# Mahalanobis Distance using Eigen-Values of the Covariance Matrix

 P: 1 Given the formula of Mahalanobis Distance: $D^2_M = (\mathbf{x} - \mathbf{\mu})^T \mathbf{S}^{-1} (\mathbf{x} - \mathbf{\mu})$ If I simplify the above expression using Eigen-value decomposition (EVD) of the Covariance Matrix: $S = \mathbf{P} \Lambda \mathbf{P}^T$ Then, $D^2_M = (\mathbf{x} - \mathbf{\mu})^T \mathbf{P} \Lambda^{-1} \mathbf{P}^T (\mathbf{x} - \mathbf{\mu})$ Let, the projections of $(\mathbf{x}-\mu)$ on all eigen-vectors present in $\mathbf{P}$ be $\mathbf{b}$, then: $\mathbf{b} = \mathbf{P}^T(\mathbf{x} - \mathbf{\mu})$ And, $D^2_M = \mathbf{b}^T \Lambda^{-1} \mathbf{b}$ $D^2_M = \sum_i{\frac{b^2_i}{\lambda_i}}$ The problem that I am facing right now is as follows: The covariance matrix $\mathbf{S}$ is calculated on a dataset, in which no. of observations are less than the no. of variables. This causes some zero-valued eigen-values after EVD of $\mathbf{S}$. In these cases the above simplified expression does not result in the same Mahalanobis Distance as the original expression, i.e.: $(\mathbf{x} - \mathbf{\mu})^T \mathbf{S}^{-1} (\mathbf{x} - \mathbf{\mu}) \neq \sum_i{\frac{b^2_i}{\lambda_i}}$ (for non-zero $\lambda_i$) My question is: Is the simplified expression still functionally represents the Mahalanobis Distance? P.S.: Motivation to use the simplified expression of Mahalanbis Distance is to calculate its gradient wrt $b$.