Is there an elementary way to show the triangular inequality for this expression ?

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

$$[0;1]\rightarrow X\qquad t\mapsto s_t,\mbox{ where } s_t(b_i,b_j)=\begin{cases}\lambda_i^t&i=j \\\ 0& i\neq j\end{cases}$$

• The paper by S. Lang, "Bruhat-Tits-Räume", Elem. Math. 54 (1999) 45 – 63 (German), contains a nice and completely elementary treatment of this example of a CAT$(0)$-space. The main work of his approach lies in establishing the fact that the matrix exponential is expanding from the symmetric matrices (with the usual metric given by the tracial scalar product) to the positive definite matrices with the metric you describe. But it admittedly looks more complicated there than one might wish for. Commented Feb 11, 2011 at 11:57

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 $$d(A,B) \le d(A,C) + d(C,B),$$ proving the desired triangle inequality. In general, we could use any symmetric gauge function to obtain a corresponding triangle-inequality.