Show that $\{x \mid A_0 + \sum x_i A_i \succcurlyeq 0 \}$ is convex

Question

Let $A_0, A_1,\dots,A_m$ be symmetric matrices. Let $x \in \mathbb R^m$ and define $$A(x) := A_0 + \sum_{i=1}^m x_i A_i$$ Show that the set $C := \{x \mid A(x) \text{ is positive semidefinite} \}$ is convex.

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For a set $C \subseteq \mathbb R^n$, I know of a few ways to show that it is convex:

1. Show that $\lambda x_1 + (1-\lambda)x_2 \in C$ for all $x_1, x_2 \in C$ and $\lambda \in [0, 1]$.

2. Show that $C$ is an intersection of convex sets (for example halfspaces).

This is easy to show using the first method, but I am struggling to show that the set is convex using the second method.

My question

The second method above uses the "outer construction" of the set which I am not comfortable with. Is there some trick to applying this method? How could I show that my set $C$ is convex using this method?

Last, are there other methods for showing a set is convex other than the two I have listed above? (I know with additional assumptions it might be easier, but I am thinking about the general case)

Answer 1

We have $B \succeq 0$ if and only if $v^\top B v \ge 0$ for all $v \in \mathbb R^n$. Thus, $$ \{x \mid A(x) \succeq 0\} = \bigcap_{v \in \mathbb R^n} \{x \mid v^\top A(x) v \ge 0\}$$ and, since $$ x \mapsto v^\top A(x) v $$ is a linear mapping, the right-hand side is an intersection of convex half spaces.

Answer 2

I want to offer a third way to prove. There are operations that preserve convexity. One of them is the inverse image of a convex set. You can take a look at section 2.3.2 of the book of Boyd and VandenbergheConvex Optimization to read more.

You can prove that a set is convex by showing that the set is an inverse image of a convex set under an affine function. Here take $f(x) = A_0 + \sum x_i A_i$. Then observe that $C$ is inverse image of Semidefinite cone $S_n^+$ under affine function $f$.