Hi Achmad,

If you apply a linear normalization, no, it does not.

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See for example the following R code where you can check that D2 = D22.

x = rnorm(100,2,3)

y = 0.4 * x + rnorm(100, 1, 3)

X = cbind(x,y)

y2 = 2*y+4

x2 = 0.5*x -3

X2 = cbind(x2,y2)

Sx

Mahalanobis distance computation uses inverse of the co-variance matrix while computing the distance which normalizes the original data by the variance and co-variance present in the original data hence there is not at all need to do the same prior to the distance computation. Variance (diagonal element of the matrix) does the standard normalization over the original data and co-variance (other than diagonal elements in the co-variance matrix) takes care of distance computation twice by highly correlated variables.

For example 1 if two variable are scaled very differently then dividing by variance bring both variable to same scale.

2. If two variables are  highly correlated say .90 then since we have already taken full distance contribution by both variables but actually these variables are the same hence we need to deduct the distance contribution proportional to the correlation of both variables. When we take inverse of the co-variance matrix we get negative terms in the other than diagonal elements since they were highly correlated terms which causes the deduction in the distance to neutralise the effect of the doubled distance contribution due to correlation.

Read this wonderful post 

http://mccormickml.com/2014/07/22/mahalanobis-distance/

Hope this helps