Hello everyone.

The matrix is

1 1 -1 2

2 2 -3 1

-1 -1 0 5

I calculated the null space of the matrix as

Yet the answer is supposedly

Why can't I express the first as part of the linear combination (i.e. as part of the span)?

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- Mar 13th 2011, 02:32 PMLord DarkinNull Space of a Matrix
Hello everyone.

The matrix is

1 1 -1 2

2 2 -3 1

-1 -1 0 5

I calculated the null space of the matrix as

Yet the answer is supposedly

Why can't I express the first as part of the linear combination (i.e. as part of the span)? - Mar 13th 2011, 03:29 PMpoirot
- Mar 14th 2011, 02:39 PMLord Darkin
Shucks! I'm sorry about putting in wrong data, but that 5 in the (3,4) spot should be NEGATIVE five.

- Mar 15th 2011, 10:06 AMHallsofIvy
You get exactly the same result. That matrix row reduces to

so that (x, y, z, u) will be in the null space if and only if x+ y- z+ 2u= 0, z+ u= 3, 6u= 0. Those last two equations give z= u= 0 and the first equation reduces to x+ y= 0 or x= -y. Every vector in the null space is (x, -x, 0, 0)= x(1, -1, 0, 0). The null space is one-dimensional and {(1, -1, 0, 0)} is a basis ((-1, 1, 0, 0) is a multiple of that so it would also be a basis vector.