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I have looked throughout the matrix cookbook and other sources, but am a bit confused by this problem. If I have a function $F = ABC$, what is the partial derivative of $F$ with respect to $B$? When it comes to matrix functions I could not find a nice answer to this problem. I did find something akin to $dF = CAdX$, but that doesn't seem to make sense as the dimensions of $C$ and $A$ would not match up for multiplication. A could be a $3\times3$, $B$ a $3\times 3$, and $ C$ a $3\times 4$. I'd appreciate help on this. Thanks.

Edit- Well I had changed the problem to be more in line with what I was originally working with below in the comments. Here is the problem: $f = tr((ABC)(ABC)^T)$ and I want the partial derivative $\frac{\partial}{\partial B} tr((ABC)(ABC)^T)$. I ended up using a modified version of copper.hat's answer. I combined $A$ and $B$ to get something like this $2(ABC)C^T$ for the gradient.

Thank you both for the help. Both answers below are technically correct, I just accepted the shorter and more convenient form.

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  • $\begingroup$ The function $G:B\longmapsto ABC$ is linear and bounded. So it is differentiable, with constant derivative equal to $G$. $\endgroup$ – Julien Apr 9 '13 at 16:01
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When dealing with functions of this sort, I find it easier to deal with the derivative applied to some perturbation.

Let $F(B) = ABC$. Since $F$ is linear, we have $F(B+\Delta) = F(B) + A\Delta C$, so it follows that $DF(B)(\Delta) = A\Delta C$. However, since $DF(B)$ is a map $\mathbb{R}^{n \times n} \to \mathbb{R}^{n \times n}$ (or $\mathbb{C}$, as the case may be), there is no particularly convenient matrix representation.

Note: Each coordinate of $DF(B)$ has a convenient representation. To see, let $\phi_{ij}(B) = e_i^T F(B) e_j = e_i^T ABC e_j$. Then, as above, we have $D\phi_{ij}(B)(\Delta) = e_i^T A\Delta C e_j$. This gives: \begin{eqnarray} D\phi_{ij}(B)(\Delta) &=& e_i^T A \Delta C e_j \\ &=& \operatorname{tr} (e_i^T A \Delta C e_j) \\ &=& \operatorname{tr} (e_j e_i^T A \Delta C ) \\ &=& \operatorname{tr} (C e_j e_i^T A \Delta ) \\ &=& \operatorname{tr} ((A^T e_i e_j^TC^T)^T \Delta ) \\ &=& \langle A^T e_i e_j^TC^T, \Delta \rangle_F \end{eqnarray} Where $\langle X, Y \rangle_F = \operatorname{tr}(X^TY)$ is the inner product induced by the Frobenius norm. With this inner product, the gradient is the given by $\nabla \phi_{ij}(B) = A^T e_i e_j^TC^T $.

Another note: If $G(B) = \operatorname{tr} ((ABC)(ABC)^T) $, then you can write $G(B) = \langle ABC, ABC \rangle_F$. Then $G(B+\Delta ) = G(B) + \langle A \Delta C, ABC \rangle_F + \langle ABC, A \Delta C \rangle_F + \langle A \Delta C, A \Delta C \rangle_F$, from which you can see that \begin{eqnarray} DG(B) (\Delta) &=& \langle A \Delta C, ABC \rangle_F + \langle ABC, A \Delta C \rangle_F \\ &=& 2 \langle ABC, A \Delta C \rangle_F \\ &=& 2 \langle A^TABC C^T, \Delta \rangle_F \end{eqnarray} From which it follows that $\nabla G(B) = 2A^TABC C^T$.

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  • $\begingroup$ Can you look at the comment I made to the answer below this? If there is no convenient solution I likely messed up elsewhere. $\endgroup$ – Robert N. Apr 9 '13 at 16:08
  • $\begingroup$ Well, there is a solution, it doesn't have a convenient representation. The point is that it is easier to consider the application of the derivative to a particular perturbation ($\Delta$ above). I don't know what you mean by a gradient for a function that is not $\mathbb{R}$-valued. $\endgroup$ – copper.hat Apr 9 '13 at 16:14
  • $\begingroup$ Well the trace function is a real valued function. The function is the trace of ABC squared. $\endgroup$ – Robert N. Apr 9 '13 at 16:20
  • $\begingroup$ @RobertN.: I don't see trace in the question? Anyway, I added a note that shows how each coordinate of $F(B)$ can be represented as a gradient. $\endgroup$ – copper.hat Apr 9 '13 at 16:36
  • $\begingroup$ @copper.hat Nice solution! $\endgroup$ – Elias Costa Apr 9 '13 at 16:38
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The derivative is just the best linear approximation and so $ \frac{dF}{dB}(X) = A X C $

In a different language, $f = L_A \circ R_C \implies df = dL_A \circ dR_C$

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  • $\begingroup$ A linear approximation? Why is there no direct solution. Maybe I screwed up elsewhere. I got to this problem from here: dF/dB = d/dB TR((ABC)transpose(ABC)). After doing one half of the derivative, I got to a point where I had to do d/dB ABC. $\endgroup$ – Robert N. Apr 9 '13 at 16:05
  • $\begingroup$ No, the derivative I gave is the direct solution to the derivative. I was trying to suggest that a derivative is a linear approximation to usually non-linear functions, but in your case, the function is linear in $B$. The best linear approximation to a linear function is the function itself. $\endgroup$ – muzzlator Apr 9 '13 at 16:07
  • $\begingroup$ What would be the course of action when working on an optimization problem where the derivative would be used as a gradient? $\endgroup$ – Robert N. Apr 9 '13 at 16:10
  • $\begingroup$ $\frac{dF}{dB}(X)$ where $F = \operatorname{Tr}(ABC) (ABC)^T$ is given by a product rule: $$\operatorname{Tr}(AXC) (ABC)^T + \operatorname{Tr}(ABC) (AXC)^T$$ This follows from the linearity of trace and bilinearity of multiplication $\endgroup$ – muzzlator Apr 9 '13 at 16:10
  • $\begingroup$ After the product rule should you not continue with the chain rule on ABC? I'll try working with this for now though. Thanks a lot! If I have any additions I'll add them here in about an hour or two, and accept the answer then. $\endgroup$ – Robert N. Apr 9 '13 at 16:17

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