# Solving $(V X V^H)^{-1}$ where $V V^H=I$ and non-square.

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 $$M. Secondly $$(.)^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!

• It's not clear what you mean by "the value of $V^{-1}$"; $V$ isn't square, so it has no inverse Commented Jun 20, 2023 at 18:53
• Yes, but instead, can I use something like a pseudo inverse since the result is an MxM matrix? Commented Jun 20, 2023 at 18:58
• With same kind of approach that I outline in the beginning of this answer, we can get away with only computing the inverse of a $2k \times 2k$ matrix, where $k = M - N$. Commented Jun 20, 2023 at 18:58
• In your case, the pseudoinverse is simply $V^+ = V^H$. I don't see a way to use the pseudoinverse here, though Commented Jun 20, 2023 at 18:59
• Yes, you are right! Commented Jun 20, 2023 at 19:00

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 $$k = N-M$$. $$CA^{-1}B = \\ \pmatrix{0 & I_{N-M}\\ WXV^H & 0}\pmatrix{V\\W}X^{-1}\pmatrix{V^H & W^H}\pmatrix{VXW^H & 0\\0 & I_{N-M}} =\\ \pmatrix{W\\WXV^HV}X^{-1}\pmatrix{V^HVXW^H & W^H} = \\ \pmatrix{WX^{-1}V^HVXW^H & WX^{-1}W^H\\ WXV^HVX^{-1}V^HVXW^H & WXV^HVX^{-1}W^H}.$$
• 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 Commented Jun 20, 2023 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 :) Commented Jun 20, 2023 at 19:54