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Given an ambient space $\mathbb{R}^d$ and two randomly oriented subspaces $A,B$ with dimensions $a,b$ respectively, how can I express the probability that $A$ and $B$ intersect non-trivially?

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This depends a great deal on how you are randomly selecting these subspaces, unless $a+b>d$, in which case the probability is 1. –  vadim123 Jun 24 '13 at 23:31
    
And if $a + b < d$ (for example, two lines in 3-space) this will be zero. The nontrivial case is $a+b = d$. –  Michael Lugo Jun 24 '13 at 23:36
    
That depends on the probability distribution. For example, let the first line be the $x$ axis, and the second be defined by $(\frac{n_1}{d_1},\frac{n_2}{d_2},\frac{n_3}{d_3})$, where the numerators and denominators are chosen by a Gaussian-like distribution on the integers (excluding denominator =0). This will have nonzero probability of coinciding with the first line. –  vadim123 Jun 25 '13 at 0:07
    
It does depend on the probability distribution (for instance if they are dependent as random variables and always guaranteed to be the same) but I think that example still has 0 probability. (It has nonzero probability density but that tells you nothing.) –  Sharkos Jun 25 '13 at 0:25

1 Answer 1

up vote 2 down vote accepted

Assuming a uniform independent distribution (say picking orthogonal vectors to span each space one at a time by uniform distributions on spheres) the probability is zero for $a+b\le d$ and one for $a+b>d$.

To see why, take $A$ to be fixed by rotation to some standard space. Then consider picking $b$ orthogonal vectors to span $B$. Note that a nontrivial intersection is equivalent to linear dependence amongst these two bases. But the probability the first introduces it is zero unless $A=\mathbb R^d$ because the angle a vector makes with a hyperplane is 0 with probability 0. Clearly the answer is one in the $a=d$ case.

But then instead of immediately picking the second vector orthogonal to the first, simply project onto the orthogonal complement. $A$ retains the same dimension, but now $b\to b-1,d\to d-1$. Thus we have the result above by induction.

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Even simpler, if $a+b\le d$, with probability 1 they are in general position in your stated distribution; hence their span is $\mathbb{R}^d$ so the two subspaces must have trivial intersection. –  vadim123 Jun 25 '13 at 0:58
    
Yes, though the point is to identify precisely what "general position" means in this context from first principles, surely? –  Sharkos Jun 25 '13 at 1:02

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