What does it mean "Distance between $k$-planes induced by the identification plane-projection matrix"? I'm reading some parts of Functions of bounded variation and free discontinuity problems by Ambrosio, Fusco, Pallara. 
At the very beginning of page 82 there's written "Let $G_k$ be the complete metric space of unoriented $k$-planes on $\mathbb R^n$, with the metric $d(\pi,\pi')=|\pi-\pi'|$ coming from the identification plane-projection matrix." 
So we want to define a distance between $k$-planes in $\mathbb R^n$, but I don't understand what "metric coming from plane-projection matrix" means. 
It is probable that this definition is written somewhere in the book and I've missed it, but I can't find it.
Could it be something like this: 
$$d(\pi,\pi')=\inf \{|x-\pi(x)|:x\in \pi'\}$$ (by $\pi(x)$ I mean the orthogonal projection of $x$ on $\pi$)?
 A: This is similar to the Grassmann metric, discussed here.  The idea is to define the distance between planes as the distance between the projection operators onto their shifted subspaces, using the operator norm.  Recall that a $k$-plane is just a shifted subspace, i.e. $\pi$ is of the form $W+a$ where $W$ is some $k$-dimensional subspace.  We then define
$$
d(\pi,\pi^\prime) = \|P_W-P_{W^\prime}\|
$$  Here $P_W,\;P_{W^\prime}$ are the projections onto to the $k$-dimensional subspaces $W=\pi-a$ and $W^\prime = \pi^\prime-a^\prime$, respectively.  This isn't particularly clear from their notation.
Here's a bad picture illustrating the idea: 

You project points onto the two (shifted) planes, then find the largest that this can be.  This picture isn't quite right because the planes should intersect the sphere, i.e. the projection points should be on the sphere.  You're esstially finding the Hausdorff distance of the intersection of these subspaces with the unit sphere -  see also this mathoverflow post.
