# Positive-definite matrix for positiveness in Mahalanobis distance

I am trying to prove that the Mahalanobis distance $d(x,y)$ is always positive, that is: $\forall{x,y \in E}\,{[0 \le d(x,y)]}$. To do so, I need to demonstrate that: $0\le\sqrt{(x-y)^TS^{-1}(x-y)}$, which is true when $S^{-1}$ exists and is positive semi-definite or p.s.d. (i.e. $z^TS^{-1}z \ge 0$ for $z\neq0$).

I was told that assumming $S^{-1}$ is p.s.d. is not enough to prove that $S^{-1}$ is also invertible (a requirement for the previous formula). The reason is that a zero matrix would produce null eigenvalues (making $S^{-1}$ not invertible) and still satisfy the p.s.d. definition.

I was adviced to use positive definite or p.d. matrices instead of p.s.d. (i.e. $z^TS^{-1}z > 0$ for $z\neq0$), but in that case, the expression $0=\sqrt{(x-y)^TS^{-1}(x-y)}$ would not be satisfied. In the same way that the logical expression $a\ge0$ means $a>0 \lor a=0$ (the OR operator only needs one condition to be true to make the whole logical expression true), could I say the positiveness property is satisfied even if $S^{-1}$ is p.d.?

Note: I know that Mahalanabis distance is not a metric, I am just trying to prove that it could be one if $S$ is p.d.

I am trying to prove the Mahalanobis distance is always positive

This distance is not always positive. For example, $d(x,x)=0$. You can prove that it is always nonnegative, that $d(x,y)\ge0$.

I was told assuming $S^{-1}$ is p.s.d. it not enough to prove it is invertible.

If you are writing $S^{-1}$, it means that $S^{-1}$ is invertible; the inverse is $S$! I think you mean to say that assuming $S$ is p.s.d. is not enough to prove $S^{-1}$ exists. This is correct, and the solution is to assume $S$ is p.d.

In this case, the expression $0=\sqrt{(x-y)S^{-1}(x-y)}$ would not be satisfied.

You do not need this to be satisfied. All you need to prove is that $d(x,y)\ge 0$, which is true when $S$, and therefore $S^{-1}$, is p.d. However, this equation is satisfied when $x=y$.

Could I say the positiveness property is satisfied even if $S^{-1}$ is p.d.?

You can say the nonnegativeness property is satisfied if $S^{-1}$ if p.d.

• Many thanks. Is there a way to prove that $x-y=\vec{0}$ is the only vector contained in the kernel of $\Sigma^{-1}$? – JFonseca Apr 24 '18 at 2:19
• @juanma2268 Yes! As long as S is positive definite, then this is the case. – Mike Earnest Apr 24 '18 at 3:07