I am stuck in a proof:
$A$ is a symmetric positive definite matrix, and $B$ is a symmetric matrix such that $||B|| ≤ ||A^{-1}||^{−1}$.
Let $B = \Psi L \Psi^T$ be the eigenvalue decomposition of matrix $B$. Remove columns of $\Psi$ that are eigenvectors of value $0$, so $L$ only contains nonsingular terms. Let $v$ be the eigenvector corresponding to the largest absolute eigenvalue of $(L^{-1} + \Psi^T A^{-1} \Psi)^{-1}$.
Then $|v^T (L^{-1} + \Psi^T A^{-1} \Psi) v| \geq |v^T L^{-1} v| − |v^T \Psi^T A^{-1} \Psi v| \geq ||L||^{−1} − ||A^{−1}||.$
How to obtain the last inequality? I am also not sure if all the information I have given before is required. Thanks a lot!
I have edited this post. My original question was how to obtain the first inequality but now, I am stuck again. Thanks in advance for helping out again!
$|x+y|\geq |x|-|y|$ because $|x|=|(x+y)-y| \leq |x+y|+|y|$.
Take $y=v^T (\Psi^T A^{-1} \Psi) v$ and $x=v^T (L^{-1}v)$.