXW^H}^{-1}. $$ From there, the inverse that we're after is the top left block of $$ \pmatrix{VXV^H &0\\0 & WXW^H}^{-1} = \pmatrix{(VXV^H)^{-1} &0\\0 & (WXW^H)^{-1}}. $$ To obtain this inverse, we can use the Woodbury formula, noting that $$ \pmatrix{VXV^H &0\\0 & WXW^H} = \pmatrix{ VXV^H & VXW^H\\ WXV^H & WXW^H} - \pmatrix{ 0 & VXW^H\\ WXV^H & 0}\\ = QXQ^H - \pmatrix{VXW^H\\0}\pmatrix{0 & I_{N-M}} - \pmatrix{0\\I_{N-M}}\pmatrix{ WXV^H&0} \\ = \underbrace{QXQ^H}_A - \underbrace{\pmatrix{VXW^H & 0\\0 & I_{N-M}}}_{B}\underbrace{\pmatrix{0 & I_{N-M}\\ WXV^H & 0}}_{C}. $$ With the Woodbury formula, we find that $$ [A - BC]^{-1} = A^{-1} - A^{-1}B(I_{2(N-M)} + CA^{-1}B)CA^{-1}. $$ where we have $A^{-1} = (QXQ^H)^{-1} = QX^{-1}Q^H$. Notably, $CA^{-1}B$ simplifies to the following $2k \times 2k$ matrix, where $k = N-M$. $$ CA^{-1}B = \\ \pmatrix{0 & I_{N-M}\\ WXV^H & 0}\pmatrix{V\\W}X^{-1}\pmatrix{V^H & W^H}\pmatrix{VXW^H & 0\\0 & I_{N-M}} =\\ \pmatrix{W\\WXV^HV}X^{-1}\pmatrix{V^HVXW^H & W^H} = \\ \pmatrix{WX^{-1}V^HVXW^H & WX^{-1}W^H\\ WXV^HVX^{-1}V^HVXW^H & WXV^HVX^{-1}W^H}. $$

Ben Grossmann
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  • Hello. Thank you for your effort. In the last equation, is the matrix at the lefthand side of the equation inverse, or am I missing something? Thank you again. The solution is very instructive. – Okan Erturk Jun 20 '23 at 19:52
  • No, it's not the equation for the inverse. It is an expression of the form $(A - UCV)$ (with $C = I$), which allows you to apply the Woodbury formula – Ben Grossmann Jun 20 '23 at 19:52
  • Okay, got it :) I will try to obtain the explicit form of $(VXV^H)^{-1}$. I hope that I can derive on my own :) – Okan Erturk Jun 20 '23 at 19:54
  • Hello again. I am looking for the solution above for a while. But I could not understand the steps above. Can you explain it a little bit more explicitly? – Okan Erturk Jun 20 '23 at 20:03
  • If you explain exactly what it is that you don't understand, I'll be happy to clarify – Ben Grossmann Jun 20 '23 at 20:34
  • Hello, I didnt get how we obtained $ \left(\begin{array}{cc} V X V^H & 0 \0 & W X W^H \end{array}\right)^{-1}$ from $QX^{-1}Q^H$.

    Secondly, in the last equation, $\left(\begin{array}{ll} V X W^H & I_{M-N} \end{array}\right)\left(\begin{array}{c} I_{M-N} \ W X V^H \end{array}\right)$ includes the product of $VXW^H I_{N-M}$ product but the dimensions are mismatches as far as I see.

     – Okan Erturk Jun 20 '23 at 20:39
  • @Okan Regarding the second part, I made a mistake; see my latest edit – Ben Grossmann Jun 20 '23 at 22:01
  • @Okan To the first question, we didn't "obtain [the matrix] from $QX^{-1}Q^H$". I'm saying that if we have this inverse, then we would be able to extract our answer. After that, I explain how the matrix can be obtained from the other – Ben Grossmann Jun 20 '23 at 22:04
  • Hello again, I got your point now. Only one question: In the last equation, It seems that $ \left(\begin{array}{cc} V X V^H & V X W^H \ W X V^H & W X W^H \end{array}\right) =QX^{-1}Q^H$ which is defined as $QXQ^H$ above. Am I mistaken? – Okan Erturk Jun 20 '23 at 22:07
  • No, I say that $\left(\begin{array}{cc} V X V^H & V X W^H \ W X V^H & W X W^H \end{array}\right)^{\color{red}{-1}} =QX^{-1}Q^H$, since $QX^{-1}Q^H = (QXQ^H)^{-1}$ – Ben Grossmann Jun 20 '23 at 22:10
  • So how the list of the last equality holds? It seems that $\left(\begin{array}{cc} V X V^H & V X W^H \ W X V^H & W X W^H \end{array}\right)$ is replaced with $QX^{-1}Q^H$, not $QXQ^{H}$. I think I am missing a point and could not got it yet. – Okan Erturk Jun 20 '23 at 22:14
  • @Okan I was in a hurry when I put the answer down earlier, I see now that I have a lot of confusing errors... I'm working on trying to make everything correct now – Ben Grossmann Jun 20 '23 at 22:32
  • @Okan I think that everything is finally correct, and I have a formula for the inverse now – Ben Grossmann Jun 20 '23 at 22:36
  • Thank you for effort again, it was very helpful. Tomorrow, I derive step by step again and inform you. Right now I really have limited internet connection and just electricity went off couple minutes ago. Than you again for helping at that late hours! – Okan Erturk Jun 20 '23 at 22:49