Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be a part of $\mathcal{M}_n(\Bbb{R})$ and $B$ the set of real eigenvalues ​​of the matrix $A$.

1) Show that if $A$ is compact then $B$ is compact as well.

2) If $A$ is closed does it implies that $B$ is closed?

Attempt :

For 1) Assume that $A$ is compact, so $A$ is a closed, bounded subset of $\mathcal{M} _n (\mathbb{R})$. We have $$\|M\| = \sup_{x \neq 0} \frac{\|Mx\|}{\|x\|} \geq \max \ \mathrm{Sp}(M).$$ Therefore $B$ is bounded.

I have to prove,now, that $B$ is closed. I tried to take a Cauchy sequence defined by the eigenvalues $​​(\lambda_n)$ and a sequence of associated matrices $(M_n)$. I tried to show that $\lambda_n \rightarrow \lambda$. I am stuck.

For 2) I feel that we have to found a counterexample.

share|cite|improve this question
A is a matrix. What does it mean that it is compact? Do you mean that A is a set of matrices? – Marm Jul 20 '14 at 12:09
Balla: probably he means a compact set of matrices. – Peter Franek Jul 20 '14 at 12:11
up vote 2 down vote accepted

(Edit: I'm not sure if I understand the question but if $A$ is a compact set of $n\times n$ matrices, then...)

Let $A$ be the set of matrices $\mathrm{diag}(n, \frac{1}{1+n})$ for $n\in\mathbb{N}$. Then $A$ is closed but the spectrum is not closed.

1) can be proved by considering the limit $\det (A_i-\lambda_i\,\mathrm{Id})=0$ for $\lambda=\lim_i \lambda_i$ and a convergent sequence $A_i\to A$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.