Hi,


We consider subspaces of $\mathbb{R}^N$. 
Suppose that we have a property called $\mbox{Prop}$ that apply to subspaces of $\mathbb{R}^N$. That is to say a function from the set of subspaces of $\mathbb{R}^N$ into $\{0,1\}$. 


The variables $X\_1,\dots,X\_N$ are gaussian taking their values in $\mathbb{R}^N$. They are supposed i.i.d. Deriving from an isotropic distribution (The covariance matrix is identity). 


Prove that ($0\leqslant n\leqslant N$)
$$
P(\mbox{Prop}(Span(X\_1,\dots,X\_n))=1)=
P(\mbox{Prop}(Span(X\_1,\dots,X\_{N-n})^\bot)=1)
$$
Without referring to the definition of $\mbox{Prop}$ other than the obvious that is 
$$
\mbox{Prop}(Span(X\_1,\dots,X\_k))
$$
is a measurable function on $\Omega$, the sample set.


In other words, the probability for some property to hold on a subspace spanned by $n$ variables chosen at random (gaussian isotropic) is equal to the probability for it to hold on the orthogonal space of the space spanned by $N-n$ variables.


This may translate to this "funny" question: $X\_1,\dots,X\_6$ are six real-valued gaussian iid variables. Prove that
$$
P\left(\sin\left(\frac{|X\_1|}{|X\_1|+|X\_2|+|X\_3|}\right)>0.2\right)=
P\left(\sin\left(\frac{|Y\_1|}{|Y\_1|+|Y\_2|+|Y\_3|}\right)>0.2\right)
$$


where $Y\_1=X\_2X\_6-X\_3X\_5$, $Y\_2=-X\_1X\_6+X\_3X\_4$ and $Y\_3=X\_1X\_5-X\_2X\_4$. (The vector $Y$ (in $\mathbb{R}^3$) is orthogonal to the subspace spanned by $(X\_1,X\_2,X\_3)$ and $(X\_4,X\_5,X\_6)$).


Thank you for your help,
 Saïd.



Equivalence between choosing a subspace and choosing its orthogonal
The Grassmannian $\mathrm{Gr}(n,N)$ is a compact manifold with a well defined canonical probability measure, its "uniform measure", characterized by the fact that it is invariant by the action of the orthogonal group $O(N)$. Starting from those $(X\_i)\_ i$, both $\mathrm{Span}(X\_1,\dots,X\_n)$ and $\mathrm{Span}(X\_1,\dots,X\_{N-n})^{\perp}$ define a uniformly distributed random element of $\mathrm{Gr}(n,N)$ (just because their probability law is $O(N)$-invariant).


