 Solution

The Euclidean matrix norm is the induced operator norm $\Vert C{\Vert }_2={\sup }_{\Vert x{\Vert }_2=1}\Vert Cx{\Vert }_2$. By the spectral theorem, a real symmetric matrix has $A=Q\Lambda Q^T$ with $Q$ orthogonal and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$. HenceOrthogonal transformations preserve the Euclidean norm, and the induced norm of a real diagonal matrix is the maximum absolute diagonal entry. Therefore, for $t\ge 0$,This gives the requested inequality with equality. If an exponent $\beta <\mu (A)$ were to work with prefactor one for every $t$, choose a unit eigenvector for $\mu (A)$. Its image has length $e^{t\mu (A)}>e^{t\beta }$ for every $t>0$, a contradiction. Thus $\mu (A)$ is the smallest possible exponent.

For symmetric matrices the spectral abscissa equals the Euclidean logarithmic norm. Symmetry is essential to this particular exact norm formula; a general non-normal matrix can have transient growth not controlled with prefactor one by its spectral abscissa.