Let $ \ A \ $ be a positive definite matrix . Prove that $ \ \frac{1}{2} \left\langle Ax-b \ , \ A^{-1}(Ax-b)\right\rangle \ $ is strictly convex,
where $ \ \left\langle u,v \right\rangle \ $ means dot product between vectors $ \ u \ $ and $ \ v \ $.
Answer:
Since $ \ A \ $ is positive definite , we have
$ \left\langle Ax, x \right\rangle \geq 0 \ $
But I do not know how to show that the inner product is strictly convex.
Let $ \ f(x)= \left\langle Ax-b \ , \ A^{-1}(Ax-b)\right\rangle \ = (Ax-b) \cdot A^{-1}(Ax-b)=(Ax-b) \cdot (x-A^{-1}b) \\ \Rightarrow f(x+y)=(Ax+Ay-b) \cdot (x+y-A^{-1}b) $
I think We have to show that $ \ f(x+y) <f(x)+f(y) \ $.
But I can not show it.
I need help doing this.