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Let $X=ZZ^\top$ where $Z$ is a full row rank $m\times n$ matrix and let $Y$ be a nonsingular $n\times n$ matrix. Consider the matrix-valued function $$ f(X) = ZYZ^\top. $$

My question: Does there exists a closed form expression for the differential $\mathrm{d}f(X)$?

Remark. I believe that my question can be reduced to finding a closed form expression of the differential $\mathrm{d}Z$ w.r.t. $X$, but I don't know if this is actually possible.

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    $\begingroup$ This doesn't make sense as stated. What does $Z^\bot$ mean? $f$ is not a function of $X$. $\endgroup$
    – copper.hat
    Dec 27, 2017 at 20:21
  • $\begingroup$ @copper.hat I believe they're using the upside-down version of $\perp$; this would be the transpose. $\endgroup$
    – Clarinetist
    Dec 27, 2017 at 20:25
  • $\begingroup$ @Clarinetist: You are correct, I meant $\top$. $\endgroup$
    – copper.hat
    Dec 27, 2017 at 20:26
  • $\begingroup$ Your function doesn't seem to be well defined since in general $X$ can be written using different matrices $Z$. $\endgroup$
    – John B
    Dec 27, 2017 at 20:45
  • $\begingroup$ @copper.hat: $\top$ is standard notation for transposition. Since $Z$ is a "factor" of $X$, $f$ is a function of $X$ as well. $\endgroup$
    – Ludwig
    Dec 28, 2017 at 8:58

2 Answers 2

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You could use vectorization to attack the problem.

First $$\eqalign{ X &= ZZ^T \cr dX &= I\,dZ\,Z^T + Z\,dZ^T\,I \cr dx &= \big((Z\otimes I) + (I\otimes Z)K\big)\,dz = A\,dz\cr }$$ where $K$ is the kronecker commutation matrix such that $${\rm vec}(Z^T) = K\,{\rm vec}(Z)$$

Next $$\eqalign{ F &= ZYZ^T \cr dF &= I\,dZ\,YZ^T + ZY\,dZ^T\,I \cr df &= \big((ZY^T\otimes I) + (I\otimes ZY)K\big)\,dz = B\,dz \cr }$$ Then assuming that $A^{-1}$ exists $$\eqalign{ df &= BA^{-1}\,dx \cr dF &= {\rm mat}(df) = {\rm mat}(BA^{-1}\,dx) \cr }$$ where mat() is the inverse of the vec() operator.

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  • $\begingroup$ Your approach seems sound, but is it possible to avoid vectorization? $\endgroup$
    – Ludwig
    Dec 28, 2017 at 9:04
  • $\begingroup$ If $Z$ is rectangular, then $A$ is as well. In which case $A^{-1}$ does not exist. So this solution only works for $(m=n)$; or whenever $(A^+A=I)$. $\endgroup$
    – lynn
    Dec 28, 2017 at 18:05
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    $\begingroup$ But since the problem statement specifies full row rank, we know that $(A^+A \ne I)$. If it had specified full column rank, then this solution would work if $A^{-1}$ were replaced by the pseudoinverse. $\endgroup$
    – lynn
    Dec 28, 2017 at 18:21
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Assuming that the function is well defined , the simplest approach is to compute as follows: $$ df(X)(WW^\top)=\lim_{h\to0}\frac1h((Z+hW)Y(Z+hW)^\top-ZYZ^\top)=ZYW^\top+WYZ^\top. $$ The linear map $df(X)$ is being applyed to a vector $WW^\top$. Now you can use coordinates to write the derivative as a matrix.

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