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enter image description hereWhich of the following are compact sets?

  1. $\{\operatorname{trace}(A): A \text{ is real orthogonal}\}$

  2. $\{A\in M_n(\mathbb{R}):\text{ eigenvalues $|\lambda|\le 2$}\}$

Well, orthogonal matrices are compact, but the trace of them may be any $x\in\mathbb{R}$, so I guess 1 is non compact. Let $x$ be an eigenvector corresponding to the eigenvalue $\lambda$; then $Ax=\lambda x$, then $\|Ax\|= |\lambda|\cdot\|x\|\le \|A\|\cdot\|x\|$ so $\|A\|\ge 2$ so $2$ is also non compact as unbounded?

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What are your thoughts on this so far? – Old John Jul 19 '12 at 17:46
Curious question: where do your questions come from? Is it a book that you are reading? – Rudy the Reindeer Jul 19 '12 at 17:57
Oh, ok. Too bad, I'd've wanted to have a look at the book because the questions seem fun. : ) – Rudy the Reindeer Jul 19 '12 at 18:02
@MattN. attached a picture for you, have fun :) this is a part of Topology section :) – Un Chien Andalou Jul 19 '12 at 18:10
Take a $2\times 2$ upper triangular matrix with 1's on the diagonal and $n$ in the upper triangular part. Is it bounded for any $n$? – copper.hat Jul 19 '12 at 18:12
up vote 2 down vote accepted
  1. The map $$\operatorname{trace}\colon\mathcal M_n(\Bbb R)\to \Bbb R$$ is linear, and from a finite dimensional vector space, hence continuous. Such mapping map compact sets to compact one, and the orthogonal group is compact, hence the first set is compact.

  2. The second set is not bounded. The matrices $A_N:=\pmatrix{0&0&\dots&0&N\\ 0&0&\dots&0&0\\ \vdots&\vdots&&\vdots&\vdots\\ 0&0&\dots&0&0}$ is in the second set, but the norm is $N$ for the norm subordinated to the supremum norm for example, is $N$. The only eigenvalue of $A_N$ is $0$ and $\{A_N,N\geq 1\}$ is not bounded hence cannot be compact. Note that we can take any norm we want, since $\mathcal M_n(\Bbb R)$ is finite-dimensional, and the choice of $2$ in the text of the exercise is not important (we can replace it by $M\geq 0$).

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would you tell me what is the matrix norm here? – Un Chien Andalou Jul 19 '12 at 19:17
and where we are using the fact $\lambda \ge 2$? – Un Chien Andalou Jul 19 '12 at 19:19
@Patience : use whatever norm you like. The norm of Davide's matrix will be $k|N|$ for some nonzero $k$. What are the eigenvalues of that matrix? – Robert Israel Jul 19 '12 at 19:52

$3.6(b)$:Since $A$ orthogonal $det(A)$=$\pm1$. So set of $3.6(b)$ is subset of $[-n,n]$. Because $\lambda_1.\lambda_2.....\lambda_n=\pm1$ then $tr(A)=\lambda_1+\lambda_2+.......+\lambda_n$ is clearly closed subset of $[-n,n]$ which is compact.

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