Spectral norm, eigenvalues range

Question

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.

Answer 1

If $A v = \lambda v$, then $\|Av\| =|\lambda| \|v\|$ and so if $\|A\|$ is the corresponding induced norm we have $\| A\| \ge |\lambda|$ for any eigenvalue (equivalently, $|\lambda| \in [-\|A\|,\|A\|]$).

Note that the spectral norm is induced by the Euclidean norm.

Since the Hessian is real symmetric, the eigenvalues are real and so for any eigenvalue $\lambda$ we have from above that $\lambda \in [-\|A\|,\|A\|]$.

Answer 2

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 $[-L,L]$.