The utility of the [Matrix Inversion Lemma](http://en.wikipedia.org/wiki/Woodbury_matrix_identity) 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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> 
> 



Matrix inversion lemma with pseudoinverses
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.


