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The utility of the Matrix Inversion Lemma has been well-exploited for several questions on MO. Thus, with some positive hope, I'd like to field a question of my own.

Suppose we pick $n$ values $x_1,\ldots,x_n$, independently sampled from $N(0,1)$ (mean 0, unit variance gaussian). Then, we form the (rank 3 at best) positive semidefinite matrix: $$A = \alpha ee^T + [\cos(x_i-x_j)],$$ where $e$ denotes the vector of all ones, and $\alpha > 0$ is a fixed scalar.

For $n \ge 3$, simple experiments lead one to conjecture that: $$e^TA^\dagger e = \alpha^{-1},$$ where $A^\dagger$ is the Moore-Penrose pseudoinverse of $A$ (obtained in Matlab using the 'pinv' function).

This should be fairly easy to prove with the right tools, such as a Matrix inversion lemma that allows rank deficient matrices or pseudoinverses. So my question is:

How to prove the above conjecture (without too much labor, if possible)?

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If you just let $x_1,\dots,x_n$ be distinct scalars instead of specifying a particular probability distribution, shouldn't you get the same result? – Michael Hardy Aug 4 '11 at 4:13
Actually, I suspected it to be true as long as all the $x$'s were such that their contribution remains independent of $ee^T$ (Mikael makes this explicit in the answer below) – Suvrit Aug 4 '11 at 16:08
Why can the rank never exceed 3? – Michael Hardy Aug 5 '11 at 21:22
@Michael: expand $\cos(x-y)=\cos x\cos y + \sin x \sin y$ to notice that $A$ is a sum of three rank-1 matrices. – Suvrit Aug 5 '11 at 23:13
up vote 10 down vote accepted

In fact more generally for any positive semidefinite matrix $A = \sum_{i=0}^k e_i e_i^T$ with $e_i$'s linearly independent, we have that $e_i^T B e_i = 1$, where $B$ is the Moore-Penrose pseudoinverse of $A$. This applies here since almost surely your matrix $A$ is of this form with $k=3$ and $e_1 = \sqrt \alpha e$.

Proof: Let $E$ be the linear span of the $e_i$'s. If I understood correctly the notion of Moore-Penrose pseudoinverse, $B$ is described in the following way: as a linear map, $B$ is zero on the orthogonal of $E$, and on $E$ it is the inverse of the restriction of $A$ to $E$. Let $\beta_{i,j}$ be defined by $B e_i = \sum_j \beta_{i,j} e_j$, so that $e_i^T B e_i = \sum_j\beta_{i,j} e_i^T e_j$. Expressing that $A B e_i = e_i$, we get in particular that $\sum_j\beta_{i,j} e_i^T e_j = 1$, QED.

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Nice clean answer Mikael. It seems so easy once somebody proves it :-) thanks! – Suvrit Aug 4 '11 at 16:09

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