0
$\begingroup$

Mahalanobis distance is given by:

$(x-\mu)^T \Sigma^{-1}(x-\mu)$

Apparently the solution to the above formula is equivalent to:

$Trace(\Sigma^{-1} (x-\mu)(x-\mu)^T)$

How can you prove that the two are indeed equivalent?

Note: This has nothing to do with university, I just saw the equivalence and was wondering where it came from

$\endgroup$
1
  • $\begingroup$ I might be tempted to use induction. You can calculate it in the $1\times 1$ and $2\times 2$ cases. Then you can show that if it holds for the $n\times n$ case then it must hold for the $(n+1)\times(n+1)$ case. $\endgroup$
    – Dave
    Jun 23 at 9:58

Your Answer

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

Browse other questions tagged or ask your own question.