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

Given B symmetric and positive definite matrix set
d(x,y)=(x-y)'B(x-y)

( 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 $(x^*-z^*)\Lambda(y^*-z^*)$

mariob6
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  • Please add the `[self-study]` tag & read its [wiki](http://stats.stackexchange.com/tags/self-study/info). 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. – gung - Reinstate Monica Mar 13 '16 at 21:12
  • If you start with the description of Mahalanobis distance I gave at http://stats.stackexchange.com/questions/62092/bottom-to-top-explanation-of-the-mahalanobis-distance/62147#62147, then this result follows immediately from the fact that Euclidean distance (in the plane) satisfies the triangle inequality. – whuber Mar 13 '16 at 21:27

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