Let $x_1 \in \mathbb{R}^{q_1}$, $x_2 \in \mathbb{R}^{q_2}$ be row vectors. Denote the $[2 \times (q_1 + q_2)]$ dimensional matrix \begin{equation} X = \begin{pmatrix} x_1 & 0 \\ 0 & x_2 \end{pmatrix}. \end{equation} Assume we have a real, symmetric $[(q_1+q_2)\times(q_1+q_2)]$ matrix $S$ that is a positive definite with bounded eigenvalues. I would like to bound the squared norm of the following product \begin{equation} ||(XSX^T)^{-1/2}X||^2 \leq \ (??) \end{equation} where $(\cdot)^{-1/2}$ denotes the square root matrix of the inverse. $||\cdot||$ is the induced $\ell^2$-norm/spectral norm.
My guess is that this should be bound by something proportional to $1/\lambda_{min}(S)$. I tried using the spectral decomposition for $S$ together with the block form but the inverse square root always throws me off. Any suggestion or help is appreciated.
Here is a quick upper bound. Since $S\ge\lambda_\min(S)I$, we have $$ XSX^T \ge X(\lambda_\min(S)I)X^T =\lambda_\min(S)XX^T $$ and in turn \begin{aligned} \|(XSX^T)^{-1/2}X\|^2 &\le\|(XSX^T)^{-1/2}\|^2\|X\|^2\\ &=\|(XSX^T)^{-1}\|\|X\|^2\\ &\le\|\lambda_\min(S)^{-1}(XX^T)^{-1}\|\|X\|^2\\ &=\lambda_\min(S)^{-1}\frac{\max\left(\|x_1\|_2^2,\|x_2\|_2^2\right)}{\min\left(\|x_1\|_2^2,\|x_2\|_2^2\right)}.\\ \end{aligned}