Compact subsets of $M(n,\mathbb R)$

Question

Consider $M(n,\mathbb R)$, the space of all $ n\times n$ matrices over $\mathbb R$.Which of the following are compact? a.$O(n)$ the set of all orthogonal matrices

b.$GL(n,\mathbb R)$ set of all non-singular matrices over $\mathbb R$

c.$SL(n,\mathbb R)$ set of all non-singular matrices over $\mathbb R$ with determinant equals one.

d.set of all nilpotent matrices.

Answer:since the determinant map is continuous and continuous image of a compact set is compact using them I have concluded (b) is not compact since image set of (b) is $(-\infty,0) \cup (0,\infty)$. Is it correct?But how to approach the others ?

If you are using the usual topology, then a set is compact if and only if it is closed and bounded. — , Oct 31 '14 at 08:02

Seirios · Accepted Answer · 2014-10-31T08:33:01.240

3

In order to show that the spaces mentionned in b), c) and d) are not compact, it is sufficient to exhib an unbounded sequence in these spaces. For example, consider respectively $$\left( \begin{matrix} n & 0 \\ 0 & n \end{matrix} \right), \ \left( \begin{matrix} n+1 & 1 \\ n & 1 \end{matrix} \right) \ \text{and} \ \left( \begin{matrix} 0 & n \\ 0 & 0 \end{matrix} \right).$$

Afterwards, you can show that $O(n)$ is compact by proving that it is closed and bounded.

Hint: if $M \in O(n)$, notice that $\|M\| \leq 1$ where $\|M\| : = \sup\limits_{ \| x \| =1 } \|Mx\|$.

edited Oct 31 '14 at 08:33

answered Oct 31 '14 at 08:21

Seirios

33,157

How does the fourth matrix be a nilpotent one? – Learnmore Oct 31 '14 at 08:28
Sorry, I read idempotent instead of nilpotent. I edited my answer, thank you for your comment. – Seirios Oct 31 '14 at 08:33
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Or $\left(\begin{smallmatrix}1/n&0\0&1\end{smallmatrix}\right)$ is a sequence of invertible matrices converging to a noninvertible one. – egreg Oct 31 '14 at 09:35
I am taking norm of a matrix M as$||M||:=(\sum_{i,j}|a_{ij}^2|)^{1/2}$ So I am not getting the Hint @Seirios – Learnmore Oct 31 '14 at 11:12
The choice of the norm does not matter: they are all equivalent. – Seirios Oct 31 '14 at 12:03

Compact subsets of $M(n,\mathbb R)$

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