TITLE: This set of matrices is open
QUESTION [6 upvotes]: I'm trying to prove that the set of the matrices whose eigenvalues have non-zero real part is an open subset of $M^n$, the set of square matrices with order $n$ which is identify with $\mathbb R^{n^2}$. 
I don't know even how to begin, this question is really different to me, I don't how to use these topological concepts in relation with the eigenvalues of matrices.
I really need help.
Thanks in advance

REPLY [4 votes]: As mentioned in other answers, the meat of this is to show that the eigenvalues of a complex matrix depend continuously on the matrix coefficients.  Note that since the eigenvalues are not naturally ordered and can come with multiplicities, one should make precise exactly what above statement means, and there may be more than one reasonable interpretation.  I will take it to mean: for all $M = (m_{ij}) \in M_n(\mathbb{C})$ and $\epsilon > 0$, there is a $\delta > 0$ such that if $M' = (m_{ij}') \in M_n(\mathbb{C})$ is such that $|m_{ij}-m_{ij}'| < \delta$ for all $1 \leq i,j \leq n$, then for every eigenvalue $\alpha$ of $M$, there is an eigenvalue $\alpha'$ of $M'$ with $|\alpha-\alpha'| < \epsilon$, and for every eigenvalue $\alpha'$ of $M'$, there is an eigenvalue $\alpha$ of $M$ with $|\alpha-\alpha'| < \epsilon$. 
(Then to answer the question, suppose $M$ has an eigenvalue $\alpha$ with nonzero real part $x$, and take $\epsilon = x$.) 
It is clear that the coefficients of the characterstic polynomial of $M$ depend continuously on the entries of $M$: indeed, they are polynomial functions in $M$.  So really we need to show:

The roots of a complex polynomial $f = a_nt^n + \ldots + a_1t + a_0$ depend continuously on the coefficients: for every $\epsilon > 0$ there is a $\delta > 0$ such that: if $g = b_n t^n + \ldots + b_1t + b_0$ is such that $|a_i-b_i| < \delta$ for all $0 \leq i \leq n$, then every root of $g$ is within $\epsilon$ of some root of $f$ and vice versa.

I posed this -- with $\mathbb{C}$ replaced by any algebraically closed normed field -- as a problem in a graduate number theory course I taught several years ago.  It was solved by Mr. David Krumm (now Dr. David Krumm).  I seem to recall that he solved it several times over (with some variant on the statement) but settled on an approach which is simple and extremely elementary.  In particular, in contrast to a statement made in another answer here, one certainly does not need Rouche's Theorem or winding number considerations (which are not available anyway in an arbitrary normed field).  His solution was so nice that I got him to write it up: it is available here.  
(Note that I asked a couple of related preliminary questions which he solved on page 1.  For this question you can start reading on page 2.  There are only two pages!)