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I don't see how to differentiate $ABA^T$ with respect to $A$ where $A$ and $B$ are $n\times n$ matrices. I know it's going to be a rank-4 tensor, but what exactly will it be?

The inspiration for this comes from having to find the derivative of the covariance matrix $\operatorname{Cov}(TX)$ with respect to $T$.

So I'll tell you all what I've done so far and maybe you can help.

I was working with the squared Bures distance $d_H^2(Cov(TX),\Sigma_v) = tr(Cov(TX) + \Sigma_v - 2(Cov(TX))^{1/2}\Sigma_v Cov(TX)^{1/2})^{1/2})$.

First I computed the derivative of $d_H^2(A,B)$ for positive matrices $A$ and $B$, which turned out to be $tr(I-A_{\#}B^{-1})$. Here we define $A_{\#}B=(AB^{-1})^{1/2}B.$

So now I was using the chain rule to compute the derivative of $d_H^2(Cov(TX),\Sigma_v)$. But in order to do that, I need to differentiate $Cov(TX)$ w.r.t. $T$. That's where I'm stuck.

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Ultimately, I'm looking to find the gradient with respect to $T$ of $$ \lambda \left\|TX-X\right\|^2 + \left\|T\right\|_{HS} + d_H^2(Cov(TX),\Sigma_v). $$ and calculate its roots.

Assuming I didn't make any mistakes, the derivatives of the first two terms are $2(TX-X)X^T$ and $T/\left\|T\right\|_{HS}$ respectively -- feel free to correct me if I'm wrong here. So the last term is what's causing problems for me when I differentiate.

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    $\begingroup$ Let $f(A) = ABA^T$ and look at $f(A+H)-f(A)$. drop the $o(\|H\|^2)$ term. $\endgroup$
    – copper.hat
    Sep 17 '20 at 17:39
  • $\begingroup$ So I have $HBA^T + ABH^T + HBH^T$ after this and I know I should drop the $HBH^T$ term now. $\endgroup$
    – Glassjawed
    Sep 17 '20 at 17:55
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    $\begingroup$ How would you represent the map $H \mapsto Df(A)(H)$ as a 4-tensor? $\endgroup$
    – copper.hat
    Sep 17 '20 at 18:02
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    $\begingroup$ Doesn't the answer here help you? $\endgroup$
    – Its_me
    Sep 17 '20 at 18:40
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    $\begingroup$ I suspect that you don't really care about the gradient of $\operatorname{Cov}(TX)$, but that this is an intermediate quantity in some larger calculation. Therefore it would be helpful if you described the larger context, since it is likely that you can complete that calculation without the need for this 4th order tensor. $\endgroup$
    – greg
    Sep 17 '20 at 19:05
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In Einstein notation,$$\begin{align}\frac{\partial(ABA^T)_{ij}}{\partial A_{kl}}&=\frac{\partial}{\partial A_{kl}}A_{im}B_{mn}A_{jn}\\&=\delta_{ik}\delta_{lm}B_{mn}A_{jn}+A_{im}B_{mn}\delta_{jk}\delta_{ln}\\&=\delta_{ik}(AB^T)_{jl}+\delta_{jk}(AB)_{il}.\end{align}$$

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Let $J$ be the all-ones matrix and $$\eqalign{ C &=(I-\tfrac 1nJ) = C^T \qquad\qquad\big({\rm Centering\,Matrix}\big) \\ B &= \Sigma_v \\ A &= {\rm Cov}(TX) \\ &= \left(\tfrac 1{n-1}\right)(TX)^TC\,(TX) \\ }$$ From this post, the Bures distance function and its differential can be simplified to $$\eqalign{ \beta(A,B) &= {\rm Tr}\Big(A+B - 2(BA)^{1/2} \Big) \\ d\beta &=\Big(I - (BA)^{-1/2}B\Big):dA \\ }$$ Now change the differentiation variable from $\;dA\to dT$. $$\eqalign{ d\beta &= \Big(I - (BA)^{-1/2}B\Big):\left(\tfrac 2{n-1}\right){\rm Sym}(X^TT^TC\,dT\,X) \\ &= \left(\tfrac 2{n-1}\right)\Big(I - (BA)^{-1/2}B\Big):(X^TT^TC\,dT\,X) \\ &= \left(\tfrac 2{n-1}\right)CTX\Big(I - (BA)^{-1/2}B\Big)X^T:dT \\ \frac{\partial\beta}{\partial T} &= \left(\tfrac 2{n-1}\right)CTX\Big(I - (BA)^{-1/2}B\Big)X^T \\ }$$


In the above derivation, the function $$\eqalign{ {\rm Sym}(M) = \tfrac 12(M+M^T) \\ }$$ was utilized, as well as the trace/Frobenius product $$\eqalign{ P:M = {\rm Tr}(P^TM) = {\rm Tr}(M^TP) = M:P \\ }$$ These have the following interaction $$\eqalign{ P:{\rm Sym}(M) = {\rm Sym}(P):M \\ }$$

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  • $\begingroup$ How come you don't have $Sym$ in the second line when calculating $d\beta$? $\endgroup$
    – Glassjawed
    Sep 18 '20 at 4:33
  • $\begingroup$ Because the matrix on the LHS is symmetric, i.e. ${\rm Sym}(M)=M$. $\endgroup$
    – greg
    Sep 18 '20 at 5:49
  • $\begingroup$ Thank you. This helps a lot. Out of curiosity, how would I get this result using the answer from @MathLearner where he has $A\otimes A$? $\endgroup$
    – Glassjawed
    Sep 21 '20 at 22:04
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The derivative of the expression given in the title can be done using vectorizations:

We have :

\begin{equation} \begin{split} M & = XYZ \\ \implies \text{vec}(M) & = \text{vec}(XYZ) \\ & = (Z^TY^T \otimes I)\text{vec}(X) \\ & = (Z^T \otimes X)\text{vec}(Y) \\ & = (I \otimes XY)\text{vec}(Z) \end{split} \end{equation}

Then, for our expression we have:

\begin{equation} \begin{split} \text{Let} \quad C & = ABA^T \\ \implies \text{vec}(C) & = \text{vec}(ABA^T) \\ \implies d(\text{vec}(C)) & = ((A^T)^TB^T \otimes I)d(\text{vec}(A)) \\ \implies \frac{d(\text{vec}(C))}{d(\text{vec}(A))} & = (AB^T \otimes I) \end{split} \end{equation} Similarly, we can differentiate w.r.t $B$

$$ \frac{d(\text{vec}(C))}{d(\text{vec}(B))} = (A \otimes A)$$

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