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Is there a way to break down or approximate an $m \times n$ matrix into a product of $m \times k$, $k \times k$ and $k \times n$ matrices, with a certain amount of error? I am not looking for SVD because here, unlike in SVD, $k> m,n$. Also, m,k,n $\neq 1$. For example, say a $4 \times 3$ matrix into $4\times 8$, $8 \times 8$ and a $8 \times 3$ matrix. I have been scouring the internet to see if there exists a matrix decomposition of sorts to do this. I would like to know if there is any such in existence.

I am basically trying to solve for X in matrix equation of the format : $AXB=C$ where X is a diagonal matrix. Matrices $A,B$ and $C$ are known. They can be square or rectangular, complex valued matrices.

Thanks.

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    $\begingroup$ This question is ill posed. 1. If we are free to choose $k$ it is trivial (take $k=n$ and use two identity matrices). 2. If we are not free to chose $k$ and this is, say, $k=1$ it is pretty obvious that such approximations are quite bad in general. $\endgroup$
    – Kurt G.
    Commented Sep 9, 2022 at 11:33
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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Sep 9, 2022 at 11:33
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    $\begingroup$ Still much too vague. Please edit the question to give us some idea of what kind of approximation you need, and why. Provide a small example. $\endgroup$
    – Ethan Bolker
    Commented Sep 9, 2022 at 11:49
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    $\begingroup$ With $k>m,n$, the problem is underdetermined, so you can just do something like the SVD and then pad with zeros as appropriate, or many other possibilities... $\endgroup$
    – Ian
    Commented Sep 9, 2022 at 11:50
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    $\begingroup$ It's just a linear system $\endgroup$
    – Lelouch
    Commented Sep 9, 2022 at 12:14

2 Answers 2

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Consider rank of the matrices on the right - it is not bigger than min of rank of kxk matrix and rank of kxn one. Hence it is not bigger than k. On the left side it can be min of {n,m}.

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As various comments said, with $k>m,n$ there are many solutions, including the simple:

$$\begin{bmatrix}a &b & c \\d&e&f\\g&h&i\\j&k&l\end{bmatrix} = \begin{bmatrix}a &b & c &0&0&0&0&0\\d&e&f &0&0&0&0&0\\g&h&i &0&0&0&0&0\\j&k&l &0&0&0&0&0\end{bmatrix} \begin{bmatrix}1 &0 & 0 &0&0&0&0&0 \\0 &1 &0 & 0 &0&0&0&0\\0 &0 &1 &0 & 0 &0&0&0\\0&0 &0 &1 &0 & 0 &0&0\\0&0&0 &0 &1 &0 & 0 &0\\0&0&0&0 &0 &1 & 0 &0\\0&0&0&0&0 &0 &1 &0 \\ 0 &0&0&0&0&0 &0 &1\end{bmatrix} \begin{bmatrix}1 &0 & 0 \\ 0 &1 &0 \\0 &0 &1 \\0&0 &0 \\0&0 &0 \\0&0 &0 \\0&0 &0 \\ 0&0 &0 \end{bmatrix}$$

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