# Is this a metric: $d\big((\mathbf{x},A_x),(\mathbf{y},A_y)\big)=\frac{1}{2}(\mathbf{x}-\mathbf{y})^\top\big(A_x+A_y\big)^{-1}(\mathbf{x}-\mathbf{y})$?

Let $\Bbb{S}_{++}^n$ be the space of $n\times n$ symmetric positive definite matrices. We define the function $d\colon(\Bbb{R}^n\times\Bbb{S}_{++}^n)\times(\Bbb{R}^n\times\Bbb{S}_{++}^n)\to\Bbb{R}$ as follows $$d\big((\mathbf{x},A_x),(\mathbf{y},A_y)\big)=\frac{1}{2}(\mathbf{x}-\mathbf{y})^\top\big(A_x+A_y\big)^{-1}(\mathbf{x}-\mathbf{y}),$$ where $\mathbf{x},\mathbf{y}\in\Bbb{R}^n$, and $A_x,A_y\in\Bbb{S}_{++}^n$.

We would like to prove or disprove that this function is a metric on $\Bbb{R}^n\times\Bbb{S}_{++}^n$. For all $(\mathbf{x},A_x),(\mathbf{y},A_y),(\mathbf{z},A_z)\in\Bbb{R}^n\times\Bbb{S}_{++}^n$, we require the following to hold true:

1. $d\big((\mathbf{x},A_x),(\mathbf{y},A_y)\big)\geq0$ (non-negativity, or separation axiom), which holds as $(A_x+A_y)^{-1}$ is still a postive-definite matrix (please correct me if I am wrong),
2. $d\big((\mathbf{x},A_x),(\mathbf{y},A_y)\big)=0$ iff $(\mathbf{x},A_x)=(\mathbf{y},A_y)$ (identity of indiscernibles, or coincidence axiom), which is not true, I think, as it suffices to be $\mathbf{x}=\mathbf{y}$ (not necessarily $A_x=A_y$; what does it actually means?),
3. $d\big((\mathbf{x},A_x),(\mathbf{y},A_y)\big)=d\big((\mathbf{y},A_y),(\mathbf{x},A_x)\big)$ (symmetry), which holds trivially, and
4. $d\big((\mathbf{x},A_x),(\mathbf{z},A_z)\big) \leq d\big((\mathbf{x},A_x),(\mathbf{y},A_y)\big) + d\big((\mathbf{y},A_y),(\mathbf{z},A_z)\big)$ (subadditivity / triangle inequality), which is not triavial, but does it hold? It seems to ressemble the Mahalanobis distance, but could we prove (or disprove) that it holds true?

To sum up, I would like to ask what is true about this function? Is it a metric on $\Bbb{R}^n\times\Bbb{S}_{++}^n$? If so, what about the comment in the coincidence axiom, and what about the triangle inequality? Does it hold? Thank you very much for your help!

## EDIT

As can be seen in the accepted answer (as well as in the comments), this function cannot be a metric. Is there any ideas on what would be an appropriate metric (at least having the above properties!), if the $n$-dimensional vectors are the mean vectors of some multivariate Gaussian distributions, and the SPD matrices are the corresponding covariance matrices? I thought of the (symmetrized) Kullback–Leibler divergence, but it seems that it's not a metric either ([4] does not hold, as @tdc shows in his/her answer). Any ideas? Thanks again!

• It is not a metric. Take $n=1$ and compute $d((1,1),(1,2))$. In general, $d((x,A_1), (x,A_2)) = 0$. – copper.hat Sep 1 '14 at 18:56

Take $n=0, n=1$, etc. This does not satisfy the third axiom of a metric that $d(x,y)=0$ iff $x=y$. Thus, it cannot possibly qualify as a metric in the standard sense.
That is, $d((x,A),(x,B))=0$.
• Thanks @AnthonyPeter, yes, it is clear now! But could you provide some insight on what would be such a metric if $\mathbf{x}$'s were mean vectors and $A$'s covariance matrices of some multivariate Gaussian distributions? I thought about Kullback-Leibler divergence (symmetrized) but it's still not a valid metric, I think... – nullgeppetto Sep 1 '14 at 19:55