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$$\rho\left(\left[\begin{matrix}0 & A \\ -A^{\dagger} & 0\end{matrix}\right]\right)=\|A\|$$ where $\rho(\cdot)$ is the spectral radius, $\|\cdot\|$ is the induced 2-norm.

Question: I am looking for either a reference or a much shorter proof for this statement.

Proof The eigenvalues obey $$\left|\begin{matrix}-\lambda I & A \\ -A^{\dagger} & -\lambda I\end{matrix}\right|=0$$ The nonzero eigenvalues (of which the largest eigenvalue is one) obey $$\left|\begin{matrix}-\lambda I & A \\ -A^{\dagger} & -\lambda I\end{matrix}\right|=\left|\begin{matrix}-\lambda I\end{matrix}\right|\left|\begin{matrix}-\lambda I-A^{\dagger} A/\lambda\end{matrix}\right|=0$$ $$\left|\begin{matrix}-\lambda^2 I-A^{\dagger} A\end{matrix}\right|=0$$ Thus $\lambda^2$ is an eigenvalue of $-A^{\dagger} A$ $$\rho\left(\left[\begin{matrix}0 & A \\ -A^{\dagger} & 0\end{matrix}\right]\right)=\sqrt{\rho\left(A^{\dagger} A\right)}$$ But $A^{\dagger} A$ is hermitian, thus its spectral radius equals its induced 2-norm $$\rho\left(\left[\begin{matrix}0 & A \\ -A^{\dagger} & 0\end{matrix}\right]\right)=\sqrt{\|A^{\dagger} A\|}$$ And $\|A^{\dagger} A\|=\|A\|^2$ $$\rho\left(\left[\begin{matrix}0 & A \\ -A^{\dagger} & 0\end{matrix}\right]\right)=\sqrt{\|A\|^2}=\|A\|$$

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Here's one way to prove it. Call your block matrix $K$. For any unit vector $v=\pmatrix{x\\ y}$, consider $\|Kv\|$ directly and show that it is bounded above by $\|A\|$. Then show that this upper bound is attainable for some $v$.

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