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 


$$[0;1]\rightarrow X\qquad t\mapsto s\_t,\mbox{ where } s\_t(b\_i,b\_j)=\begin{cases}\lambda\_i^t&i=j \\\ 0& i\neq j\end{cases}$$



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


