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Let $A$ be a nonzero symmetric $3 \times 3$ matrix. Consider the function $f(\mathbf{x})={1 \over 2}(A \mathbf{x}) \cdot \mathbf{x}$.

What is $\nabla f$?

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$\lim_{\epsilon\to 0}[f(x+\epsilon y)-f(x)]/\epsilon=(1/2)[Ax\cdot y+Ay\cdot x]=Ax\cdot y$ since $A$ is symmetric – yoyo Mar 1 '13 at 1:25
up vote 2 down vote accepted

$f(x)=\frac{1}{2} x^T A x$ and therefore, $\nabla f(x)=Ax$. see here

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Could you elaborate? – Sean Haugh Mar 1 '13 at 1:19
The formula $\frac{\partial x^T A x}{\partial x}=2Ax$ appears there explicitly, and it's true for any $n \times n$ matrix $A$. The only proof I can think of is a direct calculation, i.e. write $A=\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$, multiply and take partial derivatives. – user1337 Mar 1 '13 at 1:25

Here is one way among others to do that.

Fix $x$. Then for every $h$ $$ f(x+h)=\frac{(A(x+h),x+h)}{2}=f(x)+\frac{(Ax,h)+(Ah,x)}{2}+\frac{(Ah,h)}{2}. $$

Now, since $A$ is symmetric, we can simplify $$ L(h)=\frac{(Ax,h)+(Ah,x)}{2}=(Ax,h). $$ The latter is linear and of course bounded. It is our candidate for being the derivative of $f$ at $x$. By definition, for this to be true, it is equivalent to show that the remainder $r(h)=f(x+h)-f(x)-L(h)$ satisfies $$ \lim_{h\rightarrow 0} \frac{r(h)}{\|h\|}=0. $$

By Cauchy-Schwarz, we have $$ \lvert \frac{r(h)}{\|h\|}\rvert = \lvert \frac{(Ah,h)}{2}\rvert \leq\frac{\|Ah\|\|h\|}{2\|h\|}\leq \frac{\|A\|\|h\|}{2}. $$

So the condition is satisfied, and $$ df_x(h)=L(h)=(Ax,h). $$

By definition of the gradient, this means $$ \nabla f(x)=Ax. $$

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