random projections

Lucky Hans · **Joined:** Thu, 7 Jun 2012 11:13:33 UTC **Posts:** 4

Let P be a random orthogonal projection on R^d, chosen according to the "uniform distribution" on the set of rank k orthogonal projections. Are the squares of any two distinct entries on the diagonal uncorrelated?

(It probably doesn't matter but I assume that k is at least 2 and d at least 2k.)

Any ideas?

outermeasure · **Posted:** Fri, 29 Jun 2012 03:52:20 UTC

Lucky Hans wrote:

Let P be a random orthogonal projection on R^d, chosen according to the "uniform distribution" on the set of rank k orthogonal projections. Are the squares of any two distinct entries on the diagonal uncorrelated?

(It probably doesn't matter but I assume that k is at least 2 and d at least 2k.)

Any ideas?

Obviously $P_{i,i}=\langle Pe_i,e_i\rangle=\langle P^*Pe_i,e_i\rangle=\lVert Pe_i\rVert^2$ .

Let's do a simple calculation with d=4 and k=2, which has a reasonably nice coordinate system to work with.

The Haar measure on SO(4) is of course induced by $Spin(4)\cong Sp(1)\times Sp(1)$ . So upon identifying $\mathbb{R}^4\cong\mathbb{H}$ , we might as well take 1,i as the two orthonormal vectors, and have $\langle Pe_i,e_i\rangle\stackrel{\mathcal{D}}{\sim}\lvert P_2(Ue_iU')\rvert^2$ where U,U'

are random unit quaternions and P_2

is the projection to $\mathbb{C}\subset\mathbb{H}$ . Now, writing $U=z+\mathrm{j}w$ , $U'=z'+\mathrm{j}w'$ , we get
$\begin{aligned} \lvert P_2(UU')\rvert^2 &=\lvert zz'+\bar{w}w'\rvert^2\\ \lvert P_2(U\,\mathrm{i}U')\rvert^2 &=\lvert z\,\mathrm{i}z'+\overline{w\,\mathrm{i}}w'\rvert^2\\ &=\lvert zz'-\bar{w}w'\rvert^2 \end{aligned}$
Switching to (spherical) polars, (and for brevity $c_1=\cos\theta_1$ , $s_1=\sin\theta_1$ , etc.)
$\begin{aligned} z&=c_1+\mathrm{i}s_1c_2\\ w&=s_1s_2(c_3+\mathrm{i}s_3)\\ z'&=c_{1'}+\mathrm{i}s_{1'}c_{2'}\\ w'&=s_{1'}s_{2'}(c_{3'}+\mathrm{i}s_{3'}) \end{aligned}$
( $\theta_1,\theta_2,\theta'_1,\theta'_2\in(0,\pi)$ , $\theta_3,\theta'_3\in(0,2\pi)$ )
So
$\begin{aligned} P_{1,1}&=\lvert P_2(UU')\rvert^2\\ &=\lvert zz'+\bar{w}w'\rvert^2\\ &=\big\lvert (c_1+\mathrm{i}s_1c_2)(c_{1'}+\mathrm{i}s_{1'}c_{2'})+s_1s_2(c_3-\mathrm{i}s_3)s_{1'}s_{2'}(c_{3'}+\mathrm{i}s_{3'})\big\rvert^2\\ &=[c_1c_{1'}-s_1s_{1'}c_2c_{2'}+s_1s_2s_{1'}s_{2'}c_{3-3'}]^2+ [c_1s_{1'}c_{2'}+c_{1'}s_1c_2+s_1s_2s_{1'}s_{2'}s_{3-3'}]^2\\ P_{2,2}&=\lvert P_2(U\,\mathrm{i}U')\rvert^2\\ &=\lvert zz'-\bar{w}w'\rvert^2\\ &=\big\lvert (c_1+\mathrm{i}s_1c_2)(c_{1'}+\mathrm{i}s_{1'}c_{2'})-s_1s_2(c_3-\mathrm{i}s_3)s_{1'}s_{2'}(c_{3'}+\mathrm{i}s_{3'})\big\rvert^2\\ &=[c_1c_{1'}-s_1s_{1'}c_2c_{2'}-s_1s_2s_{1'}s_{2'}c_{3-3'}]^2+ [c_1s_{1'}c_{2'}+c_{1'}s_1c_2-s_1s_2s_{1'}s_{2'}s_{3-3'}]^2 \end{aligned}$
So now take expectations of $P_{1,1}^2, P_{1,1}^4, P_{1,1}^2P_{2,2}^2$ , and remember the (normalised) measure on S^3

is given by $\dfrac{1}{2\pi^2}s_1^2s_2\,\mathrm{d}\theta_1\,\mathrm{d}\theta_2\,\mathrm{d}\theta_3$ and you should see the covariance is nonzero. (You can cheat using Weyl's integration formula instead of passing to the explicit description of Spin(4).)

Lucky Hans · **Joined:** Thu, 7 Jun 2012 11:13:33 UTC **Posts:** 4

ok, thanks.

I have another adjacent question. I want to estimate E|p1-t p2| from below, where p1 is the squared norm of first column and p2 of the second column, respectively, and t between 0 and one. Taking the modulus hurts me. I need an estimate from below of the form c/d, where c is a positive constant that is independent on the ambient dimension d. (It may depend on the rank k of the orthogonal projections.) I am sure that it is correct. I am pretty sure that the term dE|p1-t p2| grows for growing d and fixed k..

Any ideas?

Shadow · **Posted:** Sat, 30 Jun 2012 10:01:19 UTC

Whoa, hold on dude. If you've been hurt by a modulus, consult your doctor first.

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random projections

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