Convex sets of matrices are path connected

Idea. Convexity is *exactly* the statement that straight-line paths do not leave the set.

1. Proof. Let $C\subset M_n(\mathbb{R})$ be convex and $A,B\in C$. Define
$$\gamma:[0,1]\to M_n(\mathbb{R}),\qquad \gamma(t)=(1-t)A+tB.$$
- $\gamma$ is continuous: each entry $(1-t)a_{ij}+tb_{ij}$ is an affine function of $t$.
- $\gamma(0)=A$ and $\gamma(1)=B$.
- $\gamma(t)\in C$ for every $t\in[0,1]$, because $C$ is convex.

So any two points of $C$ are joined by a path inside $C$: $C$ is path connected (the empty set and singletons are trivially path connected).

2. Four examples.
- *The whole space $M_n(\mathbb{R})$*: it is a vector space.
- *Any open or closed ball* $\{A:\|A-A_0\|<r\}$ for a norm $\|\cdot\|$: if $\|A-A_0\|<r$ and $\|B-A_0\|<r$, then by the triangle inequality $\|(1-t)A+tB-A_0\|\le(1-t)\|A-A_0\|+t\|B-A_0\|<r$.
- *Positive definite symmetric matrices*: if $x^TAx>0$ and $x^TBx>0$ for all $x\neq0$, then $x^T((1-t)A+tB)x=(1-t)x^TAx+tx^TBx>0$ (the sum is a combination with weights $1-t,t\ge0$ not both zero).
- *Stochastic matrices* (nonnegative entries, every row sums to $1$): the two conditions are preserved by convex combinations, since they are linear equalities and inequalities.

Remark. Path connectedness is strictly weaker than convexity. A circle (for instance $SO(2)$) is path connected but not convex. Most of this chapter is about sets that are path connected for *non*-obvious reasons.