# Find a matrix with S as its nullspace

1. Mar 16, 2014

### pondzo

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

Let S be the subspace of R4 given by the solution set of the equations
-b + c + d = a - 3 c and -a - 2 d = d = a - c

Find an example of a matrix for which S is the nullspace.

2. Relevant equations

Ax=0

3. The attempt at a solution

I have found that the solution space and hence the null-space can be defined as d(-3,-12,-4,1)
so now all thats left is to find a matrix A. that satisfies Ax=0, right?
I entered one such matrix, namely A= [4,-1,1,4] and the program says its incorrect, where have i gone wrong?

thanks, Michael.

2. Mar 16, 2014

### HallsofIvy

To begin with, this is wrong. From -b+ c+ d= a- 3c we can get b= 4c+ d- a, From -a- 2d= d, a= 3d. From d= a- c, c= a- d= 2d. So b= 4(2d)+ d- 3d= 6d. (a, b, c, d)= (3d, 6d, 2d, d)= d(3, 6, 2, 1)

3. Mar 16, 2014

### pondzo

This is in fact wrong, as a=-3d not a=3d

4. Mar 16, 2014

### pondzo

so if that means my basis for the nullspace was correct. Then how would i go about finding a matrix that has that nullspace?

5. Mar 16, 2014

### pasmith

A linear map $A: \mathbb{R}^4 \to \mathbb{R}^4$ is determined by the images of the standard basis vectors $\{\mathbb{e}_i : i = 1, 2, 3, 4\}$.

Here you can choose the images of three of them as you see fit (provided the images are non-zero and linearly independent), but the fourth is then determined by the condition that
$$A(-3\mathbf{e}_1 -12\mathbf{e}_2 -4\mathbf{e}_3 + \mathbf{e}_4) = -3A(\mathbf{e}_1) - 12A(\mathbf{e}_2) - 4A(\mathbf{e}_3) + A(\mathbf{e}_4) = 0.$$

6. Mar 16, 2014

### pondzo

Thank you for your answer but it doesnt make much sense to me since we havent covered this sort of material in class yet (but they ask questions on it?).

This is how i got my answer (and the program said the final matrix, A, was correct), although i dont fully understand it.
$A \begin{pmatrix} 1 & 0 & 0 & -3\\ 0 & 1 & 0 & -12\\ 1 & 0 & 1 & -4\\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$

$A= \begin{pmatrix} 1 & 0 & 0 & 3\\ 0 & 1 & 0 & 12\\ 0 & 0 & 1 & 4\\ 0 & 0 & 0 & 0 \end{pmatrix}$

7. Mar 17, 2014

### Rellek

Hey!

If you wanted to find the nullspace, you might notice you have 3 equations and 4 variables. If you were to set up a coefficient matrix and find the specific vector(s) that send this system to the zero vector, you could use those same vectors to identify the entire nullspace.

8. Mar 18, 2014

### pondzo

my problem isn't finding a basis for the nullspace, that is fairly straightforward. My problem is finding/constructing a matrix that has a specific, defined nullspace.

9. Mar 18, 2014

### vela

Staff Emeritus
The nullspace you found is a subset of the nullspace of A (assuming it's correct), but the nullspace of A contains vectors that aren't of the form d(-3,-12,-4,1).

10. Mar 19, 2014

### pondzo

Thank you Vela.

If {(-3,-12,-4,-1)} isnt the basis for the nullspace of A then how do i find a full description of the nullspace? and if this is the case why did i get full marks when i entered:
$A= \begin{pmatrix} 1 & 0 & 0 & 3\\ 0 & 1 & 0 & 12\\ 0 & 0 & 1 & 4\\ 0 & 0 & 0 & 0 \end{pmatrix}$
because isnt the null space of this matrix of the form d(-3,-12,-4,1)?

11. Mar 19, 2014

### pasmith

I think vela is explaining why [4,-1,1,4] is not the right answer.

Yes: By construction that matrix has only a one-dimensional nullspace which is spanned by (-3, -12, -4 ,1).