How to prove that $ \min_{c} || A - c I ||$ for Hermitian $A$ is its eigenvalue spread? I have the minimisation with respect to a complex number $c$:
$$ \min_{c \in \mathbb{C}} \left\Vert A - c \mathbb{1} \right\Vert, $$
for an Hermitian matrix $A$ ($\mathbb{1}$ is the identity). The matrix norm I'm considering is the standard operator norm (largest singular value).
I read in a paper that this corresponds to the spread of the eigenvalues of $A$, i.e. $ | \lambda_\mathrm{max}(A) - \lambda_{\mathrm{min}}(A)|$.
This seems reasonable, but I'm failing to show it's true. Does anybody have a hint?
 A: $A - cI$ is a normal matrix, and for a normal matrix the operator norm is the largest absolute value of an eigenvalue. If $\lambda_i$ are the eigenvalues of $A$ then $\lambda_i - c$ are the eigenvalues of $A - cI$. So we want to find $c$ such that $\text{max}_i |\lambda_i - c|$ is minimized.
This maximum is only ever attained by one of two possible eigenvalues: if the real part of $c$ is at least $\frac{\lambda_{max} + \lambda_{min}}{2}$ then it's attained by $\lambda_{min}$ (and the distance increases as the real part increases), and otherwise it's attained by $\lambda_{max}$ (and the distance increases as the real part decreases). Also removing the imaginary part of $c$ if it exists reduces the distance of $c$ to the real line and so reduces all of the above distances, since the $\lambda_i$ are all real. So the minimal value of $c$ occurs on the real line and in fact must occur at $c = \frac{\lambda_{max} + \lambda_{min}}{2}$, with minimum value $\boxed{ \frac{\lambda_{max} - \lambda_{min}}{2} }$.
