Let the set of all $n \times n$ matrices (denoted by $M_n(\mathbb R)$ ) be a metric space. Show that set of all orthogonal matrices is compact.

My attempt:

well i am beginner in real analysis. compact means every open cover has a finite subcover. so here i have to find finite subcover for set of orthogonal matrices.showing closed and bounded may help i guess. but is there a easy way to do it?

  • $\begingroup$ Yes, you can use the fact that the determinant is a continuous function on $M_n(\Bbb R)$, the fact that $M_n(\Bbb R)$ is natually homeomorphic to $\Bbb R^{n^2}$, and therefore compactness is equivalent to "closed and bounded", and the fact that the determinant of orthogonal matrices is $\pm 1$. $\endgroup$ Sep 5, 2015 at 6:48
  • $\begingroup$ Possible duplicate of Compactness of the set of $n \times n$ orthogonal matrices $\endgroup$ Oct 26, 2016 at 10:26

1 Answer 1


You can think of elements of $O_n(\mathbb{R})$ as linear operators on $\mathbb{R}^n$ with the metric given by $$ d(A,B) = \sup_{1\leq i\leq n} \|A(e_i) - B(e_i)\| $$ where $\{e_1, e_2,\ldots, e_n\}$ denotes the standard basis for $\mathbb{R}^n$ and $\|\cdot\|$ denotes the usual Euclidean norm $$ \|x\| = \sqrt{\langle x,x\rangle} $$

Now recall that a matrix is orthogonal iff its column vectors form an orthonormal basis for $\mathbb{R}^n$. So $A \in O_n(\mathbb{R})$ if and only if $$ \langle A(e_i), A(e_j)\rangle = \delta_{i,j} \quad\forall 1\leq i,j\leq n $$ Hence,

  1. $d(A,0) \leq 1$ for all $A \in O_n(\mathbb{R})$

  2. If $A_k \to A$ with $A_k \in O_n(\mathbb{R})$ for all $k$, then $$ A_k(e_i) \to A(e_i) \quad\forall 1\leq i\leq n $$ Hence, $$ \langle A(e_i),A(e_j)\rangle = \delta_{i,j} $$ whence $A \in O_n(\mathbb{R})$.

Hence, $O_n(\mathbb{R})$ is closed and bounded (as a subset of $\mathbb{R}^{n^2}$), so it is compact.

  • $\begingroup$ is there a alternative simpler way? $\endgroup$
    – Foggy
    Sep 7, 2015 at 10:08

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