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I learned from The Matrix Cookbook that the gradient of the $\log \det$ function is given by

\begin{equation} \nabla \log \text{det}(\mathbf{X}^\top \mathbf{X})=2\mathbf{X}(\mathbf{X}^\top \mathbf{X})^{-1}, \end{equation}

where $\mathbf{X}\in\mathbb{R}^{n\times r}$. I wonder which function will give the gradient

\begin{equation} 2\mathbf{A} \mathbf{A}^\top \mathbf{X}(\mathbf{X}^\top \mathbf{X})^{-1}, \end{equation}

for some matrix $\mathbf{A}\in \mathbb{R}^{n\times r}$.

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  • $\begingroup$ The gradient of a function is a vector not a matrix. Your first equation does not make sense for me. $\endgroup$ – Dog_69 Aug 4 '18 at 21:29
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    $\begingroup$ Hi, I'm arranging it in a matrix form. It is similar to the fact that the gradient of $\frac{1}{2}\|\mathbf{X}\|_{\mathrm{F}}^2$ is $\mathbf{X}$. $\endgroup$ – Wuchen Aug 4 '18 at 22:18
  • $\begingroup$ In this case the $r$ of $\mathbb R^{n\times r}$ should be $1$1. $\endgroup$ – Dog_69 Aug 5 '18 at 5:04
  • $\begingroup$ Where does this constraint come from? $\endgroup$ – Wuchen Aug 6 '18 at 14:49
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    $\begingroup$ @Dog_69 See math.stackexchange.com/questions/2807864/… $\endgroup$ – Jean-Claude Arbaut Aug 7 '18 at 6:03
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In general, there is no solution. That follows is a counter-example for the existence of $g$ s.t. $\nabla(g)=2BX(X^TX)^{-1}$ for every $X$ s.t. $rank(X)=r$ (where $B\in M_n$).

Let $n=2,r=1, B=\begin{pmatrix}a&b\\c&d\end{pmatrix},X=[x,y]^T\not= [0,0]^T$.

$\dfrac{\partial g}{\partial x}=\dfrac{2}{x^2+y^2}(ax+by),\dfrac{\partial g}{\partial y}=\dfrac{2}{x^2+y^2}(cx+dy)$. Thus

$\dfrac{\partial^2 g}{\partial x \partial y}=\dfrac{2}{(x^2+y^2)^2}(b(x^2-y^2)-2axy),\dfrac{\partial^2 g}{\partial y \partial x}=\dfrac{2}{(x^2+y^2)^2}(c(y^2-x^2)-2dxy)$.

Thus $g$ exists iff $a=d,b=-c$, that is $B=\begin{pmatrix}a&b\\-b&a\end{pmatrix}$; in particular, $B$ has never the form $AA^T$ except when $B=0$.

EDIT. The general solution of the above equation -with the $(r,\theta)$ polar coordinates- is: $g=2a\log(r)-2b\theta+$ constant.

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  • $\begingroup$ @Wuchen , did you read my answer ? $\endgroup$ – user91684 Sep 11 '18 at 22:58
  • $\begingroup$ Thank you! Sorry for the late acceptance. $\endgroup$ – Wuchen Oct 23 '18 at 19:02

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