The angle between two subspaces of $\mathbb R^n$ can not be too small Notations.
Let $A$ and $B$ be two subspaces of $\mathbb R^{2k}$ of dimensions $k$. Let's call $p_A^\perp$ the orthogonal projection onto $A$.
Let's define two quantities:
$$\psi_1(A,B)=\inf_{b\in B\setminus\{0\}} \vert\sin\widehat{(b,p_A^\perp(b))}\vert$$
and
$$\psi_\infty(A,B)=\sup_{b\in B\setminus\{0\}} \vert\sin\widehat{(b,p_A^\perp(b))}\vert.$$
The question.

Assume that it exists a constant $c$ such that
  
  
*
  
*$\psi_\infty(A,B)\geqslant c$ ;
  
*There exist $b_1,\ldots,b_k\in B$ linearly independent such that $\widehat{(b_i,b_j)}\geqslant \pi/4$ if $i\ne j$, and 
$$\forall j\in\{1,\ldots,k\},\quad \vert\sin\widehat{(b_j,p_A^\perp(b_j))}\vert \geqslant c.$$
Let's also assume that $A\cap B=\{0\}$.
Can we prove that for all $c$ sufficiently small,
$$\psi_1(A,B)\geqslant c'$$
where $c'$ is a constant depending on $c$? 

Remarks.


*

*This question is kind of linked to this other question.

*Any hints or references would be much appreciated.
 A: I guess by an orthonormal projection you mean an orthogonal projection. Let $k=2$, $A=\Bbb R^2\times \{(0,0)\}$,
$b_1=(0,1,1,0)$, and $b_2=(1,1,-1,\varepsilon)$, where $\varepsilon>0$ tends to zero. It is easy to see that  $A\cap B=\{0\}$. According to [L, p.13], for any $b_j$, among all vectors in subspace $A$, the vector $p_A^\perp(b_j)$ closest to $b_j$ is the orthogonal projection of $b_j$ on $A$. It follows $p_A^\perp(b_1)=(0,1,0,0)$ and $p_A^\perp(b_2)=(1,1,0, 0)$. Thus
$$\vert\cos\widehat{(b_1, p_A^\perp(b_1))}\vert=\frac{|(b_1, p_A^\perp(b_1))| }{\|b_1\|\cdot\| p_A^\perp(b_1)\|}=
\frac {1}{\sqrt{2} },$$ 
$$\vert\cos\widehat{(b_2, p_A^\perp(b_2))}\vert=\frac{|(b_2, p_A^\perp(b_2))| }{\|b_2\|\cdot\| p_A^\perp(b_2)\|}=
\frac {2}{\sqrt{3+\varepsilon^2}\cdot \sqrt{2}},$$
which provides the respective inequalities for sinuses. 
Also $\widehat{(b_1,b_2)}=\pi/2$, because $$\vert\cos\widehat{(b_1,b_2,)}\vert=\frac {(b_1,b_2)}{|b_1||b_2|}=0,$$
But $b_1+b_2=(1,2,0,\varepsilon)$ tends to $p_A^\perp(b_1+b_2)=(1,2,0,0)$, so $ \vert\sin\widehat{(b_1+b_2,p_A^\perp(b_1+b_2))}\vert\ge \psi_1(A,B)$ tends to zero too.
References
[L] Hung-yi Lee, Orthogonal projection.
