# Compact Subset Of $GL_n(\mathbb{C})$

Let X be a compact subset of $GL_n(\mathbb{C})$ and Y= set of all eigen values of matrices in X, we need to show Y is compact in $\mathbb{C}$, What I have done is that if $A\in X$ and $\lambda$ is an eigen value of A then $Ax=\lambda x$ for some eigen vector x, then $||Ax||=|\lambda| |x|$ but as X is compact so $||A||\le K$ for some K, so from the inequility $||Ax||\le ||A|||x|$ we finally get $|\lambda|\le K$ so Y is bounded set in $\mathbb{C}$ and from sequential argument we can say about the closedness. shall be highly pleased if you write any other way to proof.

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Maybe you can use the fact that the map from the set consists of monic complex polynomials with degree $n$ to the set consists of $n$ complex numbers without order by mapping each polynomial to its roots is continuous (I think this fact can be proved by argument principle), then use the fact that the image of a compact set under continuous map is compact. – Yuchen Liu Apr 20 '12 at 14:30
@jerrysciencemath could you explain a bit more? – Un Chien Andalou Aug 11 '12 at 22:29

Given a sequence of eigen values $y_n \in Y$ converging to $y \in \mathbb{C}$, choose matrices $A_n \in X$ such that $y_n$ is an eigen value of $A_n$. Also, take $v_n$, unitary vectors such that $A_n v_n = y_n v_n$. Since $X$ is compact, and since the set of unitary vectors is compact, there is a subsequence $n_k$, an $A \in X$ and an unitary vector $v$ such that $A_{n_k} \rightarrow A$ and $v_{n_k} \rightarrow v$.
The fact that $v_{n_k} \rightarrow v$ implies that $$A_{n_k} v_{n_k} = y_{n_k} v_{n_k} \rightarrow y v.$$ On the other hand, the fact that $A_{n_k} \rightarrow A$ implies that (since product of matrices is continuous) $$A_{n_k} v_{n_k} \rightarrow A v.$$ By the uniqueness of limits, $Av = yv$. That is, $y \in Y$.
This shows that $Y$ is closed. To show that $Y$ is bounded, just take an unbounded sequence (ie: $y = \infty$) to conclude, using the same argument as above, that $A_n$ would not have any convergent subsequence.
@Patience: In order to assure that $v_n$ would have a convergent subsequence. – André Caldas Aug 12 '12 at 2:54