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Given any $n \times k$ real matrix $M$, where $n<k$ and $rank(M)=n$, I consider the following equation (where $M'$ is the transpose of $M$):

$$ MM' = MAM' $$

Then clearly, $A = \mathbb{1}_k $, the $k\times k$ identity matrix, is a possible solution.

Is that unique, or there are other possible solutions?

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  • $\begingroup$ rank(M) cannot be larger than n... $\endgroup$
    – asaelr
    Commented Aug 14, 2016 at 14:51
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    $\begingroup$ The system of linear equations $M (A-1_k) M' = 0$ for the components of $A-1_k$ has size $n^2$ with $k^2$ unknowns. The solution space will then have dimension of at least $k^2-n^2 > 0$. $\endgroup$
    – Igor Khavkine
    Commented Aug 14, 2016 at 15:37

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Here is a counterexample.

I will use MATLAB notation for matrices, so ; ends each row of the matrix.

Let M be the 2 by 3 matrix M = [1 2 3;4 5 6]. Then A = [ 1.25 0 0.25;0 0 0;0.25 0 1.25].

MM' = MAM'= [14 32;32 77].

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  • $\begingroup$ Thanks, and how have you found $A$? Is there a general result that allows to find the solutions? $\endgroup$
    – Ulderique Demoitre
    Commented Aug 14, 2016 at 15:07
  • $\begingroup$ It's s solution to an underdetermined set of linear equations. There are an $\infty$ number of solutions A with this M. $\endgroup$
    – Mark L. Stone
    Commented Aug 14, 2016 at 15:10
  • $\begingroup$ Here is the solution which maximizes trace(A) subject to the 2-norm of A being <= 100. A=[17.5 -33 16.5;-33 67 -33;16.5 -33 17.5]. I just "made up" this criterion, just as a way of selecting another of the $\infty$ of solution. I found this as a solution of a senidefinite optimization problem. $\endgroup$
    – Mark L. Stone
    Commented Aug 14, 2016 at 15:16
  • $\begingroup$ Letting M be a 1 by 2 matrix, such as [1 2], is sufficient to get an $\infty$ of solutions A, in accordance with comment ot the question by @Igor Khavkine . $\endgroup$
    – Mark L. Stone
    Commented Aug 14, 2016 at 15:37

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