Inequality on Mahalanobis distance

Question

If $x^Tx \geq y^Ty$, what are the conditions on $R$ to be $x^TRx \geq y^TRy$ for a positive semidefinite $R$ ?

Answer 1

Let us consider an example where $\textbf{x}= [2 \quad 0]^{\top}, \textbf{y}= [0 \quad 1]^{\top}$ and $\mathbf{R}= \pmatrix{1 & 0 \\ 0 & 6}$. In that case, $\lVert \textbf{x} \rVert^2 = 4$ and $\lVert \textbf{y} \rVert^2 = 1$, but $\textbf{x}^{\top} \mathbf{R} \textbf{x} = 4$ and $\textbf{y}^{\top} \mathbf{R} \textbf{y} = 6$.

So, even if $\mathbf{R}$ is PD that does not guarantee your desired inequality.

---

From the example, we can conclude that the scaling must be uniform in all directions, that is, eigenvalues must be the same.

Let us consider that $\mathbf{R}$ is a symmetric matrix. We can write as follows: $$\mathbf{R}=\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{\top}=\lambda\mathbf{Q}\mathbf{I}\mathbf{Q}^{\top}=\lambda\mathbf{I}.$$ Now, $$\textbf{x}^{\top}\mathbf{R}\textbf{x}=\lambda \lVert \textbf{x} \rVert^2.$$ In order to maintain the inequality, we need $\lambda > 0$.

[Note: The above analysis is valid only when $\mathbf{R}$ is symmetric.]