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.


1 Answer 1


$\newcommand\si\sigma\newcommand\al\alpha\newcommand\be\beta$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 $\angle(u,v)=\pi/2$. $\quad\Box$


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