Hermitian Matrix M(x) is continuous on x. Is its eigenvalue also continuous on x?

Question

Each element of matrix $M(x)$ is a continuous function of $x$. Does this imply that all the eigenvalues are continuous function of $x$ too?

Answer 1

Yes, continuity should be guaranteed, as noted by Ilmari Karonen. But if you care about smoothness (which is what people usually want), this is a more complicated question. If the eigenvalues are distinct for all the $x$ you are interested in, then the eigenvalues are all smooth functions of $x$. However, if your eigenvalues can become degenerate for certain values of $x$, then there usually exists an ordering of the eigenvalues (dependent on the value of $x$) such that they are smooth. Take for example $$ \begin{bmatrix} 1-x & x \\ x & 1-x \end{bmatrix} $$ The eigenvalues are $\lambda = 1,1-2x$. These are obviously a smooth function of $x$, but the largest (most positive) eigenvalue jumps from $1-2x$ to $1$ as $x$ increases past $1\over 2$.