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Consider real square matrices $X$ and $A$ of same size, where $A$ is known to be symmetric positive definite. I came across the matrix equation $XX^{\top} = AX^{\top}$, which solved for $X$ gives either $X = 0$, or $X=A$, or det$(X) = 0$ and det$(X-A)=0$. It is the third case ($X$ is neither equal to 0, nor equal to $A$) I seek to understand more.

Can something more be said about a singular matrix $X$ that solves det$(X-A) = 0$? I am hoping to use the additional information (symmetric positive definiteness) known about $A$, but not sure how to go about it.

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    $\begingroup$ Otherwise stated, for a given symmetric positive definite $A$, you ask about nonzero singular matrices $X$ and $Y$ such that $A=X+Y$. $\endgroup$
    – Wolfgang
    Dec 4, 2015 at 9:15

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For any matrix $A$, if you make $X$ by keeping one column of $A$ and filling up with zeros, it trivially holds. For $n=2,3$ it may become a little bit more interesting if you request $X$ and $Y:=A-X$ not to contain an all-zero vector as a row or column. But for $n\ge4$ you may choose each of $X$ and $Y$ with two identical columns equal to any vector.

So for the question as stated, there cannot be said much more about $X$.

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