# Eigenvectors of a Wishart matrix

I have been trying to find a good source (or clarifications) to help me understand this point. I am very new to random matrix theory so any pointers will be appreciated. Here is what I think I have understood so far

Consider N, P-dimensional random vectors drawn from a multivariate normal distribution and stacked into a matrix $X$ ($X$ will have dimension $N \times P$).The covariance matrix $S = \frac{1}{N}X^{*}X$ will have Wishart distribution with $\Sigma$, the $P \times P$ covariance matrix of the underlying multivariate normal referred to as the support. The question is how do the eigenvectors of this sample covariance matrix look?

Under certain conditions (which I am trying to understand as part of this question) ALL eigenvectors of the random sample covariance matrix are expected to be uniformly distributed on a unit sphere (see http://projecteuclid.org/download/pdfview_1/euclid.aop/1181334252 on pg 6/41 and in this thesis https://rucore.libraries.rutgers.edu/rutgers-lib/40727/ on pg 26)

(1) Could someone provide a reference to this result and

(2) is this true for any covariance matrix? Most studies are interested in only the diagonal support covariance matrix case and it makes me wonder if the result is applicable for more general supports.

(3) Even in the diagonal case, they take $\Sigma = \sigma^{2}I$. Will the result hold in the diagonal support case if the entries on the diagonal are not the same i.e. $\Sigma = diag[\sigma_{1},\sigma_{2}...]$?

• Hope the question makes more sense now. The way I am trying to understand it is the distribution (say) of the angle one of these sample eigenvectors makes from any given vector. I think the result says the distribution should be uniform. Is this the right way to think about it?
– Sid
Commented Mar 19, 2015 at 23:39
• I am not sure what this supposed to mean. If $n$ is very very large, then Wishart random matrices $S$ will be very close to $\Sigma$, right? So the eigenvectors will be very close to the eigenvectors of $\Sigma$ (which can be anything -- it's a parameter), just deviating slightly from them on the surface of the unit sphere. What can a statement about "uniform distribution" mean in this case? Commented Mar 19, 2015 at 23:54
• Honestly, I am not quite sure myself. Nobody seems to explicitly state the conditions. There are some instances where they mention this result with N fixed and some insist on (P/N) fixed. When N is fixed, if I chose N = $10^{9}$ we get back the case you are discussing. math.stackexchange.com/questions/656805/… is another instance where such a fact is stated in passing without mentioning the conditions...I know I am missing something, but I am not quite sure what
– Sid
Commented Mar 20, 2015 at 0:23
• Perhaps it refers to the $\Sigma=I$ case? Then variables are uncorrelated with unit variance, so there are no "preferred" directions in the $p$-dimensional space, and the eigenvectors of Wishart random matrices can only be uniformly random. Commented Mar 20, 2015 at 0:28

Assume $$\Sigma = I_p$$. The eigenvectors of $$X^TX$$ are the right singular vectors of $$X$$, e.g., $$v_1,...,v_r$$ with $$r=\min(n,p)$$ if the SVD of $$X$$ is $$X=\sum_{i=1}^r s_i u_i v_i^\top$$.
You wish to prove that $$v_i$$ is uniformly distributed on the sphere. The key is that for any deterministic or independent random rotation matrix $$Q\in O(p)$$, $$X$$ and $$XQ$$ are equal in distribution (this follows immediately for instance by observing that the rows of $$X$$ and $$XQ$$ are both iid $$N(0,I_p)$$).
This implies that for every $$i=1,...,r$$, the $$i$$-th singular vector $$v_i$$ is equal in distribution to the $$i$$-th singular vector of $$XQ$$ which is no other than $$Q^\top v_i$$ since the SVD of $$XQ$$ is $$XQ = \sum_i s_i u_i v_i^\top Q = \sum_i s_i u_i (Q^\top v_i)^\top.$$ Hence $$v_i$$ is equal in distribution to $$Q^\top v_i$$. Note that we made no assumption on $$Q$$ except that it belongs to $$O(p)$$ and is independent of $$X$$. In particular, we would like to choose a random $$Q\in O(p)$$ such $$Q^\top u$$ is uniformly distributed on the sphere for every deterministic $$u$$ with $$\|u\|=1$$; if such $$Q$$ exists, then in our SVD problem, $$Q^\top v_i$$ (which is equal in distribution to $$v_i$$) is thus uniformly distributed on the sphere.
Of course, the follow up question is how to construct such random matrix $$Q$$. This distribution is called the Haar measure of the orthogonal group $$O(p)$$.
If $$\Sigma\ne I_p,$$ this is not true. For instance, if $$\Sigma= a a^T$$ is rank one, the only right singular vector of $$X$$ is nonrandom and equal to $$a/\|a\|$$.