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Given a vector $y$ in a Hilbert space (possibly infinite-dimensional) and the subspace $\mathcal{T}$ spanned by a set of linearly dependent vectors $(x_1, \dots, x_p)$, I want to write the orthogonal (with respect to the inner product $\langle , \rangle$ of the Hilbert space) projection of $y$ on $\mathcal{T}$.

By generalizing the formula for linearly independent vectors (see, e.g., this question), I think it should be given by $$ \Pi ( y | \mathcal{T} ) = \sum_{ij} \langle y , x_i \rangle \left(G(x)^{-}\right)_{ij} x_j $$ where $G(x)_{ij} = \langle x_i , x_j \rangle$ is the singular Gram matrix of the vectors and $G(x)^{-}$ denotes its Moore-Penrose pseudoinverse.

Any comment or reference is greatly appreciated.

justmyfault
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  • An alternative construction: if $A$ is the map from $A:\Bbb C^p \to \mathcal H$ defined by $A(e_j) = x_j$ where $e_1,\dots,e_p$ denotes the canonical basis, then $AA^+$ will be the projection onto your subspace, where $A^+$ denotes the M-P pseudoinverse. – Ben Grossmann Jan 23 '20 at 08:41
  • Your construction seems reasonable but I can't quickly see if this will work. I'll think about it more if I find the time. – Ben Grossmann Jan 23 '20 at 08:43
  • Your result is correct iff $$ y \mapsto \left[\sum_{i} \langle y,x_i \rangle (G(x)^-){ij}\right]{j=1}^p $$ is the pseudoinverse of my map. It would suffice to check the four conditions given here, but there's almost certainly a cleaner approach. – Ben Grossmann Jan 23 '20 at 08:48
  • @Omnomnomnom thanks a lot for your comments! I am not super comfortable using the pseudo-inverse of an operator (not simply a matrix), but I'll look into that! My thinking was that all the geometry of the vectors is "encoded" in their Gram matrix, and since it's a finite-dimensional span it should not be very different from the finite-dimensional case. But I couldn't find this formula explicitly written in a book and haven't been able to prove it rigorously. – justmyfault Jan 23 '20 at 08:50
  • @Omnomnomnom OK I found the statement that $A^{+} = (A^* A)^{+} A^$ in Luenberger's book (p. 165 http://sites.science.oregonstate.edu/~show/old/142_Luenberger.pdf ). This is exactly what I need, since $A^ A$ is exactly the Gram matrix of the vectors (I might have inverted some index in my previous "attempt", because I was considering vectors on the real field and real valued inner products)... I'll try to write a tidy answer for future reference. – justmyfault Jan 23 '20 at 09:35
  • Another thought is that in order to show that your map is a projector, it suffices to check that it behaves correctly on all $x_j$ and on $\mathcal T^\perp$. – Ben Grossmann Jan 23 '20 at 09:50

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