Hello,
Let $X'X$ be a positive definite matrix and let $\mathbf{1}$ denote the vector of ones.
I'm hoping to construct a positive, diagonal matrix $W$ such that
$$(W X'X W) \mathbf{1} = \mathbf{1}$$
$X$ and $W$ are all assumed to have real-valued entries, and $X'$ denotes the transpose of $X$.
I don't, yet, have a proof that such a matrix $W$ always exists, but strongly suspect it. Any ideas on algorithms, proofs, or counter-examples would be gratefully received.
The problem arises from work in statistics.
thanks,
David.
Best Answer
Consider the simplex of nonzero diagonal matrices W with nonnegative entries up to scaling, and the simplex of nonzero vectors V with nonnegative entries up to scaling.
There is a map, $V=\max(WX′XW\mathbf 1,0)$, from the first simplex to the second, with $\max(a,0)$ interpreted entrywise. This is well-defined because $WX'XW\mathbf 1$ always has some positive entry, because the sum of its entries is $1'W X' X W1$, with $W1$ a nonzero vector and $X'X$ positive-definite.
This map clearly sends each k-cell of the first simplex into the corresponding k-cell of the second simplex, since if some of the coordinates of $W$ are $0$ then some of the coordinates of $V$ are $0$.
Every such map on simplices must be surjective. This is because the map from the boundary sphere of one simplex to the boundary sphere of the other is degree one, because every such map on simplices has a boundary-preserving homotopy to the standard isomorphism between those simplices, by induction.
So there is some $W$ such that $\max(WX′XW\mathbf 1,0)=\mathbf 1$. So $WX'XW\mathbf 1=\mathbf 1$.