On linear automorphism on positive definite matrices. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:14:08Z http://mathoverflow.net/feeds/question/115798 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115798/on-linear-automorphism-on-positive-definite-matrices On linear automorphism on positive definite matrices. HJ 2012-12-08T15:05:47Z 2012-12-08T16:44:30Z <p>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?</p> http://mathoverflow.net/questions/115798/on-linear-automorphism-on-positive-definite-matrices/115808#115808 Answer by Robert Bryant for On linear automorphism on positive definite matrices. Robert Bryant 2012-12-08T16:44:30Z 2012-12-08T16:44:30Z <p>I assume that you mean 'positive definite <em>symmetric</em> matrices'. Here's one proof, though it's certainly not the most clean.</p> <p>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 <em>nonnegative</em> 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 $L$.</p>