# Is there a simple way to find a matrix whose null space is the span of a given set of vectors?

The problem, and my solution is outlined below. I think my solution is correct, but I feel as if I went about my solution in an awkward way, and that there may be a better/cleaner way to solve the given problem. Would appreciate some feedback on what I did.

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Looks OK to me. The nullspace and the row space are orthogonal complements to each other, so the nullspace of a matrix whose row space is $U$ will be the row space of a matrix whose nullspace is $U$. – Gerry Myerson Sep 29 '13 at 1:04
Agreed, but is there any way to avoid the part where I generated a bunch of conditions on aij, and then manually used those conditions to construct E, where Ex=0? I thought that was a bit cumbersome, and am wondering if there is a better way to construct E? – dlaser Sep 29 '13 at 2:01
I've posted an answer which you may find to be a slicker way to construct $E$. – Gerry Myerson Sep 29 '13 at 5:27

I'll take up the story at the point where we have the matrix $$B=\pmatrix{1&0&0&-4&\phantom{-}3\cr0&1&0&-2&\phantom{-}4\cr0&0&1&\phantom{-}0&-1\cr}$$ and we want to find a basis for the nullspace. The non-pivot columns are the 4th and 5th. The nullspace will have a vector with 4th component 1, 5th component zero; by inspection, that vector is $(4,2,0,1,0)$. It will also have a vector with 4th component zero, 5th component 1; by inspection, that's $(-3,-4, 1,0,1)$. So we can take $E$ to be $$\pmatrix{\phantom{-}4&\phantom{-}2&0&1&0\cr-3&-4&1&0&1\cr}$$
Note that the rows of my $E$ add up to the 1st row of OP's $E$, while 3 times row 1 plus 4 times row 2 of my $E$ gives 4 times row two of OP's $E$, so both matrices have the same row space.