Matrices with real positive spectrum

Idea. Same trick as for Hurwitz matrices: $\mathcal{S}$ is star-shaped about $I$.

Step 1: eigenvalues along the deformation. For $A\in\mathcal{S}$ and $t\in[0,1]$ let $A_t=(1-t)A+tI$. If $\lambda_1,\dots,\lambda_n$ are the eigenvalues of $A$ with multiplicity (all real and positive), then $A_t$ has eigenvalues $(1-t)\lambda_i+t$. (To see this, triangularise $A$ over $\mathbb{C}$: $A=PTP^{-1}$ with $T$ upper triangular with diagonal $\lambda_i$; then $A_t=P((1-t)T+tI)P^{-1}$ and $(1-t)T+tI$ is upper triangular with diagonal $(1-t)\lambda_i+t$.)

Step 2: they are real and positive. For $t\in[0,1]$, $(1-t)\lambda_i+t$ is a convex combination of the positive numbers $\lambda_i$ and $1$, hence real and $>0$. So $A_t\in\mathcal{S}$.

Step 3: conclusion. $A_0=A$, $A_1=I$, and $t\mapsto A_t$ is continuous. Every element of $\mathcal{S}$ is joined to $I$ inside $\mathcal{S}$, so $\mathcal{S}$ is path connected.

Remark. The same argument shows that $\{A:\text{all eigenvalues real and} \ge0\}$ and $\{A:\text{all eigenvalues in an interval }J\}$ (for $J$ convex and $1\in J$) are star-shaped. The set of matrices with all eigenvalues real is *not* obviously connected: its connectedness (Problem 26 handles the simple-spectrum part) needs more work.