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Consider matrices $Y\in\mathbb{R}^{n\times n}$ and $X\in\mathbb{R}^{n\times m}$ where $m\geq n$. $X$ is unknown but $Y=XX'$, which implies that $Y$ is positive definite (I see no reason why this couldn't alternately be expressed as a positive semi-definite problem with $Y=X'X$, a different $Y\in\mathbb{R}^{m\times m}$ would still be known).

What the easiest method to find $X$? I was thinking of minimizing the Frobenius norm, but wasn't sure if there was some relatively straightforward thing that I'm missing.

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    @MichaelHardy Thanks.2012-09-12

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Consider $X = I$, then $X' = I$ and $Y = I$. On the other hand, if $X = -I$, then $X' = -I$, but $Y = I$ still. Are there any other assumptions you can make about $X$?

In fact, for any $X$, $(-X)(-X)' = XX'$, so there are always at least two solutions.

Additionally, if $X_1 = \left(\begin{array}{cc} a_1 & b_1 \\ 0 & 0\end{array}\right),\quad X_2 = \left(\begin{array}{cc} a_2 & b_2 \\ 0 & 0\end{array}\right)$

with $a_1^2 + b_1^2 = a_2^2 + b_2^2$ then $Y = X_1 X_1' = X_2 X_2'$.

I think maybe something needs to be said about the singular values of $X$ to get better uniqueness - maybe that they're all non-degenerate?

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    @John Well, as Robert Israel pointed out, in the square case, if $Y$ is invertible, $X$ is invertible, so eigenvalues are non-zero, and the non-uniqueness is characterized as above.2012-09-12
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You also need $Y$ to be symmetric. An algorithm which does this if you want $X$ to be triangular is the Cholesky decomposition.

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    You might need full pivoting if the matrix is singular. ($Y = XX'$ already implies $Y$ is symmetric.)2012-09-11