Getting rid of not invertible matrices

Question

I recently came across the follwing formulation:

$CYC'=M$

where all letters are matrices and the ' stands for transposed. I was wandering if there is a clever way to "isolate" $Y$ (in the sense of having $Y=\dots$) even if matrix $C$ is not invertible.

$Y$ is also diagonal and positive semidefinite.

I probably should mention that (due to the specific case I am working with), with a little bit of work I could reduce the problem to the form $x'Yz=k$, where $x$ and $z$ are different vectors and $k$ is a scalar. I don't know if this is relevant to the solution, and still I'd be more interested in knowing if there is an aswer for the general case.

Thanks for the help (and please let me know if I am doing something wrong regarding guidelines, I am kind of new)

Answer 1

With the vectorization operator, we have $$ CYC' = M \implies (C \otimes C)\operatorname{vec}(Y) = \operatorname{vec}(M), $$ where $\otimes$ denotes a Kronecker product. In the case that $C$ is invertible, the equation can be solved to yield $$ \operatorname{vec}(Y) = (C \otimes C)^{-1}\operatorname{vec}(M) = (C^{-1} \otimes C^{-1})\operatorname{vec}(M). $$ From there, $\operatorname{vec}(Y)$ could be "unvectorized" to yield $Y$.

In the case that $C$ is not invertible, the solution is non-unique and may not exist. However, we can obtain a least-squares solution using the MP Pseudoinverse.

Answer 2

Well even in the reduced case you mention there might be many possibilities for $Y$. For example, let's work with $2 \times 2$ matrices and say that $$ x=z=\begin{pmatrix}1 \\ 1 \end{pmatrix} \  \text{ and } \ k=4 $$ Then, at least all of the following options for $Y$ $$ \begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}, \begin{pmatrix}2 & 0 \\ 0 & 2 \end{pmatrix}, \begin{pmatrix}0 & 2 \\ 2 & 0 \end{pmatrix}, \begin{pmatrix}4 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix}0 & 4\\ 0 & 0 \end{pmatrix}, \begin{pmatrix}0 & 0 \\ 4 & 0 \end{pmatrix}, \begin{pmatrix}0 & 0 \\ 0 & 4 \end{pmatrix} $$ satisfy $x'Yz=4$. In particular, there's more than one option that's diagonal and positive semidefinite.