# Dealing with connectness and compactness of matrices.

Consider the set of all $n\times n$ matrices with real entries, considered as the space $\mathbb{R^{n^2}}$ What can we say about connectedness and compactness of the following sets?

1. The set of all orthogonal matrices.
2. The set of all matrices with trace equal to unity.
3. The set of all symmetric and positive definite matrices.

I need help to understand basic concepts of solving these kind of problems. I mean how to show connectedness or compactness when we have set of matrices with certain specific properties. As mentioned above. Thanks

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general ideas: use the determinant, try to connect members of your set to the $E_n$ or another matrix, use similiarity etc and learn about lie groups, for compactness think closed and bounded –  Blah May 14 '12 at 22:13
you should have a look at your math classes, and if you don't have one articles on wikipedia to find all the way you can solve such a problem. For compactness, if I remember what I've learn 8 years ago, in a finite space ou can easily use the sequential definition to prove that it's closed and bounded. (for example with n=1, the set of all symmetric positive matrices is similar to R+ => not bounded and therefore not compact) –  Ricky Bobby May 14 '12 at 22:16
@RickyBobby I am familiar with definition like closed and boundedness for compactness in euclidean space. But i feel difficulty when i deal with matrices. I know i have to apply same definition but how? –  srijan May 14 '12 at 22:19
@srijan You should try to familiarize yourself with different common norms on matrices, and functions (like the determinant). Then you will have most of the tools to solve this kind of problems. –  Ricky Bobby May 14 '12 at 22:22
@RickyBobby Can you please provide me any link or source related with that? –  srijan May 14 '12 at 22:24

Some orthogonal matrices have determinant $+1$ and others $-1$. The determinant is a continuous function. Therefore the set of orthogonal matrices is not connected. It is compact, since it's closed and bounded in the set of all $n\times n$ matrices. It's closed because it's the inverse-image of a one-point set under the continuous function $M\mapsto MM^T$. For boundedness, consider the norm $\|M\|=\sup\{\|Mx\| : \|x\|=1, x\in\mathbb{R}^{n\times 1}\}$. With this norm, every orthogonal matrix has norm $1$. Show that the topology induced by that norm is the same as that induced by summing the squares of the entries, and you've got boundedness.
The diagonal matrix with entries $x,1-x,0,0,0,\ldots$ has trace $1$ and does not approach a limit as $x\to\infty$, so that set cannot be compact.
If $M$ is positive-definite, then so is $cM$ for positive scalars $c$, and as $c\to\infty$, there is no limit, so that set is not compact.
@srijan : The context in which it's being used is what you see above, plus the Heine-Borel theorem, and I'm just viewing it as an $n^2$-dimensional Euclidean space. The sense in which it's being used is the usual one for $n^2$-dimensional Euclidean spaces. –  Michael Hardy May 14 '12 at 22:34