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Which of the following sets are compact in $\mathbb{M}_n(\mathbb{R})$
(a) The set of all upper triangular matrices all of whose eigenvalues satisfy $|\lambda|≤2 $.
(b) The set of all real symmetric matrices all of whose eigenvalues satisfy $|\lambda|≤2 $.
(c) The set of all diagonalizable matrices all of whose eigenvalues satisfy $|\lambda|≤2 $.

(a) & (c) are not true. But I am not sure about (b).can anybody help me please.

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Is $\mathbb{M}_{n}(\mathbb{R})$ the set of $n\times n$ real valued matrices with the usual topology? – T. Eskin Jan 19 '13 at 18:20 is the usual topology.can you please tell me that what should be the answer. – poton Jan 19 '13 at 18:30
up vote 5 down vote accepted

Let $S$ be the set of all real symmetric matrices all of whose eigenvalues satisfy $|\lambda| \leq 2$.

The main idea is that $\rho(M)=||M||$ for any real symmetric matrix $M$ (see spectral radius), with $||M||= \sup\limits_{||v||=1} ||Mv||$. It is easy to prove using that any real symmetric matrix is diagonalizable in an orthonormal basis.

So $S$ is bounded: $||M|| = \rho(M) \leq 2$ for every $M \in S$; and $S$ is closed: if $M_n \to M$ with $M_n \in S$, $\rho(M)=||M||=\lim\limits_{n \to + \infty} ||M_n||= \lim\limits_{n\to + \infty} \rho(M_n) \leq 2$.

In fact, $S$ is compact.

EDIT: Some basic facts about the spectral radius.

Because the space of matrices is finite dimensional, all norms are equivalent; for convenience, take the operator norm $||\cdot || : M \mapsto \sup\limits_{||v||=1} ||Mv||$ associated to the euclidean norm on $\mathbb{R}^n$.

We define the spectral radius of a matrix $M$ as the maximum $\max |\lambda|$ taking on the eigenvalues $\lambda$ of $M$. Then:

Theorem: Let $M$ be a real symmetric matrix. Then $\rho(M)=||M||$.

Proof: Let $\lambda_{max}$ be an eigenvalue of $M$ such that $|\lambda_{max}|=\rho(M)$ and $v_{max}$ an associated eigenvector of norm $1$. Then $||M|| \geq ||Mv_{max}||=||\lambda_{max}v_{max}||=|\lambda_{\max}|=\rho(M)$.

Because $M$ is a real symmetric matrix, there exists an orthonormal basis of eigenvectors $v_1,...,v_n$ of $M$, with associated eigenvalues $\lambda_1,...,\lambda_n$.

Let $w \in \mathbb{R}^n$ be such that $||w||=1$. We can write $\displaystyle w= \sum\limits_{i=1}^n w_ie_i$, so $\displaystyle Mv= \sum\limits_{i=1}^n \lambda_iw_ie_i$, hence $$||Mv||^2 = \sum\limits_{i=1}^n|\lambda_iw_i|^2 \leq |\lambda_{max}|^2 \sum\limits_{i=1}^n |w_i|^2= \rho(M)^2||w||=\rho(M)^2$$

Consequently, $||M|| \leq \rho(M)$ and finally $\rho(M)=||M||$. $\square$

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This answer is false, as evidenced by the matrices $M=2I$ and $2\cdot M=4I$. – Georges Elencwajg Jan 19 '13 at 18:58
@GeorgesElencwajg: Of course you are right. I changed my answer, thank you. – Seirios Jan 19 '13 at 19:55
Ah, this time it is perfect: +1 – Georges Elencwajg Jan 19 '13 at 20:20
@Panu: The matrix $A_n=\begin{bmatrix} 2 & n \\ 0 & 1 \end{bmatrix}$ is diagonalisable and is an upper triangular matrix whose eigenvalues are $\leq 2$. Taking $||\cdot|| : A \mapsto \max |a_{ij}|$, $||A_n|| \to \infty$. So the sets in (a) and (c) are not bounded. – Seirios Jan 20 '13 at 15:22
@Panu: It's done. – Seirios Jan 20 '13 at 21:53

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