Pseudoinverse of columns of a matrix

First, some background:

I'm working on an implementation in C# of Lemke's algorithm (for solving linear complementarity problems) based on this Matlab implementation: http://ftp.cs.wisc.edu/math-prog/matlab/lemke.m

The implementation calculates $d = B / b_e$ in a tight inner loop. (Here $/$ is the Matlab "backslash" command, which calculates the Moore-Penrose pseudoinverse least squares "solution"). So as the implementation has it I'd have to decompose a matrix every iteration to compute the answer, which would be way to slow for practical use.

However, looking at it closer, the $B$ matrix is actually a permutation of a subset of the columns of a larger matrix which doesn't change from iteration to iteration, which has the form (in block notation) $[M, -I, c]$, where $M$ is a supplied square matrix, $I$ is the identity matrix of the same size as $M$, and $c$ is a calculable column vector. I'll call this larger matrix $B_s$, and $B_s P = B$, where $P$ is like a permutation matrix except that it actually excludes some columns (so that the result, $B$, is square).

As for $b_e$, it's actually just a column from $B_s$.

Which means that I can effectively pre-calculate all possible values for $d$ before I start the loop. If I let $B_s^{+}$ represent the Moore-Penrose Pseudoinverse of $B_s$ (likewise with $P$), I get:

$d = P^+ B_s^+ b_e$, where $b_e =$ column $e$ from $B_s$.

Now the question

Given a matrix $B_s$ and it's Moore-Penrose pseudoinverse $B_s^+$, is there any special significance to calculating $B_s^+b_e$ where $b_e$ is a column of $B_s$?

If $B_s$ were square and of full rank, the pseudoinverse would just be the normal matrix inverse. In which case:

1. $b_e = B_s e$ (where $e$ is a column vector with only one entry set to $1$ and all others set to $0$).
2. $B_s^{+} b_e = B_s^{-1} (B_s e) = (B_s^{-1} B_s) e = (I) e = e$

However, I don't think that's true for the general pseudoinverse of rectangular, possibly rank deficient, matrices.

I'm not aware of anything about $B^{+}\_{s}b_{c}$ that would be helpful here. It's not a column of the identity, because $B^{+}B$ isn't generally $I$. You can easily confirm this by looking at a random B matrix.
• Ah actually I answered my own question. I can't distribute the pseudoinverse from $B^+ = (B_s P)^+$ since $P P^T \not= I$. Back to the drawing board :/ – Jay Lemmon Jan 19 '11 at 19:02