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. 



How to characterize singular matrix $X$ that solves det$(X−A)=0$, where $A$ is symmetric positive definite?
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 $X$.


