# Thread: Null Space of a Matrix

1. ## 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

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

Yet the answer is supposedly

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

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

2. 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

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

Yet the answer is supposedly

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

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)

3. Shucks! I'm sorry about putting in wrong data, but that 5 in the (3,4) spot should be NEGATIVE five.

4. You get exactly the same result. That matrix row reduces to
$\displaystyle \begin{bmatrix}1 & 1 & -1 & 2 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 6\end{bmatrix}$
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.