Linear Algebra, null space

In summary: You can choose a different pair of variables to be arbitrary, but I'll just choose x_3 and x_4 in this example. Solve the equations to see that you get these two solutions:$$\begin{align}a_1 &= 2x_4 - x_3 \\a_2 &= 2x_3 + 4x_4 \\a_3 &= x_3 \\a_4 &= x_4\end{align}$$This gives you the two vectors you need, in terms of x_3 and x_4:$$\begin{align}v_1 &= \begin{bmatrix} 2x_4 - x_3 \\ 2
  • #1
mpittma1
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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?
 
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  • #2
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
mpittma1 said:

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 said:
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?

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).
 
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1. What is a null space in linear algebra?

The null space of a matrix A is the set of vectors that, when multiplied by A, result in a zero vector. In other words, it is the set of all possible solutions to the equation Ax = 0, where x is a vector of appropriate dimensions.

2. How is the null space related to the rank of a matrix?

The dimension of the null space is equal to the number of columns in the matrix minus its rank. In other words, the null space contains all the vectors that are orthogonal to the row space, and therefore the dimension of the null space is equal to the number of linearly independent rows subtracted from the total number of columns.

3. Can a matrix have a null space of dimension greater than 1?

Yes, a matrix can have a null space of any dimension greater than or equal to 0. This means that there can be an infinite number of solutions to the equation Ax = 0, or that the null space can contain more than one vector.

4. How can the null space be used in practical applications?

The null space is often used in solving systems of linear equations. It can also be used in least squares regression and in solving optimization problems. Additionally, the null space can provide insight into the structure and properties of a matrix.

5. Is the null space unique for a given matrix?

No, the null space can have an infinite number of bases, meaning that there can be multiple sets of vectors that span the null space. However, the dimension of the null space remains the same regardless of the choice of basis.

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