# The set of traces of orthogonal matrices is compact

Is the following set compact: $$M = \{ \operatorname{Tr}(A) : A \in M(n,\mathbb R) \text{ is orthogonal}\}$$ where $\operatorname{Tr}(A)$ denotes the trace of $A$?

In order to be compact $M$ has to be closed and bounded.

$\|A\|=\sqrt {\sum_{i,j} {a_{ij}}^2}=\sqrt n$ and hence bounded. So $\operatorname{Tr}(A)<\sqrt n$. Hence $M$ is bounded.

Now we have to prove that $M$ is closed. Let $\operatorname{Tr}(A_n)$ be a sequence of matrices converging to $\operatorname{Tr}(A)$ where $A_n$ is a sequence of orthogonal matrices. The only thing remaining to show is that $A$ is orthogonal.

• Since you were given this question, I would imagine that you're allowed to use the result that the set of orthogonal matrices is compact. – Omnomnomnom Nov 18 '15 at 17:23
• $A$ is not a set, so $A$ is not compact. The set of orthogonal matrices is compact under the usual topology. – Thomas Andrews Nov 18 '15 at 17:34
• There's a flaw in your argument. You cannot assume that $A_n \to A$ such that $tr(A_n) \to tr(A)$. In general you have to show that for any sequence of matrices $(A_n)$ such that $\lim tr(A_n) = l$ for some number $l$, then $l = tr(A)$ for some orthogonal matrix $A$. There's a difference. – Najib Idrissi Nov 18 '15 at 19:51
• It's not true that $\text{Tr}(A) \le \sqrt{n}$. The correct bound is $n$. – Robert Israel Nov 18 '15 at 20:05

Besides using compactness of the set of orthogonal matrices, you can show directly that the set of possible traces is $[-n,n]$. Note that $I$ and $-I$ are orthogonal with $\text{Tr}(I) = n$ and $\text{Tr}(-I) = -n$. On the other hand, each matrix element of an orthogonal matrix has absolute value at most $1$, so $\text{Tr}(T) = \sum_{i=1}^n T_{ii}$ has absolute value at most $n$.

To get orthogonal matrices with every trace value from $-n$ to $n$, consider those made from diagonal blocks of the form $$\pmatrix{\cos \theta & \sin \theta\cr -\sin \theta & \cos\theta}$$ with an additional diagonal entry of $+1$ or $-1$ in case $n$ is odd.

• Sir ,I did not get your last paragraph "to get orthogonal..." for example how to get a $3\times 3$ orthogonal matrix with $\operatorname{Tr A}=0.6$ say – Learnmore Nov 19 '15 at 2:23
• it will be great if you could explain – Learnmore Nov 19 '15 at 2:23
• $\pmatrix{\cos \theta & \sin\theta & 0\cr -\sin\theta & \cos\theta & 0\cr 0 & 0 & 1\cr}$ where $\cos \theta = -0.2$. – Robert Israel Nov 19 '15 at 3:30

Hint: To show that $A$ is orthogonal, consider $\lim_{n \to \infty} \left\| A_n^TA_n - I \right\|$, noting that the function $$f(X) = \left\| X^TX - I \right\|$$ is continuous.

The orthogonal matrices are compact, as I show below. The trace function is continuous, so the image of the orthogonals under this function must be compact as well.

To see that the orthogonals are compact, first note that the condition $A^TA = I$ is a closed condition. It is the preimage of a single point (closed) under a continuous map, as long as you define the continuous map properly. Further, they are bounded, since their columns are all of norm one.

• Nitpick: having bounded eigenvalues is not sufficient to prove that the matrices themselves are bounded (e.g. transvections). – D. Thomine Nov 18 '15 at 17:33
• is the trace function continuous because every linear map on a finite dimensional space is continuous – Learnmore Nov 18 '15 at 17:33
• Or because the maps $f_{ij}$ which send $A\mapsto A_{ij}$ are continuous, and the linear combination of them is continuous. @learnmore – Thomas Andrews Nov 18 '15 at 17:36
• In fact, learnmore shows that the orthogonal matrices are bounded by noting that they all satisfy $\|A\| = \sqrt n$ – Omnomnomnom Nov 18 '15 at 17:43
• @Eric Auld they are bounded because every row/column vector has norm 1 – Learnmore Nov 18 '15 at 17:43