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I want to optimize a function $E = \sum_i(Y_i-f(Z_i))^2$ w.r.t. $T$ where $f(Z)=ZTV^t$, and all are matrices (not necessarily square). Here $T$ acts as weights as in regression for example. $Y \in R^{n \times p}$, $Z \in R^{n \times k}$, $T \in R^{k \times k}$, and $V \in R^{p \times k}$. $p,n,k$ are arbitrary integers ( $>1$).

How can I do this?

From a gradient based algorithm point of view, do you need to compute the $\partial E/\partial T_{a,b}$ where $T_{a,b}$ is an element of $T$? If so, how can you do that and after computing, do you just simultaneously perform $T_{a,b} = T_{a,b} - \alpha * \partial E/\partial T_{a,b} $ for all $a,b$ and learning rate $\alpha$ as in the usual gradient descent?

($A^t$ denotes the transpose of matrix $A$)

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  • $\begingroup$ You can represent matrix multiplication with Kronecker products. Then it becomes a normal least squares problem in the vectorization you choose. $\endgroup$ – mathreadler Apr 1 '18 at 7:09
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Let's use a colon to denote the trace/Frobenius product $$A:B={\rm tr}(A^TB)$$ and let's define the additional matrix variables $$\eqalign{ F &= ZTV^T,\,\,\,M = (F-Y) \cr }$$ Write the function in terms of these new variables, then find its differential and gradient $$\eqalign{ {\mathcal E} &= M:M \cr\cr d{\mathcal E} &= 2M:dM = 2M:dM \cr&= 2M:Z\,dT\,V^T \cr &= 2Z^TMV:dT \cr\cr \frac{\partial {\mathcal E}}{\partial T} &= 2Z^TMV \cr&= 2Z^T(F-Y)V \cr }$$ Set the gradient to zero and solve $$\eqalign{ Z^TYV &= Z^TFV = Z^T(ZTV^T)V \cr\cr T &= (Z^TZ)^{-1}Z^TYV(V^TV)^{-1} \cr&= Z^+Y(V^+)^T \cr }$$ where $M^+$ denotes the pseudoinverse of $M$.

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