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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?2012-07-19
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    @OldJohn I have written my thought2012-07-19
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    Ask yourself if $\mathbb{tr}$ is continuous, and if so, what does that mean if the real orthogonal matrices form a compact set. Also, why don't you consider, say, upper triangular matrices for Part 2.2012-07-19
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    Curious question: where do your questions come from? Is it a book that you are reading?2012-07-19
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    @MattN. question paper, not a book.2012-07-19
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    Oh, ok. Too bad, I'd've wanted to have a look at the book because the questions seem fun. : )2012-07-19
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    @copper.hat thank you for the hint, so $1$ is compact. still $2$ is not clear.2012-07-19
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    @MattN. attached a picture for you, have fun :) this is a part of Topology section :)2012-07-19
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    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 $n$?2012-07-19
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    thank you @copper.hat, I have already shown unboundedness2012-07-19
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    Well, if its unbounded, it can't be compact. But you haven't shown unboundedness above, only that $\|A\| \geq 2$.2012-07-19
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    Haha, great, thank you, these are awesome : )2012-07-19

2 Answers 2

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  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?2012-07-19
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    and where we are using the fact $\lambda \ge 2$?2012-07-19
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    @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?2012-07-19
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$3.6(b)$: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.