TITLE: Eigenvalues are continuous?
QUESTION [10 upvotes]: I have two matrix $A$ and $B$ and consider $C(t)=A+tB$, with $t\in [0,1]$.
Are the eigenvalues of $C(t) $: $\lambda_i:[0,1]\rightarrow \mathbb{C}$ continuous functions?
I guess that the answer is yes, but why?

REPLY [7 votes]: If $t\mapsto A(t)$ is a continuous function from an interval $I$ of the real line in $\mathbb{C}^{n\times n}$, then there exist $n$ continuous functions $\lambda_i:I\to \mathbb{C}, i=1,\ldots,n$ such that for each $t\in I$ the spectrum of $A(t)$ is equal to $\{\lambda_1(t),\ldots,\lambda_n(t) \}$. For some $t_0\in I$
and $i\neq j$ we can have $\lambda_i(t_0)=\lambda_j(t_0)$.
One can see proofs of this Theorem in:
[1] T. Kato: A short introdunction to perturbation theory for linear operators, Springer-Verlag, 1982. Pages 126-127, Theorem 5.2.
[2] R. Bhatia: Matrix analysis, Springer, 1997. Pages 154-155, Theorem VI.1.4 and Corollary VI.1.6.