Angle between subspaces

In some of the subsequent discussions, it is convenient to discuss how a subspace may converge to another subspace.

Consider two subspaces $U$ and $V$. We can define two orthogonal projects $P_U$ and $P_V$ unto $U$ and $V$. Then the generalized angle or distance between these subspaces is defined by

$$\Vert P_U-P_V{\Vert }_2$$

We can make sense of this definition on a simple example. Consider $x$ and $y$ 2 unit vectors. What is the distance between span$\{x\}$ and span$\{y\}$?

Denote by $\theta \ge 0$ the angle between $x$ and $y$, and denote by $c=\cos (\theta )$, $s=\sin (\theta )$. Then, there exists a unit vector $z$ orthogonal to $x$ in span$\{x,y\}$ such that

$$y=cx+sz$$

Then

$$\begin{array}{r}x{x}^T-y{y}^T=\left(\begin{array}{cc}x& z\end{array}\right)\left(\begin{array}{cc}1-c^2& -cs\\ -cs& -s^2\end{array}\right)\left(\begin{array}{c}{x}^T\\ z^T\end{array}\right)\\ =s\left(\begin{array}{cc}x& z\end{array}\right)\left(\begin{array}{cc}s& -c\\ -c& -s\end{array}\right)\left(\begin{array}{c}{x}^T\\ z^T\end{array}\right)\end{array}$$

Since the matrix

$$\left(\begin{array}{cc}x& z\end{array}\right)$$

is orthogonal, we have that:

$$\Vert x{x}^T-y{y}^T{\Vert }_2=s\Vert \left(\begin{array}{cc}s& -c\\ -c& -s\end{array}\right){\Vert }_2=s$$

since

$$\left(\begin{array}{cc}s& -c\\ -c& -s\end{array}\right)$$

is orthogonal.

So the distance between these subspaces is equal to $s=\sin (\theta )$. That distance is 0 if the subspaces are equal. The maximum distance is 1.