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Suppose I have X random variable have form $\langle x1,0,x2\rangle$ and Y random variable have form $\langle y1,y2,0\rangle$. These variables have 1 dimension in common. Is it possible to determine this fact when a rotation has been applied to the data?

To make this precise: let $X$ and $Y$ are independent random in $\mathbb{R}^d$ with $E[X]=E[Y]=0$, and $$\text{rank}E[X'X]=\text{rank}E[Y'Y]=r$$

Let A be $A$ a linear transformation $A$ such that $AX$ and $AY$ are non-zero in $r$ dimensions $x_1,\ldots,x_r$ and $y_1,\ldots,_r$ with $$x_1,\ldots,x_i=y_1,\ldots,y_i=1,\ldots,i$$

Let $A$ be chosen to maximize $i$, how do I compute $i$?

In my application $r\approx 100$, $d\approx 1000$, number of samples $\approx$ 1M

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  • $\begingroup$ "Have one dimension in common" appears to be a mischaracterization, because it implicitly relies on the primacy of a given basis $\{(1,0,0),(0,1,0),(0,0,1)\}.$ Upon rotation the meaning of that basis disappears. All you really have going on is that $X$ is confined to a 2D subspace and $Y$ is confined to a potentially different 2D subspace. It's a theorem of linear algebra that when both subspaces have full dimension, their intersection has dimension 1: your "one dimension in common." $\endgroup$ – whuber Sep 16 '19 at 11:35
  • $\begingroup$ @whuber Not sure if a concidence, but in simulations I'm able to recover d from trace(Z)/norm(Z) where Z is a solution to cov(X) Z + Z cov(X) = 2 cov(Y) -- wolframcloud.com/obj/yaroslavvb/newton/lyapunov.nb . IE, I sample random variables of form <x1,...,x{d}, x{d+1}, ..., x{rank-d},0, 0,..., 0> and <y1,...,y{d}, 0, ...., 0, y{d+1}, y{rank-d}> and try to get d $\endgroup$ – Yaroslav Bulatov Sep 16 '19 at 18:05
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    $\begingroup$ Linear algebra tells us the dimension of the intersection equals $$\operatorname{rank}(X^\prime X) + \operatorname{rank}(Y^\prime Y) - \operatorname{rank}((X+Y)^\prime (X+Y)).$$ $\endgroup$ – whuber Sep 16 '19 at 18:28

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