Show that convex hull of set of all positive determinant matrices is $\mathcal{M}_n(\mathbb{R})$.

Question

I know that set of positive determinant matrices is not a convex set as we can take $A = \mbox{Diag}(1, ..., 1)$ and $B = \mbox{Diag}(-1, -1, 1, 1, ..., 1)$. We can see $\frac{1}{2} A + \frac{1}{2}B$ is not a positive determinant matrix.

Then what is it's convex hull ? Is it $\mathcal{M}_n\mathbb{R}$?

Answer 1

The convex hull is the whole matrix space when $n\ge2$. Since every square matrix $A$ is the average of two nonsingular matrices (consider $A=\frac12\left[(A+cI)+(A-cI)\right]$ for a sufficiently large $c$), it suffices to prove that every matrix with a negative determinant is the average of two matrices with positive determinants.

Let $D=\operatorname{diag}(-1,1,\ldots,1)$ and $P=\pmatrix{-1&-2\\ 2&1}\oplus I_{n-2}$. Then $\det(P)>0$ and $D=\frac12(P+P^T)$. So, if $\det(A)<0$, then $A=DDA=\frac12(PDA+P^TDA)$ is the average of two matrices with positive determinants.

Answer 2

Sure. Take any matrix $A$. Now let $M$ be a very big real number. Claim $ A + MI $ and $ A - MI $ are invertible. Indeed, if $ M $ is larger than the largest eigenvalue this is true by definition.