Let $$A = \begin{pmatrix} I & B\\ B^H & I\end{pmatrix}$$ with $\|B\|_2 < 1$. Show that $$\|A\|_2 \cdot \|A^{-1}\|_2 = \frac{1 + \|B\|_2}{1 - \|B\|_2}$$
My attempt:
I know that $$||A||_2||A^{-1}||_2=\frac{\sigma_\max}{\sigma_\min}$$ where $\sigma_\max$ and $\sigma_\min$ are the largest and smallest singular values. But I don´t know how to use the fact that $||B||_2<1.$
Someone could help me?
Let $b_1\ge\ldots\ge b_n\ge0$ be the singular values of $B$, where $n$ is the smaller dimension of $B$. Then $A$ is unitarily similar to $\pmatrix{1&b_1\\ b_1&1}\oplus\cdots\oplus\pmatrix{1&b_n\\ b_n&1}\oplus I$. The condition that $\|B\|_2<1$ ensures that the smallest eigenvalue of $A$, namely $1-b_1$, is positive. The result now follows easily.