Let $m>n$ and consider the Set
$$S\_{m,n}=\{A \in \mathbb{R}^{m \times n}\lvert A^TA=I\_n \}.$$
Does the function $d\colon S\_{m,n} \times S\_{m,n} \rightarrow \mathbb{R}$ defined by
$$d(A,B)=\sqrt{1-\det(A^TB)}$$
define a pseudometric on $S\_{m,n}$? (A pseudometric satisfies all conditions of a metric except that two elements can also have distance zero.) 
Consider the equivalence relation $A \sim B$ if there exist an orthogonal $Q \in \mathbb{R}^{n \times n}$ with $A=BQ$. The set $S\_{m,n}$ together with the equivalence relation can be identified with the grassmannian manifold $Gr(n,\mathbb{R}^m)$. Does $d$ define a metric on $Gr(n,\mathbb{R}^m)$? This question interests me because im trying to approximate (interpolate) functions which take values in the grassmanian manifold and the above metric would open up a possibility for approximating such functions.
The difficult part is the triangle-inequality, i.e. for all $A,B,C \in S\_{m,n}$ we need to prove that
$$\sqrt{1-\det(A^TC)}\leq \sqrt{1-\det(A^TB)}+\sqrt{1-\det(B^TC)}.$$


Thanks for any help in advance.



Is this a metric on the Grassmannian Manifold?
**EDIT** Actually, Cauchy-Binet suffices as the OP notices in the comments. I'll leave my overkill proof here for your amusement.




---


The proof below appeals to a famous result of Schoenberg (I've simplified the statement a bit), and basic linear algebra.



> 
> **Schoenberg's theorem** (see e.g., [Prop. 3.2, 1]). Let $X$ be a nonempty set and $\psi: X \times X \mapsto \mathbb{R}$ be positive definite kernel. Then, there exists an RKHS $H$ and a map $\varphi : X \to H$ such that
>  \begin{equation\*}
>  \|\varphi(x)-\varphi(y)\|\_H^2 = \frac{1}{2}[\psi(x,x)+\psi(y,y)] - \psi(x,y).
> \end{equation\*}
> 
> 
> 


We show that the function $\psi(A,B) = \det(A^TB)$ is positive definite, which as a result of Schoenberg's theorem shows that
\begin{equation\*}
 1-\det(A^TB) = \|\varphi(A)-\varphi(B)\|\_H^2,
\end{equation\*}
from which the triangle inequality is immediate. 


To prove the positive definiteness of $\psi$, we show that it is an inner-product by invoking the [Cauchy-Binet formula](http://en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula) (using Wikipedia's notation, except that for us $A$ is $m \times n$):
\begin{equation\*}
 \det(A^TB) = \sum\_{S \in \binom{[m]}{n}} \det(A^T\_{[n],S})\det(B\_{S,[n]}) = \sum\_{S \in \binom{[m]}{n}} \det(A\_{S,[n]})\det(B\_{S,[n]}) = \langle \phi(A), \phi(B)\rangle.
\end{equation\*}


**[1]** C. Berg, J. P. R. Christensen, and P. Ressel. *Harmonic Analysis on Semigroups: Theory of Positive Definite and Related Functions*, Springer GTM 100, 1984.


