Show that $$T = \{ \mbox{tr} (A) : A \in O_n (\mathbb{R}) \}$$ is compact.
I tried to show this set is compact. I could not. Any hint would suffice. Thanks in advance.
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GA316
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1The trace is equal to the sum of eigenvalues. What do you know about the eigenvalues of orthogonal matrices? – Arthur Jan 08 '15 at 11:30
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thanks. unit modulus. How to use this here? – GA316 Jan 08 '15 at 11:32
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There are $n$ eigenvalues, and they all have unit modulus. The sum of them therefore has to be between...? – Arthur Jan 08 '15 at 11:37
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2The trace is continuous. The orthogonal group is compact. – user1551 Jan 08 '15 at 11:50
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@user1551 nice answer. can you please post this as answer, so that I can accept it? thanks. – GA316 Jan 08 '15 at 11:56
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@Arthur of modulus less than or equal to n? is this what you are saying or am i wrong? – GA316 Jan 08 '15 at 11:57
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The trace is a continuous function (this should be obvious). The orthogonal group is compact (one can easily check that it is closed and bounded). Since continuous functions map compact sets to compact sets, the result follows.
user1551
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Consider the inner product $\langle A,B \rangle=tr(AB^t)$ in $M_{n}(\mathbb{R})$. If $A$ is orthogonal then $AA^t=Id_{n}$ and $|tr(A)|=|tr(A Id_n)|\leq\sqrt{tr(AA^t)}\sqrt{tr(IdId^t)}=tr(Id)=n$.
Thus, your set $T$ is a subset of $[-n,n]$. Now, let's see tha $T=[-n,n]$.
If $n$ is even. Consider the following matrix in $O_n(\mathbb{R})$:
$A_n(\theta)= \begin{pmatrix}cos(\theta)&-sin(\theta)&0&0&\ldots&0&0\\ sin(\theta)&cos(\theta)&0&0&\ldots&0&0\\ 0&0&cos(\theta)&-sin(\theta)&\ldots&0&0\\ 0& 0& sin(\theta)&cos(\theta)&\ldots&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&0&\ldots&cos(\theta)&-sin(\theta)\\ 0& 0& 0& 0&\ldots&sin(\theta)&cos(\theta)\\ \end{pmatrix}$
Notice that $tr(A_{n}(\theta))=ncos(\theta)$. Since $-1\leq cos(\theta)\leq 1$ then $-n\leq tr(A_{n}(\theta))\leq n$.
If $n$ is odd. Consider the following matrix in $O_n(\mathbb{R})$: $B_n(\theta,\pm)=\begin{pmatrix}A_{n-1}(\theta)& 0_{n-1\times 1}\\ 0_{1\times n-1} & \pm 1\end{pmatrix}$.
Notice $-n+2=-(n-1)+1\leq tr(B_n(\theta,+))\leq (n-1)+1=n$ and $-n=-(n-1)-1\leq tr(B_n(\theta,-))\leq (n-1)-1=n-2$.
Thus, $T=[-n,n]$.
Daniel
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