Let $f(x) := x^T A x + b^T x + c$. If we only know $A \in \mathbb{R}^{n \times n}$, why does convexity of $f$ require that $A + A^T$ be positive semidefinite (PSD) and why does strong convexity require $A + A^T \succeq \mu I_n$ for $\mu > 0?$
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You mean $A \in \mathbb R^{n\times n}$? The term $x^TAx = \frac12 x^T\left(A + A^T\right) x$ so you only need to the $A + A^T$ – Kroki May 11 '23 at 18:00
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The skew-symmetric part of $A$ contributes nothing to $f$ – Rodrigo de Azevedo May 11 '23 at 18:10
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From section 3.1.4 of Boyd & Vandenberghe's Convex Optimization:
We now assume that $f$ is twice differentiable, that is, its Hessian or second derivative $\nabla^2 f$ exists at each point in $\textbf{dom} f$, which is open. Then $f$ is convex if and only if $\textbf{dom} f$ is convex and its Hessian is positive semidefinite: for all $x \in \textbf{dom} f$, $$\nabla^2 f (x) \succeq 0$$
where $\textbf{dom} f$ denotes the domain of the function $f$.
Note that the Hessian of $$f ({\bf x}) := \frac12 {\bf x}^\top \left( {\bf A} + {\bf A}^\top \right) {\bf x} + {\bf b}^\top {\bf x} + c$$ is $$\nabla^2 f ({\bf x}) = \color{blue}{{\bf A} + {\bf A}^\top}$$
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Convexity requires that $A+A^T$ is positive semidefinite because if the matrix is not positive semidefinite, the function simply is not convex. You can already see this for the case $n=1$. Then your function is simply $$ f(x) = ax^2 + bx + c, $$ which is not a convex function of $a<0$. It is not strictly convex if $a\le 0$.
You can then easily generalize this to higher dimensions.
