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Let $A \in \mathbb{R^n}$ be a real $n \times n$ matrix.

How can I prove that the function $$q:\mathbb{R}^n \to \mathbb{R} \text{ with } q(x):= x^TAx$$

is totally differentiable on $R^n$ and find its total derivative in every point?

I know that I can use the Cauchy-Schwarz-inequality.

So I would have:

$\vert\langle x|y\rangle\vert = x^Ty = \sum_{i=1}^n x_iy_i \leq (\sum_{i=1}^n x_i^2)^{1/2} (\sum_{i=1}^n y_i^2)^{1/2} = \left\lVert x \right\rVert_2 \left\lVert y \right\rVert_2$ for all $x,y \in \mathbb{R}^n$.

I have also found the following: Differentiate $f(x)=x^TAx$

But in that thread total derivation isn't proven as well as using Cauchy-Schwarz-inequality.

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I do not know what you mean by totally differentiable, but you can consider the composite function $q = B \circ \psi$ where $\psi(x) =(x, x)$ and $B(x,y) = y^\intercal A x.$ Apply the chain rule bearing in mind $B'(x, y) \cdot (h, k) = B(x, k) + B(h, y).$

EDIT: The chain rule states that if $f$ and $g$ are defined in open sets $U$ and $V$ such that $f(U) \subset V$ and they are both differentiable then $g \circ f$ is differentiable and $(g \circ f)'(x) = g'(f(x)) \circ f'(x),$ where the right hand side is composition of linear functions.

Now, the function $q = B \circ \psi$ so that $q'(x) = B'(\psi(x)) \circ \psi'(x).$ And if you want to evaluate in a vector $h,$ we have $q'(x) \cdot h = B'(\psi(x)) \circ \psi'(x) \cdot h = B'(x,x) \cdot (h, h) = B(h, x) + B(x, h) = h^\intercal A x + x^\intercal A h.$

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  • $\begingroup$ Sorry, but I don't really understand. Can you enhance the answer a bit? $\endgroup$ – user616187 Nov 15 '18 at 20:29

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