1
$\begingroup$

Let $L_1$, $L_2$ be two subspaces in $\mathbb{R}^n$ and $\dim(L_1) = \dim(L_2) = s<n$. Let $$0 \leq \theta_1 \leq \dotsb\leq \theta_{s} \leq \pi/2$$ be the principal angles between $L_1, L_2$. Let there be two unit vectors $u_1\in L_1$ and $u_2 \in L_2$ and the principal angle between $\operatorname{span}\{u_1\}$ and $\operatorname{span}\{u_2\}$ is $\alpha \in [0,\pi/2]$. Is there any known upper bound on $\alpha$ in terms of $\theta_{s}$?

Here is my try: Let $U_1, U_2 \in \mathbb{R}^{n \times s}$ be two orthogonal matrices such that their columns are orthonormal bases of $L_1$ and $L_2$ respectively. Now any unit vector in $L_j$ is essentially of the form $U_j \alpha_j$ where $\lVert \alpha_j\rVert_2 =1 $ for $j \in \{1,2\}$. Also, the singular values of $U_1^\top U_2$ are $\{\cos \theta_1, \dotsc, \cos \theta_s\}$. Now, I don't have any idea how to proceed any further. Any help will be appreciated.

$\endgroup$

1 Answer 1

2
$\begingroup$

$\newcommand\si\sigma\newcommand\al\alpha\newcommand\be\beta$Let $U:=L_1$ and $V:=L_2$. Without loss of generality, $s=\dim U=\dim V\ge1$. By the section Angles between subspaces of the Wikipedia article, there are orthonormal bases $(a_1,\dots,a_s)$ and $(b_1,\dots,b_s)$ of $U$ and $V$, respectively; nonnegative integers $\si$ and $\al$; and $\be_1,\dots,\be_\al$ in the interval $(0,\pi/2)$ such that for any $u\in U$ and $v\in V$ we have $$u\cdot v=\sum_{i=1}^\si u^i v^i+\sum_{i=\si+1}^{\si+\al} u^i v^i\cos\be_i,$$ where $u\cdot v$ is the dot product of $u$ and $v$; the $u^i$'s are the coordinates of $u$ in the basis $(a_1,\dots,a_s)$; and the $v^i$'s are the coordinates of $v$ in the basis $(b_1,\dots,b_s)$.

So, if $\si+\al<s$ or $\si+\al>1$, then $u\cdot v=0$ and hence $\angle(u,v)=\pi/2$ for some nonzero $u\in U$ and $v\in V$.

Otherwise, if $\si+\al\ge s$ and $\si+\al\le1$, then $s=1$.

Thus, excluding the trivial case $s=1$, the maximal angle between $\text{span}\{u\}$ and $\text{span}\{v\}$ for nonzero vectors $u\in U$ and $v\in V$ is always $\pi/2$.


This conclusion can also be obtained more elementarily, as follows. Suppose that $s\ge2$ and let $(a_1,\dots,a_s)$ be any basis of $U$. Take any nonzero $v\in V$. Then for for some nonzero $(u^1,\dots,u^s)\in\mathbb R^s$ and $u:=\sum_{i=1}^s u^i a_i$ we have $u\ne0$ and $u\cdot v=\sum_{i=1}^s u^i(a_i\cdot v)=0$ and hence $\angle(u,v)=\pi/2$. $\quad\Box$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.