Spectral norm, eigenvalues range I stumbled upon a property in solutions of some exercises which stated that if a hessian of a possibly non-convex function f(x) is bounded in spectral norm then its eigenvalues lie in the interval.
$$ ||\nabla^2f(x)||_2 \leq L $$
$$ eigenvalues \in [-L, L]$$
I fail to understand or more I am unable to find where this property comes from, I looked through many materials about spectral norm, spectral radius and I think at this point I am completely confused. I know that spectral norm is the maximal singular value of a matrix. In this case does it mean that hessian is symmetric so eigenvalues == singular values? How do we go further with that to get the interval? I get the upper bound of the interval, it's obvious but why the lower bound. Thank you in advance for pointing me to right sources or directly answering.

The spectral norm of a matrix is, by definition, the largest absolute value of its eigenvalues. If the largest absolute value of the eigenvalues is less than or equal to $L$, then all the eigenvalues have absolute value less than or equal to $L$, so if they are real, they all lie in the interval $[-L,L]$.
