# Complex least-squares problem

Having a matrix $\mathbf{A} \in \mathcal{C}^{m\times n}$ I solve following least-squares problem $$Re(\mathbf{A}^H \mathbf{A})x=Re(\mathbf{A}^H\mathbf{b}).$$ If the matrix $\mathbf{A}$ was a real matrix, the solution to the equation above could have been written as $$\mathbf{x} = \sum_{i=1}^{rank(\mathbf{A})} \frac{\mathbf{u}_i^T\mathbf{b}}{s_i}\mathbf{v}_i,$$ where $\mathbf{u}_i$ and $\mathbf{v}_i$ are corresponding left and right singular vectors and $s_i$ is an $i$th singular value.

My question is whether the solution to the least-squares problem stated in the first equation can be written in a similar way given the SVD of $\mathbf{A} = \mathbf{U}\mathbf{S}\mathbf{V}^H$?

I know that one could split matrix $\mathbf{\tilde A} = [Re(\mathbf{A});~Im(\mathbf{A})]$ and equivalently solve the real problem, but the condition to stay within the complex SVD of the original matrix is of main concern here.

Thank you.

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That should be $u_{i}^{H}b$ rather than $u_{i}^{T}b$. –  Brian Borchers Feb 19 at 22:33
@BrianBorchers: thanks, but somehow I can't edit it... Have you got any idea how to write this equation given the complex SVD so that it solves stated normal eqs? –  Alexander Feb 20 at 7:32

I can't see how it can be done, but others might be able to help you further. I get, with $A=USV^H$, $A^H=VS^H U^H$,

$Re(A^H A)x = Re(A^H b) => Re(V S^H U^H U S V^H) x = Re(V S^H U^H b) => Re(V S^H S V^H) x = Re(V S^H U^H b)$

Normally, you would now reduce $V S^H$ away, but since they are inside the real part, and since $Re(V S^H S V^H) \neq Re(V S^H) Re(S V^H)$, I am not sure how to proceed. Can anyone else get an idea?

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