0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.