Prove that the minimum value of $q(x)$ is $-\left \langle b,A^{-1}b \right \rangle$

Question

Prove that the minimum value of $q(x)$ is $-\left \langle b,A^{-1}b \right \rangle$

I know that If $A$ is symmetric, then the gradient of the function $q(x)=\left \langle x,Ax \right \rangle-2\left \langle x,b \right \rangle$ at $x$ is $2(Ax-b)$, how can I use this to prove what I want? Thank you very much.

Answer 1

Let us assume that there is such a minimum and that $A$ is invertible.

If $x_0$ is a local extrema of $q$ then $Ax_0-b=0$ i.e $x_0=A^{-1} b$. As you know there is a minimum this (unique) local extrema is thus the minimum of $q$.

By plugging the expression of $x_0$ in $q$: $$q(x_0)=\langle A^{-1}b, b \rangle -2 \langle A^{-1}b, b \rangle=-\langle A^{-1}b, b \rangle$$