Find a matrix with S as its nullspace

[pondzo]

Homework Statement

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.

Homework Equations

Ax=0

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 that's 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.

 

[HallsofIvy]

Homework Statement

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.

Homework Equations

Ax=0

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)

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)

so now all that's 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.

 

[pondzo]

From -a- 2d= d, a= 3d

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

 

[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?

 

[pasmith]

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

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

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
[tex]
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.
[/tex]

 

[pondzo]

Thank you for your answer but it doesn't make much sense to me since we haven't 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 don't 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}
##

 

[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.

 

[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.

 

[vela]

Homework Statement

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.

Homework Equations

Ax=0

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 that's 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.

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).

 

[pondzo]

Thank you Vela.

If {(-3,-12,-4,-1)} isn't 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 isn't the null space of this matrix of the form d(-3,-12,-4,1)?

 

[pasmith]

Thank you Vela.

If {(-3,-12,-4,-1)} isn't the basis for the nullspace of A then how do i find a full description of the nullspace?

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

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 isn't the null space of this matrix of the form d(-3,-12,-4,1)?

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