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.

  • $\begingroup$ That should be $u_{i}^{H}b$ rather than $u_{i}^{T}b$. $\endgroup$ Feb 19, 2013 at 22:33
  • $\begingroup$ @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? $\endgroup$
    – Alexander
    Feb 20, 2013 at 7:32

2 Answers 2


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?


Check out the LAPACK routines ZGELLS or ZGELLD. They solve the LS problem using SVD. See the official LAPACK documentation for the routine.

  • $\begingroup$ But OP only wants to solve for the real part? $\endgroup$
    – OscarB
    Feb 21, 2013 at 13:37
  • $\begingroup$ @OscarB: that is not what I read in the OP's post. He has a matrix with complex entries and wants to solve the complex LS problem... $\endgroup$
    – GertVdE
    Feb 21, 2013 at 16:36

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