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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.

MAS
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    (1) $A$ is positive definite, so that $\left<Ax,x\right>\ge 0$, and $\left<Ax,x\right>=0$ if and only if $x=0$. This would be helpful to get "$<$" instead of "$\le$". (2) Your target is to show that $f(tx_1+(1-t)x_2)<tf(x_1)+(1-t)f(x_2)$ holds for all $x_1\ne x_2$ and all $t\in\left(0,1\right)$. – hypernova Apr 29 '18 at 05:02
  • or consider the Hessian – LinAlg Apr 30 '18 at 17:34

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