# Is $\{A \in E: A \text{ has distinct eigenvalues}\}$ dense in $E = \{A \in M_n(\mathbb R): \max_i \text{Re}(\lambda_i(A))=0\}$?

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?

• Looks good to me. – user1551 Nov 10 '18 at 5:40

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.

• 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? – user1101010 Nov 11 '18 at 6:13
• 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). – loup blanc Nov 11 '18 at 14:35
• Thanks. If possible, could you take a look at this question I asked math.stackexchange.com/questions/2993572/…? They are very similar. – user1101010 Nov 11 '18 at 17:43