Null Space of a Matrix

• Mar 13th 2011, 03:32 PM
Lord Darkin
Null 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

$Span((-1,1,0,0)^T,(-5,0,-3,1)^T)$

$(-1,1,0,0)^T, Span(-5,0,-3,1)^T$

Why can't I express the first as part of the linear combination (i.e. as part of the span)?
• Mar 13th 2011, 04:29 PM
poirot
Quote:

Originally Posted by Lord Darkin
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

$Span((-1,1,0,0)^T,(-5,0,-3,1)^T)$

$(-1,1,0,0)^T, Span(-5,0,-3,1)^T$

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

the answer is simply span( (-1,1,0,0)). Try inputting (-5,0,-3,1) and it comes out as (0,0,10), not (0,0,0)
• Mar 14th 2011, 03:39 PM
Lord 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, 11:06 AM
HallsofIvy
You get exactly the same result. That matrix row reduces to
$\begin{bmatrix}1 & 1 & -1 & 2 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 6\end{bmatrix}$
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.