4
$\begingroup$

Suppose $V$ is a $N\times n$ matrix the columns of which are independently distributed uniformly on $\mathbf S^{N-1}$ the surface of the unit sphere in $\mathbf R^N$. I conjecture that $V^TV$ approaches the identity matrix in norm (say, norms that are equivalent to the Frobenius norm), as $N\to\infty$ in expectation or even almost surely. I can prove it for $n=2$. I need one for arbitrarily given $n$. In general, what is the joint distribution of $V^TV$ for given finite $n$ and $N$?

Is the conjecture correct? Do we need the theory of random matrix to obtain the answers?

Improve this question
$\endgroup$
10
  • $\begingroup$ You know that you can represent them as $\frac Z {||Z||}$ where the Z are N dimensional normal with identity covariance & I think correlation is exactly same as for the normals $\endgroup$
    – user83457
    Commented Jan 27, 2017 at 15:29
  • $\begingroup$ @michael: I realized there was an ambiguity in what I meant by correlation matrix. I have edited the question. Please review it. However, your comment is still relevant and informative. Thank you. $\endgroup$
    – Hans
    Commented Jan 27, 2017 at 18:19
  • $\begingroup$ This is right, it follows from concentration of measure on the sphere (in fact, it's basically rephrasing this): a randomly chosen vector is almost certainly near the equator in high dimensions. $\endgroup$
    – Christian Remling
    Commented Jan 28, 2017 at 1:12
  • $\begingroup$ @ChristianRemling: I understand this is true for $n=2$. Would you please write the detail of the proof, perhaps with reference, as an answer? Thank you. $\endgroup$
    – Hans
    Commented Jan 28, 2017 at 2:08
  • $\begingroup$ @ChristianRemling is right. Hans, you want to think of this concentration happening not for fixed $n$, but rather, as $n$ tends to infinity, most of the surface area of a high dimensional sphere is by the equator. A quick proof of this is just to consider your favorite coordinate of a vector picked uniformly over sphere (this coordinate will be about $1/\sqrt{n}$). If you don't like working directly with your distribution, you could get a better idea via Gaussians as suggested [very nice distribution] or Fourier analysis (Talagrand's inequality may be relevant). $\endgroup$
    – Pat Devlin
    Commented Jan 28, 2017 at 2:51

1 Answer 1

Reset to default
2
$\begingroup$

The columns are distributed as $X^N/||X^N||$, where $X^N = (X_1^N,\dots,X^N_N)$ is the standard Gaussian vector in $\mathbb{R}^N$. The product of two different columns, $$ R_N:=\frac{\sum_{i=1}^N X_i^NY_i^N}{\left(\sum_{i=1}^N (X_i^N)^2\sum_{i=1}^N (X_i^N)^2\right)^{1/2}} =\frac{\frac1N\sum_{i=1}^N X_i^NY_i^N}{\left(\frac1N\sum_{i=1}^N (X_i^N)^2 \frac1N\sum_{i=1}^N (X_i^N)^2\right)^{1/2}}, $$ converges to $0$ almost surely$^*$ by SLLN. Alternatively, you can understand this product as an empirical correlation function of the two samples: $X$ and $Y$.

The distribution is generalized chi-square, if this helps.

$^*$ It is not very clear what almost surely means here, since the joint characteristics for matrices for different $N$ are not given. But I believe that this is true for any joint distribution, that is, for any $C>0$, $\sum_{N\ge 1}\mathbb{P}(|R_N|>C)<\infty$. This should be possible to extract from properties of Gaussian distribution or from concentration inequalities.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .