# Linear Algebra, null space

1. Mar 19, 2014

### mpittma1

1. The problem statement, all variables and given/known data

Construct a matrix whose null space consist of all linear combinations of:

v1 = (Column matrix) <1 -1 3 2>

v2 = (Column matrix) <2 0 -2 4>

2. Relevant equations

NS(A) = {x ε Rn I Ax =0}

w = k1v1 + k2v2

3. The attempt at a solution

I know that I'm looking for a 4 x 4 matrix and I'm only given 2 vectors to form "all linear combinations".

Do I need to first construct a matrix with 2 free variable columns showing the linear combination of the two given vectors, and then plug that into the null space equation and solve for the homogeneous system?

2. Mar 19, 2014

### mpittma1

I think I figured it out,

Let A = [a1a2a3a4]
[b1b2b3b4]

B = {(v1v2)}

then AB = [a1a2a3a4]
[b1b2b3b4]
* {(v1v2)} = [0 0]
[ 0 0]

Then I transpose BTAT

And the two vectors (a1a2a3a4)
(b1b2b3b4) have to be in the nullspace of BT

So all I have to do is find two vectors that span NS(BT)

So, BT = [1 -1 3 2] [1 0 1 -2]
[2 0 2-4] ≈ [0 1 -2 -4]

Then solving for my free variables,

x1 = 2x4 - x3
x2 = 2x3 + 4x4

2x4 - x3 -1 2
2x3 + 4x4 2 4
x = x3 = x3 1 + x4 0
x4 0 1

Hence,

∴ A = [-1 2 1 0]
[2 4 0 1] .

Any one care to confirm?

3. Mar 20, 2014

### Staff: Mentor

You have a mistake. For your matrix A, A * <1, -1, 3, 2>T = <0, 0>T, so that's fine, but A * <2, 0, -2, 4> = <-4, 8>T, which is not OK.

The simplest matrix that will work for you is a 1 X 4 matrix, like this:
$$A = \begin{bmatrix}a_1 & a_2 & a_3 & a_4 \end{bmatrix}$$

You could use it in this product:
$$\begin{bmatrix}a_1 & a_2 & a_3 & a_4 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & 0 \\ 3 & -2 \\ 2 & 4\end{bmatrix} = \begin{bmatrix} 0 & 0 \end{bmatrix}$$

This will give you two equations in four unknowns, which means that two of your variables are arbitrary (you can set them to whatever values you choose).