# Let $\Phi(x) = \frac{1}{2}x^TAx - x^Tb$ where A is positive definite matrix. Prove that $\nabla\Phi = Ax - b$ and that minimum of $\Phi$ is $A^{-1}b$

so I'm dealing with the next problem: Let$\ \Phi(x) = \frac{1}{2}x^TAx - x^Tb$ where A is positive definite matrix. Prove that $\nabla\Phi = Ax - b$ and that minimum of $\Phi$ is $A^{-1}b$ on whole $\mathbb{R}^n$.

I really don't know how determine the gradient. I do know I have to prove that $\Phi(A^{-1}b)<\Phi(x),\ \forall x\in\mathbb{R}^n,x\neq A^{-1}b$.

Any suggestion would be appreciated.

Thanks.

• What is the sufficent and necessary condition that minimas have/needs? Gradient: You can write it as <x,Ax>+<b,x> were <.,.> is the standard scalar product, and go from there. – user160069 May 4 '17 at 17:54

$$x^TAx=\sum_{i}x_i\sum_{j}a_{ij}x_j=\sum_{i,j}x_ia_{ij}x_j.$$
Now proceed to calculate the gradient. For example $\partial_i$ would give:
$$\partial_k (x^TAx)=\sum_{j\neq k}a_{kj}x_j+\sum_{i\neq k}a_{ik}x_k+2a_{kk}x_i=\sum_{j=1}^na_{kj}x_j+\sum_{i=1}^na_{ik}x_k.$$
Notice that the r.h.s. is equivalent to $[Ax]_k+[x^TA]_k=[Ax]_k+[A^Tx]_k,$ (since transposes of vectors don't affect the $k$'th coefficient).
So, the full gradient is $x^T(A+A^T)$ (to make it a row vector). Since $A$ is symmetric, $A=A^T$, which will simplify it.