Compact Sets in $\mathbb{R^{n^2}}$

Question

I have a question of multivariable analysis and I don't know how to resolve this.

The $n \times n$ orthogonal matrices form a compact subset of $\mathbb{R^{n^2}}$?

I will be very grateful for the help.

Answer 1

The map $$ F: \mathbb{R}^{n\times n} \to \mathbb{R}^{n\times n},\, F(X)=XX^T $$ is continuous (it is actually differentiable), and $$ O(n)=\{A\in \mathbb{R}^{n\times n}:\, AA^\top=I_n\}=F^{-1}(I_n). $$ Since $\{I_n\}$ is closed and bounded in $\mathbb{R}^{n\times n}$, and $F$ is continuous therefore $O(n)$ is closed and bounded because each entry $a_{ij}$ of $A\in O(n)$ is. In fact, for every $i=1,2,\ldots,n$ we have $$ \delta_{ij}=(AA^\top)_{ij}=\sum_{k=1}^na_{ik}a_{jk} \Longrightarrow \sum_{j=1}^na_{ij}^2=1 \Longrightarrow |a_{ij}|\le 1 \quad \forall i,j=1,2,\ldots,n. $$ Hence $O(n)$ is compact.

Answer 2

Basic idea: Suppose you have, for each $m\in \mathbb {N},$ an orthonomal basis $\{u_1(m), \dots , u_n(m)\}$ of $\mathbb {R}^n.$ Suppose you also know that for each fixed $k\in \{1,\dots ,n\},$ $u_k(m) \to v_k$ in $\mathbb {R}^n$ as $m\to \infty.$ Isn't it likely that $\{v_1,\dots , v_n\}$ is also orthonormal in $\mathbb {R}^n$ ?

Answer 3

A matrix $Q$ is orthogonal if its columns and rows are orthogonal unit vectors. A subset of $\mathbb{R}^{n^2}$ is compact if it is closed and bounded.

Bounded: If $Q$ is orthogonal, its columns are unit vectors, so they all have norm $1$. What do you know about the norm of a matrix, with respect to the norms of its columns? Hint: All norms on $\mathbb{R}^{n^2}$ are equivalent, and $||\cdot||_1$ is the norm that takes the maximum column sum of the input matrix.

Closed: Let $Q_k\rightarrow Q$ be a convergent sequence of orthogonal matrices, and let $q_k^j$ denote its columns, $j=1,\dots,n$. Since each $Q_k$ is orthogonal, $(q_k^i,q_k^j)=0$ for $i\neq j$, and $(q_k^i,q_k^i)=1$, for all $k$. What does that tell you about the limit, $Q$? The same type of argument applies to the rows. Hint: $(q_k^i,q_k^j)\rightarrow (q^i,q^j)$ for each pair $(i,j)\in\{1,\dots,n\}^2$..