Given a general matrix $A(t), t>0$, with real entries, I would like to know if the eigenvalues of $A(t)$ are continuous functions of $t$. These eigenvalues may be real or complex.
What about the spectral radius?
A classical result from complex analysis states that the roots of a polynomial vary continuously with the coefficients. Can we use the theorem directly to prove the above? or there are other cases where the eigenvalues are actually discontinuous?
What I'm actually doing is trying to prove that there exists a $t$ for which the spectral radius of $A(t)$ is in $(0,1)$, and I'm doing that by proving that the spectral radius is 1 if $t\rightarrow 0$ and 0 if $t\rightarrow \infty$ (which I already know). Then I would invoke the continuity of the spectral radius to say that there must exist a value of $t$ for which the spectral radius is in $(0,1)$