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Define matrix $\mathbf{A} \in \mathbb{C}^{n \times k}$, matrix $\mathbf{B} \in \mathbb{C}^{m \times k}$, and vector $\mathbf{c} \in \mathbb{C}^{k \times 1}$. Also, assume that $\mathbf{A}$ and $\mathbf{c}$ are independent of $\mathbf{B}$. What is the derivative of $\mathbf{A}\mathbf{B}^T\mathbf{B}\mathbf{c}$ respect to $\mathbf{B}$.

Jeff
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  • I've doubts this is differentiable, i.e. the main question is the differntiability of Transpose, which doesn't look to me a continuous function for a first glance. – Michael Medvinsky Oct 21 '15 at 13:41

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$\def\C{\mathbf C}\def\M#1#2{\operatorname{Mat}_{#1,#2}(\C)}$Denote by $f \colon \M mk \to \C^n$ the function in question, that is $$ f(B) = AB^tBc $$ We have for any $B, H \in \M mk$ that \begin{align*} f(B+H) - f(B) &= A(B^t + H^t)(B + H)c - AB^tBc\\ &= AH^tBc + AB^tHc + AH^tHc \end{align*} Now, note that $\def\norm#1{\left\|#1\right\|}$by submultiplikativity and continuity of $B \mapsto B^t$, we have $$ \norm{AH^tHc} \le \norm A \norm H^2 \norm c = o(\norm H), \qquad H \to 0$$ That is, the (Frechet) derivative of $f$ at $B$ is the function $$ f'(B) \colon \M mk \to \C^n, \qquad H \mapsto AH^t Bc + AB^tHc $$

martini
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  • can you please explain or provide a reference to why transpose is continuous and differentiable function? – Michael Medvinsky Oct 21 '15 at 13:47
  • The transpose is a linear map $(-)^t \colon \M nm \to \M mn$, every linear map $T$ between finite dimensional vector spaces is continuous (and hence differentiable with $T'(x) = T$, all $x$), see for example here http://math.stackexchange.com/questions/112985/every-linear-mapping-on-a-finite-dimensional-space-is-continuous ... and as far as it matters for $f$, we've shown it above. – martini Oct 21 '15 at 13:49
  • thanks, note in the second equation, last term is fixed should be $AH^tHc$ – Michael Medvinsky Oct 21 '15 at 13:57
  • @MichaelMedvinsky Thx, corrected. – martini Oct 21 '15 at 13:58
  • ahem.... the transpose is not a linear map http://math.stackexchange.com/questions/1143614/is-matrix-transpose-a-linear-transformation – Michael Medvinsky Oct 21 '15 at 14:00
  • The transpose is linear for sure, as $(\mu A+\lambda B)^t = \mu A^t + \lambda B^t$, which is the definition of being linear. – martini Oct 21 '15 at 14:02
  • "linear map" and linear is not the same thing, linear map would mean that there is a matrix $M$ such that for all matrix $A$ , $A^T=MA$ – Michael Medvinsky Oct 21 '15 at 14:04
  • also the link you have provided states the continuity of a linear map, but not the differentiability. – Michael Medvinsky Oct 21 '15 at 14:26