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 $(x^*-z^*)\Lambda(y^*-z^*)$
[self-study]
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