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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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    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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    @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.