Here is one cute analysis application. Consider a continuously differentiable function $A(x)$ on a real argument and taking values in the $N \times N$ matrices. Furthermore suppose that $\lambda(0)$ is a simple eigenvalue of $A(x)$, i.e. $dim(\ker( (A - \lambda(0))^N )) = 1$. Then there exists a small interval $I$ containing $0$ and a function $\lambda(x)$ such that $\lambda(x)$ such that $\lambda(x)$ is an eigenvalue of $A(x)$.
The proof is just applying the implicit function theorem to $$f(x,E) = \det(A(x) - E),$$ since $\lambda(0)$ being simple implies that $\partial_E f(0,\lambda(0)) \neq 0$.
Furthermore, using the characterization of eigenvalues using the determinant, one can get some geometric information on how the curve $\lambda(x)$ looks.