Is the set of all matrices in $M(n; R)$ all of whose eigenvalues satisfy the condition $|λ| ≤ 2.$ compact?
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1$\begingroup$ Hint: take an upper triangular matrix with distinct eigenvalues less than 2, and let one of the off diagonal entries go to infinity. $\endgroup$– Cheerful ParsnipJan 15, 2014 at 1:50
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No. Using the standard topology on $M(n,\Bbb R) \cong \Bbb R^{n^2}$, consider unbounded subset of upper-triangular matrices $$A_k = \begin{bmatrix}\lambda_1 & k \\ & \lambda_2 & k \\ & & \ddots & k \\ & & & \lambda_n\end{bmatrix}$$ with $k\to\infty$, $|\lambda_i|\le 2$ as you required, and all other entries $0$.
answered Jan 15, 2014 at 1:50
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$\begingroup$ What is the meaning of saying an upper triangular matrix unbounded? $\endgroup$ Jan 15, 2014 at 1:59
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$\begingroup$ As I said, you want to think of matrices as vectors with $n^2$ components, and use the usual vector space structure and norm on $\Bbb R^{n^2}$. So I'm using the Euclidean norm on $\Bbb R^{n^2}$, and the Euclidean length of these matrices $A_k$ is at least $k\sqrt{n-1}$, which goes to infinity as $k\to\infty$. $\endgroup$ Jan 15, 2014 at 2:08
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$\begingroup$ No "sir" needed :) You're welcome. If you're content with the answer, you should accept it so that your question will not stay on the "unanswered questions" list. $\endgroup$ Jan 15, 2014 at 2:39