# Proof That the Mahalanobis Distance is $\ge 0$

I was just introduced to the Mahalanobis distance between two vectors $$\mathrm{\mathbf{X}}$$ and $$\mathrm{\mathbf{Y}}$$ of random variables:

$$|| \mathrm{\mathbf{X}} - \mathrm{\mathbf{Y}}||_{\Sigma} = ((\mathrm{\mathbf{X}} - \mathrm{\mathbf{Y}})^T \Sigma^{-1}(\mathrm{\mathbf{X}} - \mathrm{\mathbf{Y}}))^{1/2},$$

where $$\Sigma$$ is the covariance matrix.

As I understand it, the 4 properties that a function $$d(x,y)$$ must satisfy in order to be a metric are as follows:

1. $$d(x, y) \ge 0$$
2. $$d(x, y) = 0 \Longleftrightarrow x = y$$
3. $$d(x, y) = d(y, x)$$
4. $$d(x, z) \le d(x, y) + d(y, z)$$

I only have an introductory-level knowledge of statistics, so I'm wondering how it is that the Mahalanobis distance satisfies property 1? Ignoring the square root, why is it that $$(\mathrm{\mathbf{X}} - \mathrm{\mathbf{Y}})^T \Sigma^{-1}(\mathrm{\mathbf{X}} - \mathrm{\mathbf{Y}})$$ can't be negative?

I would greatly appreciate it if people could please take the time to clarify this.

This is because the $$\Sigma^{-1}$$ matrix (inverse of the covariance matrix) is symmetric definite positive.
Once that you have a symmetric positive definite (SPD) matrix $$S$$, it is easy to define:
• a scalar product $$\langle v,w\rangle_S=\langle v,Sw\rangle$$ (where $$\langle.,.\rangle$$ is the usual scalar product)
• an associated distance: $$d(v,w)=\|v-w\|_S^2=\langle v-w,v-w\rangle_S=\langle v-w,S(v-w)\rangle$$