The question I am working on is as follows. A is an $n\times n$ matrix. A is diagonalizable by a similarity transformation s.t. $T^{-1}AT=B ={\rm diag}(\lambda_1,..,\lambda_n). $ $\tilde{A}=A+{G}. $ If $\tilde{\lambda}$ is an eigenvalue of $\tilde{A},$ show $$ \|(B-\tilde{\lambda}I)^{-1}T^{-1}{\bf G}T\|_p\geq 1. $$
My strategy was to use algebra to solve for the left side matrix and then calculate all of the possible norms for it. $\tilde{\lambda}$ and $\lambda$ happened to cancel out nicely, but this does not incorporate using the opposite to show $\tilde{A}-\tilde{\lambda}I$ is invertible.
The hint is generally confusing and am not sure to do what with it.
To incorporate your professor's hint: if the opposite is true, then there is a value of $p$ for which we have $$ \|(B-\tilde{\lambda}I)^{-1}T^{-1}{\bf G}T\|_p < 1 $$ However, note that $(\tilde A - \tilde \lambda I)$ is invertible if and only if $T^{-1}(\tilde A - \tilde \lambda I)T$ is invertible, and $$ T^{-1}(\tilde A - \tilde \lambda I)T = \\ T^{-1}((A + G) - \tilde \lambda I)T = \\ (T^{-1}AT^{-1} - \tilde \lambda I) + T^{-1}GT = \\ (B - \tilde \lambda I) + T^{-1}GT $$ This matrix will be invertible if and only if the matrix $$ (B - \tilde \lambda I)^{-1}[(B - \tilde \lambda I) + T^{-1}GT] = \\ I + (B - \tilde \lambda I)^{-1}T^{-1}GT $$ is invertible. Now, use the fact that $I + M$ will be invertible whenever $\|M\| < 1$ for some norm $\|\cdot\|$.