I came across this question when I self study my friends'note. He took data mining class in university and the question is like below and I cannot find answers from Google:
I know that to prove that function is strictly convex, I need to calculate the Hessian matrix and if Hess(x) is positive definite, then the function f is strictly convex. However, I have some problem deriving the Hessian matrix. Any help or hint is much appreciated!
You have $f(x)=0.5(\langle Ax,x\rangle-\langle b,x\rangle -\langle Ax,A^{-1}b>+\langle b,b\rangle )$. So it is sufficient to show that $\langle A(tx+(1-t)y,tx+(1-t)y\rangle -\langle b,tx+(1-t)y\rangle -\langle A(tx+(1-t)y),A^{-1}b\rangle <$ $<t(\langle Ax,x\rangle -\langle b,x\rangle -\langle Ax,A^{-1}b\rangle )$ $+(1-t)(\langle Ay,x\rangle -\langle b,y\rangle -\langle Ay,A^{-1}b\rangle )$ for each $t$ from $(0,1)$. Eliminating equal terms on both sides leaves us with $\langle A(tx+(1-t)y,tx+(1-t)y\rangle <t\langle Ax,x\rangle +(1-t)\langle Ay,y\rangle $. So now it is about $\langle Ax,x \rangle$ being striktly convex for a positive definite $A$ which is a well known fact.