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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?

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maybe you mean $t \in [0,1]$ instead of $y$ ? – sigmatau Nov 7 '13 at 21:36
These functions aren't even defined, I don't see how they could be continuous. What is true is that the set of eigenvalues is continuous (for the right topology on the power set). – Najib Idrissi Nov 7 '13 at 21:38
Do you want to express that there are continuous functions $\lambda_i\colon [0,1]\to \mathbb C$ such that $\{\lambda_1(t),\ldots,\lambda_n(t)\}$ are the eigenvalues of $C(t)$? – Hagen von Eitzen Nov 7 '13 at 21:40
You are right Hagen von Eitzen. How can I convince me of this? – yemino Nov 7 '13 at 22:04

The answer is yes, and this is dependent on the fact that the roots of a polynomial vary continuously with its coefficients.

We have the following theorem taken from A Brief Introduction to Numerical Analysis by Tyrtyshnikov.

Theorem: Consider a parametrized batch of polynomials $$p(x,t) = x^n + a_1(t)x^{n-1} + \cdots + a_n(t),$$ where each $a_i(t)$ is a continuous function on the interval $[\alpha,\beta]$. Then there exists continuous functions $$x_1(t),\ x_2(t),\ \cdots,\ x_n(t),$$ on $[\alpha, \beta]$ such that for each $x_i(t)$ we have $$p(x_i(t),t) = 0,\ \ \ t\in[\alpha,\beta].$$ $\square$

With $C(t)=A+tB$, each entry of the matrix is a linear polynomial in $t$ and hence the characteristic polynomial will be parametrized in the form above with $t\in[0,1]$. The theorem then directly implies that the roots of the characteristic polynomial, i.e. the eigenvalues of the matrix $C(t)$, are expressible as continuous functions of $t$.

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This is way to heavy for me on this Thursday afternoon, but nobody gave you an upvote yet, so here it is...) – imranfat Nov 7 '13 at 22:04
@imranfat At least it's not a Monday morning :) – EuYu Nov 7 '13 at 22:11
Thanks EuYu!! I have that book, may you tell me in what page is that theorem. I've searched but have not found. – yemino Nov 7 '13 at 22:48
@yemino The theorem is presented in Section 3.9 on page 26. I don't have the book at hand so I used Google Books. Here is a link. – EuYu Nov 7 '13 at 22:58
@Adam Yes, the functions $x_i(t)$ are continuous. It was stated so in the source I linked to in my previous comment, but I forgot to include it in the answer. I've updated the answer now. Thanks for pointing that out. – EuYu Mar 27 at 21:03

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