# Equivalence between choosing a subspace and choosing its orthogonal

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.

## 1 Answer

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).