Prove that $f$ is strictly convex iff $A$ is positive definite

Question

I would like to have a help with this question.

Let $$f(X) = \frac{1}{2}X^TAX + b^TX$$ with $A=A^T$. Prove that $f(X)$ is strictly convex if and only if $A$ is positive definite.

Regards!

Answer 1

Hint

The hessian of the function becomes $$\nabla^2f(X)={1\over 2}(A+A^T)=A$$now what does strict convexity imply here and impose on $\nabla^2f(X)$?

Answer 2

Your question is actually on minute 29:02 in this YouTube lecture.

If $A$ is symmetric and PD, then what you say is true. You got that $f(X)$ is differentiable with respect to $X$ hence it suffices to prove the second order condition, namely,

$$\nabla^2 f(X) = A \succ 0$$

Hence $f(X)$ is stricly convex.