# compact subsets of $\mathbb{R^{n^2}}$

Consider the set of all $n \times n$ matrices with real entries as the space $\mathbb{R^{n^2}}$ . Which of the following sets are compact?

1. The set of all orthogonal matrices.
2. The set of all matrices with determinant equal to unity.
3. The set of all invertible matrices.

I am stuck on this problem. Can anyone help me please...............

how can we check closedness and boundedness of matrices???

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Again, do you have any thoughts on the problem? – Michael Albanese Dec 27 '12 at 14:18
Do you have any examples you have gone through previously? Your question (and your previous question) isn't really about the sets you've listed, but the ability to check compactness of a given set using this definition; your post should reflect that fact. – Michael Albanese Dec 27 '12 at 14:23
You are right that a topological space is compact if every open cover has a finite subcover. When you are working in a Euclidean space $\mathbb{R}^N$, however, it is usually easier to use the condition that set is compact if and only if it is closed and bounded. Maybe you can try thinking about that.... – froggie Dec 27 '12 at 14:36
@ froggie sir, how can we check closed and boundedness of matrices???????? – user45799 Dec 27 '12 at 14:40
@Prasanta Any set in finite dimensional Euclidian space defined as the solution to a polynomial is closed (the polynomial defines a continuous map to $\Bbb R$, and the solution is the inverse image of the closed set $\{0\}$). Boundedness is checked with Pythagoras. Admittedly, this won't help you all the way, but it's enough to get you started. – Arthur Dec 27 '12 at 14:55

Let's think about $1)$. A matrix is orthogonal if it consists of rows and columns of orthogonal unit vectors (in this case, of $\mathbb{R}^n$). Let $A \in \mathbb{R}^{n^2}$ be such a matrix. Pick a row or column from $A$, say $(a_1, ..., a_n)$. By assumption, $|(a_1, ..., a_n)| = 1$, from which we have that each $|a_j| \le 1$. This establishes the boundedness.

Next, let $\{A_n\}_{n=1}^{\infty}$ be a sequence of convergent orthogonal matrices. We need to show that its limit, say $A$ is orthogonal. Use the continuity of the norm and inner product to conclude that $A$ is orthogonal.

For $2)$, I'll work on the case of $n =2$ and you may generalize. Fix any $n$ and consider the matrix $\left( \begin{array}{ccc} n & n-1 \\ 1 & 1 \\ \end{array} \right)$

This matrix has determinant $1$ but arbitrarily large $\mathbb{R}^{n^2}$ norm. Is the set of such matrices compact then?

For the third one, what do you know about invertible matrices? Apply part $2)$.

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option 2 should be bounded, because n can't be infinity......... but how can we check its closedness – user45799 Dec 27 '12 at 15:05
. please guide me... @ anonymous sir........ – user45799 Dec 27 '12 at 15:09
@Prasanta: This answer shows that option 2 is not bounded. – froggie Dec 27 '12 at 16:47
@Prasanta : I showed that option 2 is not bounded because we have matrices of determinant 1 and arbitrarily large norm. It should be quite easy to see that the matrix I provided has norm at least $n$. So there is no such positive number $R$ so that for any matrix $A$ of unit determinant we have that $|A| \le R$. This is what it means for the matrices to not be bounded. – anonymous Dec 27 '12 at 17:28

To check that a set of $n\times n$ matrices $\subset \mathbb{R}^{n^2}$ is compact, it is probably easiest to check that it is closed and bounded.

I'm going to give an example that is not any of the three you have in the problem, but will hopefully help you with them. Let $S$ be the collection of $n\times n$ symmetric matrices. We can then ask: is $S$ closed? Is $S$ bounded? Is $S$ compact?

I claim that $S$ is in fact closed. To see this, suppose that $M_n$ is a sequence of elements in $S$, i.e., a sequence of symmetric matrices, and suppose $M_n$ converges to a matrix $M$. To prove that $S$ is closed, we must show that $M$ is also a symmetric matrix. Write $M_n = (a_{ij}^{(n)})$, and $M = (a_{ij})$. The fact that $M_n\to M$ says exactly that $a_{ij}^{(n)}\to a_{ij}$ as $n\to \infty$ for each $i$ and $j$. But $M_n$ is symmetric, so $a_{ij}^{(n)} = a_{ji}^{(n)}$, so we get $$a_{ij} = \lim_n a_{ij}^{(n)} = \lim_n a_{ji}^{(n)} = a_{ji}.$$ This proves $M$ is symmetric, and that $S$ is closed.

I claim next that $S$ is not bounded. $S$ is bounded if and only if the entries of all the matrices in $S$ are bounded uniformly, so to prove my claim I need only find elements of $S$ with arbitrarily large entries. For each $c\in \mathbb{R}$, the matrix $M$ all of whose entries are $c$ lies in $S$, proving that the matrices in $S$ do not have uniformly bounded entries. This proves $S$ is not bounded.

Now is $S$ compact? $S$ is compact if and only if it is closed and bounded. Since it is not bounded, it is not compact. I hope this helps.

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option 2 should be bounded, because n can't be infinity......... but how can we check its closedness – – user45799 Dec 27 '12 at 15:12
@ froggie sir . please guide me. – user45799 Dec 27 '12 at 15:18