I'm trying to demonstrate that the set of orthonormal matrices $\mathcal{O}(n,m) =\{ A | A^TA = I_m\} $ is closed. I am in a metric space $\mathbb{R}^{n*p}$ that uses the measure $d^2(X,Y) = trace((X-Y)^T(X-Y))$. I don't really understand how to interpret the notion of closed when it comes to this example since it's with matrices and not singular values, functions, or open/closed balls.


1 Answer 1


Let's denote $\mathcal M(n,m)$ the linear space of real matrices of dimension $ n \times m$. The map

$$\begin{array}{l|rcl} \phi : & \mathcal M(n,m) & \longrightarrow & \mathcal M(m,m) \\ & A & \longmapsto & A^tA \end{array}$$

is continuous as $A^tA$ entries are polynomial functions of $A$ entries.

The singleton $\{I_m\}$ is a closed subset of the normed vector space $(\mathcal M(m,m), \Vert \cdot \Vert_2)$ where $\Vert \cdot \Vert_2$ stands for the norm defined by

$$\Vert A \Vert_2^2 = \sum_{i,j} a_{ij}^2.$$ Therefore, $\mathcal O(n,m)$ which is the inverse image of $\{I_m\}$ under $\phi$ is closed for the topology induced by the distance $d$ as

$$d^2(X,Y) = \Vert X-Y \Vert_2^2.$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .