I'm trying to prove that the Mahalanobis distance is an actual distance, more in general

Given B symmetric and positive definite matrix set

( the ' means Transpose)

Of course the only difficoult thing to prove is the triangoular inequality

Any insights?

EDIT: this is what I've tried so far... given B, using the spectral thm, B=QBQ', thus $ d(x,y)=(Q'(x-y))' \Lambda (Q'(x-y)) $

then $Q'(x-y)=(x^*-y^*)$ and it's sufficient to prove the inequality for $z^*=Qz$.

Anyways, when i write $(x^*-y^*)=(x^*-z^*)+(z^*-y^*)$ and put it into the distance formula, I cannot decide the sign of $(x^*-z^*)\Lambda(y^*-z^*)$

  • $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$ Mar 13, 2016 at 21:12
  • $\begingroup$ If you start with the description of Mahalanobis distance I gave at stats.stackexchange.com/questions/62092/…, then this result follows immediately from the fact that Euclidean distance (in the plane) satisfies the triangle inequality. $\endgroup$
    – whuber
    Mar 13, 2016 at 21:27


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.