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Let the set $\mathcal E$ be defined \begin{align*} \mathcal E = \{A \in M_n(\mathbb R): \max_i \text{Re}(\lambda_i(A)) = 0\}, \end{align*} i.e., the largest real part of all eigenvalues is $0$. Let \begin{align*} \mathcal F = \{A \in \mathcal E: A \text{ has distinct eigenvalues}\}. \end{align*} I want to prove $\mathcal F$ is dense in $\mathcal E$.

The proof I have in mind: For any $A \in \mathcal E$, we take the real Schur form of $A$ \begin{align*} A = SJS^{-1} = S \begin{pmatrix} J_1&&* \\&\ddots\\&&J_k\end{pmatrix} S^{-1}, \end{align*} where we order the blocks such that $\max_i \text{Re} (\lambda_i(J_1) )= 0$. It follows that $J_1$ is either a scalar $0$ or a block \begin{align*} \pmatrix{0 & -b \\ b & 0} \end{align*} for some $b > 0$. Now consider \begin{align*} J_n = \begin{pmatrix} J_1&* & * \\& \hat{J}_2 & *\\ & & \ddots \\& & &\hat{J}_k\end{pmatrix}, \end{align*} where $\hat{J}_i$ is defined as \begin{align*} \hat{J}_i = \begin{cases} -\frac{i}{n} && \text{if } J_i =0, \\ (1-\frac{i}{n})J_i && \text{otherwise}. \end{cases} \end{align*} Take $A_n = S J_n S^{-1}$ and for sufficiently large $n$, $A_n$ has distinct eigenvalues and $A_n \in \mathcal E$ with $A_n \to A$.

Is this statement and the proof correct? Is there some other way, for example using characteristic polynomials, to show this statement?

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    $\begingroup$ Looks good to me. $\endgroup$ – user1551 Nov 10 '18 at 5:40
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As user1551 wrote, it looks good ; however it takes at least 15 minutes to convince ourselves.

Remark 1. Only the eigenvalues with $0$ real part may be a problem. Indeed we must attack them on their left.

Remark 2. It is useless to detail a convergent sequence. Rather, take an epsilon.

We can write that follows

We may assume that $A=diag(U_p,V_{n-p})$ (up to a real change of basis) where $U\in \mathcal{E}_1=\{B\in\mathcal{E}(p);spectrum(B)\subset \{z;re(z)<0\}\},V\in \mathcal{E}_2=\{B\in\mathcal{E}(n-p);spectrum(B)\subset \{z;re(z)=0\}\}$.

$B\in \mathcal{F}(p)$ (with the correct dimension) is characterized by $discrim(\det(B-xI),x)\not= 0$; then $\mathcal{F}(p)$ is Zariski open dense in $\mathcal{E}_1$, that is not the case for $\mathcal{E}_2$ (if we move a little an element of $\mathcal{E}_2$, then the real part of some eigenvalue can become $>0$).

We may assume that $V$ is block triangular with diagonal blocks in the form $2\times 2$: $\begin{pmatrix}0&b\\-b&0\end{pmatrix}$ (with $b\not= 0$) or $1\times 1$: $[0]$ (a total of $k$ blocks -distinct or not-). Let $\epsilon >0$ and let $(\alpha_i)_{i\leq k}$ be distinct in $(-\epsilon,0)$. It suffices to change the $0$ diagonal of index $i$ with the $\alpha_i$ diagonal.

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  • $\begingroup$ Thanks for your answer. There is one part of your remark I am confused of: in your definition, $A = diag(U, V)$ where $U, V$ are defined as subsets of $\mathcal E$. Do you really mean they are just square matrices of dimension $<n$ that satisfy the spectrum conditions? $\endgroup$ – user1101010 Nov 11 '18 at 6:13
  • $\begingroup$ One must adapt the dimensions; $dim(U)=p,dim(V)=n-p$ where $p\in [[0,n]]$ and $U,V$ are elements of $\mathcal{E}(p),\mathcal{E}(n-p)$. The eigenvalues of $U$ (resp. $V$) are the eigenvalues of $A$ with $<0$ real part (resp. with $0$ real part). $\endgroup$ – loup blanc Nov 11 '18 at 14:35
  • $\begingroup$ Thanks. If possible, could you take a look at this question I asked math.stackexchange.com/questions/2993572/…? They are very similar. $\endgroup$ – user1101010 Nov 11 '18 at 17:43

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