How Do You Construct a Matrix with a Specific Null Space?

[mpittma1]

Homework Statement

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>

Homework Equations

NS(A) = {x ε Rⁿ I Ax =0}

w = k₁v₁ + k₂v₂

The Attempt at a Solution

I'm not sure where to start with this problem.

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?

 

[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 B^TA^T

And the two vectors (a₁a₂a₃a₄)
(b₁b₂b₃b₄) have to be in the nullspace of B^T

So all I have to do is find two vectors that span NS(B^T)

So, B^T = [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?

 

[Mark44]

Homework Statement

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>

Homework Equations

NS(A) = {x ε Rⁿ I Ax =0}

w = k₁v₁ + k₂v₂

The Attempt at a Solution

I'm not sure where to start with this problem.

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?

I think I figured it out,

Let A = [a₁a₂a₃a₄]
[b₁b₂b₃b₄]

B = {(v₁v₂)}

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

Then I transpose B^TA^T

And the two vectors (a₁a₂a₃a₄)
(b₁b₂b₃b₄) have to be in the nullspace of B^T

So all I have to do is find two vectors that span NS(B^T)

So, B^T = [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?

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).