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I have a square matrix $D$ whose size is $m \times m$ multiplied with another $m \times m$ square matrix $C$, I need to optimize the matrix $C$ to have a unitary matrix $DC$. I mean optimize the matrix $C$ such as $DC$ is a unitary matrix.

In my opinion, that can be formulated as below:

\begin{align} \min_C \|DC - Y\|^2_F&& \text{s.t.}&&(DC)(DC)^H = I_m \end{align}

where $\|\cdot\|_F$ is the Frobenius norm operator and $Y$ is any unitary matrix.

So, I don't know if we can deal with the above equality as a variant of the Procrustes problem or that's not possible. Is it possible to optimize that above equation based on $C$ following my way ? or is there another way we can set the matrix $C$ to have $(DC)(DC)^H = I_m$ ?

NB: All the matrices are real and $det(D) = 0$.

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  • $\begingroup$ If $det(D) \ne 0$ D is invertible and you can just use $C = D^{-1}$. Than you have $D \cdot C = I$ and $I$ sure is unitary. $\endgroup$
    – Hilmar
    Jun 5 at 15:20
  • $\begingroup$ @Hilmar What's about if $det(D) = 0$ ? $\endgroup$
    – Fatima_Ali
    Jun 5 at 15:26
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    $\begingroup$ Then your problem isn't solvable: If $det(D) = 0$ then $det(D\cdot C) = 0$. That means the product can't be unitary regardless of C. I'll turn this into an answer, since I think that's all there is to it $\endgroup$
    – Hilmar
    Jun 5 at 16:33
  • $\begingroup$ Hey Fatima, I think Hilmar's answer is factually correctly answering your question as stated, but I'm not sure it's what you meant. Is it possible $\text{mean}_C$ is supposed to mean $\min\limits_C$? $\endgroup$ Jun 5 at 19:11
  • $\begingroup$ and when you say "$Y$ is any unitary matrix", does it mean you can freely choose $Y$ as long as it's unitary, or is $Y$ externally given? $\endgroup$ Jun 5 at 19:19
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Could it be that you are indeed looking for the closest orthogonal matrix $Y$? Then, there is a solution which involves computing the square root of $ D^TD$ . If $E=(D^TD)^{1/2}$ were invertible, the solution would be its inverse. Yet, it is not invertible here. Then, there is a trick. If I remember well, you have to perform an eigenvalue/eigenvector decomposition of $E$, replace the null eigenvalues by $1$, you then get a novel matrix $E^*$ which is invertible, and its inverse is the (unique) solution.

If what I wrote is correct enough, I might come back with details. Meanwhile, you can look at:

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  • $\begingroup$ Yes Exactly, I am looking for the closest unitary matrix $Y$. The matrix $E$ in my case is not invertible. Could you please explain how to solve that matrix $E$? ... Thanks again $\endgroup$
    – Fatima_Ali
    Jun 6 at 4:35
  • $\begingroup$ Yes, please let me some time. Did you check the two links, did you understand the basic steps? $\endgroup$ Jun 6 at 5:46
  • $\begingroup$ I am reading it, but I couldn't get its exact idea. I try to implement when I understand from it in Matlab but something wrong is happening. Please, your explanation will be really appreciated. $\endgroup$
    – Fatima_Ali
    Jun 6 at 8:05
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    $\begingroup$ Second, based on my understanding, $E$ might be complex matrix. However all other matrices are real. I don't know if it can be set to real or no. Anyway, awaiting for you explanation, thanks again. $\endgroup$
    – Fatima_Ali
    Jun 6 at 12:02
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    $\begingroup$ Sure, as long as it is clear that we do this on our free time $\endgroup$ Jun 10 at 22:04
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If D is not singular, i.e. $\operatorname*{det}(D) \ne 0$ than simply $C = D^{-1}$ will do the trick, since $D \cdot C = I$, which is obviously unitary.

If D is singular, i.e. $\operatorname*{det}(D) = 0$ than the product is also singular, i.e. $\operatorname*{det}(D \cdot C)=0$ which means the product cannot be unitary for all possible matrices $C$.

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