This set of matrices is open 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
 A: As you can prove, for example with Rouché's Theorem, the roots of a complex polynomial vary continuously with the coefficients. 
To state this more precisely: Suppose $f(z) = z^n+a_{n-1}z^{n-1}+\dots+a_0$ has roots $w_1,\dots,w_r$. Then, given $\varepsilon>0$, there is $\delta>0$ so that whenever $|b_j-a_j|<\delta$ for $j=0,\dots,n-1$, the roots of $g(z)=z^n+b_{n-1}z^{n-1}+\dots+b_0$ are all contained in $\bigcup\limits_{k=1}^r B(w_k,\varepsilon)$.
A: Pick a norm (preferrably the $\Vert \cdot \Vert_\infty$) and write down, what 
$$B_\epsilon(M) := \{A \in M^n | \Vert M-A \Vert < \epsilon\}$$
looks like. Then prove that every matrix in $O := \{ A\in M^n | \Re(\lambda) \neq 0 \ \forall \lambda \in \sigma(A)\}$ has an $\epsilon > 0$ with
$$B_\epsilon(M) \subset O$$
For that you can use theorems about the bevahiour of eigenvalues under small disturbances.
A: Here's another approach: show that the complement is closed. So, given a convergent sequence $M_1,M_2,\dots\to M$ of matrices (in some metric equivalent to the usual metric on $\mathbb R^{n^2}$) which each have an eigenvalue whose real part is zero, we want to show that $M$ has an eigenvalue whose real part is zero.
There are eigenvalues $\lambda_i$ and eigenvectors $v_i$ such that $\mathrm{Re}(\lambda_i)=0$ and $M_iv_i=\lambda_iv_i$ for all $i$; furthermore we can normalise so that $\|v_i\|=1$ in the $L_2$ norm. So $\lambda_i=\langle M_iv_i,v_i\rangle$ where $\langle x,y\rangle$ denotes the usual Hermitian inner product ($\sum_k x_k\overline{y_k}$). Because the set of unit vectors is compact, passing to a subsequence if necessary, we may assume that $v_i$ converges to some unit vector $v$. The vectors $M_iv_i= \langle M_iv_i,v_i\rangle v_i$ then converge to $\langle Mv,v\rangle v$ by continuity of multiplication, and the real part of $\langle Mv,v\rangle$ is clearly zero.
A: 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!)
A: This is a direct consequence of the following three facts:


*

*The inverse image of an open set under a continuous function is open.

*$(\mathbb{C}\setminus i\mathbb{R})^n$ is open.

*The eigenvalues of a real or complex matrix are continuous functions of the matrix entries. See, e.g. James M. Ortega, Numerical Analysis: A Second Course, p.45, Theorem 3.1.2. Since coefficients of the characteristic polynomial of a matrix are polynomials in the matrix entries, and eigenvalues are roots of the characteristic polynomial, it suffices to prove that the roots of a generic polynomial $p(z)$ over $\mathbb{C}$ are continuous functions of the coefficients of $p$. And this is exactly what the proof in Ortega infers. As noted by Ted Shifrin in another answer here, this can be proved easily with Rouché's Theorem.

