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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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    @jerrysciencemath could you explain a bit more?2012-08-11

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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.

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    thank you, and they are compact as Unitary matrices are compact?2012-08-14