On linear automorphism on positive definite matrices.

Question

I saw a statement in [Murakami, On automorphisms on Siegel domains] that every linear automorphism $\phi$ on the set of positive definite matrices can be represented as conjugation: i.e. there is a matrix $B\in GL(n,\mathbb{R})$ such that $\phi(A)=B^t A B$. It seems an easy statement but I couldn't prove it. Can somebody help me?

Answer 1

I assume that you mean 'positive definite symmetric matrices'. Here's one proof, though it's certainly not the most clean.

A linear automorphism of the space of all symmetric matrices that preserves the cone of positive definite matrices will preserve the closure of that cone, i.e., the nonnegative symmetric matrices. The extreme rays of that closed cone are the rank $1$ nonnegative symmetric matrices, i.e., the matrices of the form $s = vv^T$ where $v\in\mathbb{R}^n$ (thought of as columns of height $n$). Now, for any pair of linearly independent vectors $v_1$ and $v_2$ in $\mathbb{R}^n$, the $3$-plane spanned by the matrices of the form $(av_1+bv_2)(av_1+bv_2)^T$ will contain a cone of rank $1$ elements and the automorphism will have to carry that $3$-plane to a $3$-plane that has a cone of rank $1$ elements, and it is easy to see that this implies that this image $3$-plane must be spanned by matrices of the form $(aw_1+bw_2)(aw_1+bw_2)^T$ for some linearly independent pair $w_1$ and $w_2$ in $\mathbb{R}^n$. The upshot of this is that, because all collineations of $\mathbb{RP}^{n-1}$ are projectivizations of linear automorphisms of $\mathbb{R}^n$, there will have to be a linear map $L:\mathbb{R}^n\to\mathbb{R}^n$ such that the automorphism carries $vv^T$ to a positive multiple of $(Lv)(Lv)^T$ for all nonzero $v\in \mathbb{R}^n$. Composing the given automorphism with the automorphism induced by $L^{-1}$, we are reduced to the case of an automorphism of symmetric matrices that carries $vv^T$ to a multiple of $vv^T$ for all nonzero $v\in \mathbb{R}^n$. Of course, since the symmetric matrices have a basis made of rank $1$ elements, it follows that this automorphism must be a (positive) multiple of the identity, so even that can be absorbed into $L$.