If A and B are mxm matrices and $f$ is defined as $f: \Bbb{R}^m \rightarrow \Bbb{R}$ by $f(x): A\overrightarrow x \bullet B\overrightarrow x$ . How would one go about finding the Jacobian matrix $J(f)(\overrightarrow c)$ ? Thank you!
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Does $\bullet$ denote here the inner product? And also, did you mean $f(x)=Ax\cdot Bx$? – Berci Feb 15 '19 at 17:49
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yes sorry, my bad on notation mistakes – hoya2021 Feb 16 '19 at 20:37
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Hint: The gradient (as a column vector) of $\mathbf{x}^T M \mathbf{x}$ (where $M$ is a fixed matrix) is $\color{blue}{(M + M^T)\mathbf{x}}$. Try and prove this yourself, or see this link for hints: Gradient of $x^{T}Ax$). Transpose this to get the Jacobian matrix as a $1\times m$ matrix.
To use this fact for your problem, use the fact that $\mathbf{z}\cdot \mathbf{w} = \mathbf{z}^T \mathbf{w}$ (and a fact about the transpose of a product).
Minus One-Twelfth
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Does this look correct to you? $$A\overrightarrow x \bullet B\overrightarrow x = A^T\overrightarrow x B\overrightarrow x$$ and then I use that to get the gradient of $$(B\overrightarrow x + B^T\overrightarrow x) A\overrightarrow x$$ and then transpose this to the a 1xm Jacobian $$J(f)(\overrightarrow c) = [ (B(\overrightarrow c) + B^T(\overrightarrow c) ) A(\overrightarrow(c) ] $$ – hoya2021 Feb 17 '19 at 17:04
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Unfortunately this isn't quite correct. For the first step, remember that $(A\mathbf{x})^T = \mathbf{x}^T A^T$. So $(A \mathbf{x})\cdot (B\mathbf{x})$ becomes $\mathbf{x}^T (A^T B)\mathbf{x}$. – Minus One-Twelfth Feb 17 '19 at 18:59
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so then is the gradient $$(A^T B + B^T A)\mathbf{x}$$ and then the Jacobian $$J(f)(\overrightarrow c) = [ ( (A^T B + B^T A)\overrightarrow c)^T]$$ ? – hoya2021 Feb 17 '19 at 19:32
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