Consider the space $X$ of all scalar products on $\mathbb{R}^n$. For a scalar product $s$ and a base $B:=b_1\ldots,b_n$ let $M_{s,B}$ denote the matrix, whose $(i,j)$-th entry is $(s(b_i,b_j))$ . Given two scalar products $s,s'$ one can find by PCA a orthonormal basis $B$ of $s$ such that $M_{s',B}$ is a diagonal matrix. By positive definiteness, all diagonal entries $\lambda_,\ldots,\lambda_n$ are positive. Let $d(s,s'):=\sqrt{\sum_{i=1}^n\log(\lambda_i)^2}$. I want to show, that this defines a metric on the set of all scalar products. Symmetry and Definiteness are clear. But why does it satisfy the triangular inequality ?
To be honest I already know, that it is a metric. This distance function comes from a Riemannian metric on the set of all scalar products. But I am looking for a simpler way (without computing the Levi-Civita connection and showing, that the geodesics satisfy the ODE and so on).
Furthermore the resulting space should be a $CAT(0)$-space. It would be nice if one could show the CAT(0) inequality directly. The geodesic from $s$ to $s'$ is given by
$$[0;1]\rightarrow X\qquad t\mapsto s_t,\mbox{ where } s_t(b_i,b_j)=\begin{cases}\lambda_i^t&i=j \\\ 0& i\neq j\end{cases}$$
Best Answer
Does the following proof for triangle inequality for the Riemannian metric for posdef matrices help?
(I am currently travelling, so the proof is missing exact references; will add them later when I am near my books.)
The proof that recall below is one of my favorites, and I first saw it in a paper by R. Bhatia. I think this proof must be also available in his book: Positive Definite Matrices.
Let $A$ and $B$ be strictly posdef matrices. We wish to prove that the function $$ d(A,B) = \left(\sum\nolimits_i (\log\lambda_i(AB^{-1}))^2\right)^{1/2}, $$ is a metric; here $\lambda(X)$ is the vector of eigenvalues of $X$ sorted in decreasing order. Clearly, the only non-trivial part is the triangle-inequality.
Definition: Let $x$ and $y$ be vectors with entries sorted in decreasing order. We say $x$ is majorized by $y$, written $x \prec y$, if $$ \sum\nolimits_i^k x_i \le \sum\nolimits_i^k y_i,\quad 1 \le k \le n,\quad \sum\nolimits_i^n x_i = \sum\nolimits_i^n y_i. $$
Definition: Let $\varphi : R^n \to R$. We say $\varphi$ is Schur-convex, if $x \prec y$ implies that $\varphi(x) \le \varphi(y)$.
Lemma 1 (Lidskii): Let $A$ and $B$ be posdef matrices. The following majorization holds $$ \log \lambda(AB) \prec \log\lambda(A) + \log\lambda(B). $$ I think this Lemma is proved in Chapter 3 of R. Bhatia's Matrix Analysis.
Triangle-Inequality: Using Lemma 1 we have \begin{align*} \log \lambda(AB^{-1}) &= \log\lambda(C^{-1/2}AC^{-1/2}C^{1/2}B^{-1}C^{1/2} \\\\ &\prec \log\lambda(C^{-1/2}AC^{-1/2}) + \log\lambda(C^{1/2}B^{-1}C^{1/2})\\\\ &= \log\lambda(AC^{-1}) + \log\lambda(CB^{-1}). \end{align*} Since $\|x\|_2$ is Schur-convex, and satisfies the triangle-inequality, on applying it to both sides of the above majorization we obtain \begin{equation} d(A,B) \le d(A,C) + d(C,B), \end{equation} proving the desired triangle inequality. In general, we could use any symmetric gauge function to obtain a corresponding triangle-inequality.