Which of the following are compact sets? Which of the following are compact sets?


*

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

*$\{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?
 A: *

*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. 

*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 $M\geq 0$). 
A: $3.6(b)$: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. 
