Any help with a Hessian Headache? Let $f(x) = \frac{1}{2}\langle Ax,x\rangle - \langle b,x \rangle + c$ with $A\in \mathbb{R}^{n\times n}$ and $b\in \mathbb{R}^n$, $c\in \mathbb{R}$. Assume that $A$ is symmetric and positive definite. Show that $f$ has a unique global minimum at some point $x_{*}$ and determine $f(x_{*})$ in terms of $A,b,c$.

So my idea was to show that $f$ is strongly convex. Then there would exist a unique global minimum for $f$.
I computed the gradient and the hessian of $f$:
$\nabla f(x)$ = \begin{bmatrix}
\sum^{n}_{i=1}a_{1i}x_i + a_{11}\sum^{n}_{i=1}x_i + b_1 \\
\vdots \\
\sum^{n}_{i=1}a_{ni}x_i + a_{nn}\sum^{n}_{i=1}x_i + b_n \\
\end{bmatrix}
$D^2f|_x$ = \begin{bmatrix}
a_{11}+a_{11} & a_{12}+a_{11}&.&.&.&.& a_{1n}+a_{11} \\
a_{21}+a_{22} & a_{21}+a_{22}&.&.&.&.& a_{2n}+a_{22} \\
\vdots \\
a_{n1}+a_{nn} & a_{n1}+a_{nn}&.&.&.&.& a_{nn}+a_{nn} \\
\end{bmatrix}
But I've been having a lot of trouble showing that $D^2f|_x$ is positive definite (if it even is). Does anyone have an hints for this problem.
 A: $$
f(x) = \frac{1}{2}\sum_{ij}x_i A_{ij}x_j - \sum_i b_i x_i + c
$$
The gradient has components
\begin{eqnarray}
\frac{\partial f}{\partial x_\alpha} &=& \frac{1}{2}\sum_{ij}(x_iA_{ij}\delta_{\alpha j } + \delta_{\alpha i}A_{ij}x_j) - \sum_i b_i \delta_{\alpha i} \\
&=& \frac{1}{2}\sum_{i}x_i A_{i\alpha}  + \frac{1}{2}\sum_{j}A_{\alpha j}x_j - b_\alpha \\
&=& \frac{1}{2} \sum_iA_{\alpha i} x_i + \frac{1}{2}\sum_i A_{\alpha i}x_i - b_\alpha = \sum_{i}A_{\alpha i}x_i - b_\alpha \\
&=& (Ax - b)_\alpha
\end{eqnarray}
In other words
$$
\nabla f = Ax - b \tag{1}
$$
As for the hessian
\begin{eqnarray}
\frac{\partial^2 f}{\partial x_\alpha \partial x_\beta} &=& \sum_{i}A_{\alpha j}\delta_{\beta i} = A_{\alpha \beta}
\end{eqnarray}
So the Hessian is 
$$
Hf = A \tag{2}
$$
From (1), the critical point is $x^* = A^{-1}b$, and is a minimum because $H$ is positive definite
A: Its easier to complete the square, but you can compute derivatives if you liked.
You need to know the following identities (using that $A$ is symmetric positive definite):
$$\nabla_x  \langle b,x\rangle = b\\ 
\nabla_x \frac12\langle x,Ax \rangle\cdot  h =  \langle x,Ah\rangle =  \langle h,Ax\rangle \\
\nabla^2_x\frac12 \langle x,Ax\rangle (h,v) = h^T A v = v^TAh = \langle h,Av\rangle  $$
so the hessian is the constant matrix $A$ assumed to be positive definite. 
Your computation seems to have a mistake, $$\partial _{x_i} \langle x,Ax \rangle = \partial_i (x_j A_{kj}x_k) = \delta_{ij} A_{kj}x_k + \delta_{ik} x_jA_{kj} = 2 Ax$$

Completing the square: Set $B^2 = A$ (exists by pos. def.). Then we have $\langle Ax,x\rangle = \|Bx\|^2$, 
$$ f(x) = \frac12 \langle Ax,x\rangle - \langle b,x\rangle + c= \frac 12 \|Bx-B^{-1}b\|^2+c -\frac12 \|B^{-1} b\|^2$$
