Understanding the Nullspace of a Matrix for Subspace U in R4

[negation]

Homework Statement

Let U be the subspace of R4 given by:
U = the nullspace of the matrix

[0 0 2 4
0 3 -4 2]

The Attempt at a Solution

let v = (v1,...v4) and w = (w1...w4)
(0,0,2,4) = (λ1v1 + λ2v2 + λ3v3 + λ4v4)
(0,3,-4,2) = (λ1w1 + λ2w2 + λ3w3 + λ4w4)



I haven't come across such problem before nor were taught what a nullspace of a matrix is.
Could someone provide some explanation as to the question?

 

[HallsofIvy]

Surely, you could look up "nullspace" in the index of your textbook? The "nullspace" of a matrix is the set of all vectors that the matrix maps into the 0 vector (i.e. makes "null"). It is easy to show that the set of all such vectors is a subspace (if Au= 0 and Av= 0 then $A(\alpha u+ \beta v)= \alpha Au+ \beta Av= 0$).

So here you are asked to find all $\begin{bmatrix}x \\ y \\ z \\ u\end{bmatrix}$ such that
$$\begin{bmatrix}0 & 0 & 2 & 4 \\ 0 & 3 & -4 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\\ u\end{bmatrix}= \begin{bmatrix}2z+ 4u \\ 3y- 4z+ 2u\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$$

Solve the equations 2z+ 4u= 0 and 3y- 4z+ 2u= 0. (There is not a unique answer.)

 

[negation]

Surely, you could look up "nullspace" in the index of your textbook? The "nullspace" of a matrix is the set of all vectors that the matrix maps into the 0 vector (i.e. makes "null"). It is easy to show that the set of all such vectors is a subspace (if Au= 0 and Av= 0 then $A(\alpha u+ \beta v)= \alpha Au+ \beta Av= 0$).

So here you are asked to find all $\begin{bmatrix}x \\ y \\ z \\ u\end{bmatrix}$ such that
$$\begin{bmatrix}0 & 0 & 2 & 4 \\ 0 & 3 & -4 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\\ u\end{bmatrix}= \begin{bmatrix}2z+ 4u \\ 3y- 4z+ 2u\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$$

Solve the equations 2z+ 4u= 0 and 3y- 4z+ 2u= 0. (There is not a unique answer.)

It's not in the content of the book. Search engine showed rather dissappointing results too.

 

[negation]

Surely, you could look up "nullspace" in the index of your textbook? The "nullspace" of a matrix is the set of all vectors that the matrix maps into the 0 vector (i.e. makes "null"). It is easy to show that the set of all such vectors is a subspace (if Au= 0 and Av= 0 then $A(\alpha u+ \beta v)= \alpha Au+ \beta Av= 0$).

So here you are asked to find all $\begin{bmatrix}x \\ y \\ z \\ u\end{bmatrix}$ such that
$$\begin{bmatrix}0 & 0 & 2 & 4 \\ 0 & 3 & -4 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\\ u\end{bmatrix}= \begin{bmatrix}2z+ 4u \\ 3y- 4z+ 2u\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$$

Solve the equations 2z+ 4u= 0 and 3y- 4z+ 2u= 0. (There is not a unique answer.)

0x + 0y +2z +4u = 0
0x + 3y -4z +2u = 0

0 0 2 4 | 0
0 3 -4 2 | 0

R1<->R2

0 3 -4 2 | 0
0 0 2 4 | 0

z and u are free variables. There exists infinite solutions to the SOLE.
(x,y,z,u) = (x,4/3z-2/3u,-2u,u)

The questions askes for a spanning set for U matrix. So if the null matrix (0,0,0,0) spans(x,y,z,u), then, (0,0,0,0) is a linear combination of (x,y,z,u).

Edit: λ1(x)+λ2(4/3z-2/3u) +λ3(-2u) + λ4(u) = (0,0,0,0)

 

[Mark44]

0x + 0y +2z +4u = 0
0x + 3y -4z +2u = 0

0 0 2 4 | 0
0 3 -4 2 | 0

R1<->R2

0 3 -4 2 | 0
0 0 2 4 | 0

The matrix above is not completely reduced. In fact, all you did was switch the two rows.

z and u are free variables. There exists infinite solutions to the SOLE.
(x,y,z,u) = (x,4/3z-2/3u,-2u,u)

The questions askes for a spanning set for U matrix. So if the null matrix (0,0,0,0) spans(x,y,z,u), then, (0,0,0,0) is a linear combination of (x,y,z,u).

(0, 0, 0, 0) is the zero vector in R⁴. It does NOT span anything other than itself. In your sentence that starts "So if the null matrix ...", there is not one thing that is true.

Edit: λ1(x)+λ2(4/3z-2/3u) +λ3(-2u) + λ4(u) = (0,0,0,0)

You need to pay more attention to definitions, especially those for "nullspace" and "span". Look in the index (at the back of the book) to see where these terms appear in the book. If a term is used on more than one page, it usually is defined on the first page of the list. The table of contents at the front of the book is probably less helpful, as it has a list of the chapters without much detail on what's in the chapter.

 

[negation]

The matrix above is not completely reduced. In fact, all you did was switch the two rows.
(0, 0, 0, 0) is the zero vector in R⁴. It does NOT span anything other than itself. In your sentence that starts "So if the null matrix ...", there is not one thing that is true. You need to pay more attention to definitions, especially those for "nullspace" and "span". Look in the index (at the back of the book) to see where these terms appear in the book. If a term is used on more than one page, it usually is defined on the first page of the list. The table of contents at the front of the book is probably less helpful, as it has a list of the chapters without much detail on what's in the chapter.

0 0 2 4 | 0
0 3 -4 2 | 0

R1<->R2

0 3 -4 2 | 0
0 0 2 4 | 0

RREF...

0 3 0 -6| 0
0 0 2 4 | 0

Leading variables: y, z
Non-leading: x,u

2z + 4u = 0
z = -2u
3y -6u = 0
y = 2u

(x,y,z,u) = (x,2u,-2u,u)

 

[Mark44]

0 0 2 4 | 0
0 3 -4 2 | 0

R1<->R2

0 3 -4 2 | 0
0 0 2 4 | 0

RREF...

0 3 0 -6| 0
0 0 2 4 | 0

The 4th number in the top row is wrong.

Leading variables: y, z
Non-leading: x,u

2z + 4u = 0
z = -2u
3y -6u = 0
y = 2u

(x,y,z,u) = (x,2u,-2u,u)

You should check your work.
This matrix product ...
$$ \begin{bmatrix}0 & 0 & 2 & 4 \\ 0 & 3 & -4 & 2 \end{bmatrix} \begin{bmatrix}x \\ 2u \\ -2u \\ u \end{bmatrix}$$
should result in the column vector <0, 0>. Does it?

 

[negation]

The 4th number in the top row is wrong.


You should check your work.
This matrix product ...
$$ \begin{bmatrix}0 & 0 & 2 & 4 \\ 0 & 3 & -4 & 2 \end{bmatrix} \begin{bmatrix}x \\ 2u \\ -2u \\ u \end{bmatrix}$$
should result in the column vector <0, 0>. Does it?

0, 3, 0, 10 | 0
0, 0, 2, 4| 0

Leading variables: y, z
Non-leading variables: x, u

2z + 4u = 0
2z = -4u
z = -2u

3y + 10u = 0
y = -10u/3

(x,y,z,u) = (x,-10u/3,-2u,u)

[0, 0, 2, 4; 0, 0, 2, 4] [x; -10u/3; -2u;u] = [0;0]

 

[Mark44]

It's easy to make mistakes when you're reducing a matrix, so it's important to check your work (as I suggested in my previous post). Also, it's helpful to indicate in your work what you're doing. Off to the left of my matrices, if I'm adding -3 times row 2 to row 1 (for example), I put -3 to the left of row 2, and 1 to the left of row 1, with an arrow going from row 2 to row 1.

That way, if my answer doesn't check out, I can go back and look at my individual steps. Having an audit trail, so to speak, makes it easier to quickly check my row-reduction work.

 

[Mark44]

0, 3, 0, 10 | 0
0, 0, 2, 4| 0

Leading variables: y, z
Non-leading variables: x, u

2z + 4u = 0
2z = -4u
z = -2u

3y + 10u = 0
y = -10u/3

(x,y,z,u) = (x,-10u/3,-2u,u)

[0, 0, 2, 4; 0, 0, 2, 4] [x; -10u/3; -2u;u] = [0;0]

That's more like it! Congrats!

Your solution could be written as <x, 0, 0, 0> + <0, -(10/3)u, -2u, u>, or better yet, as x<1, 0, 0, 0> + u<0, -10/3, -2, 1>, where x and u are arbitrary real numbers.

Since there are two vectors, the nullspace of your matrix is a two-dimensional subspace of R⁴. In other words, it is a plane in four dimensions. The vectors {<1, 0, 0, 0>, <0, -10/3, -2, 1>} span the nullspace, and since they are linearly independent (which I can tell by inspection), they form a basis for the nullspace.

 

[negation]

It's easy to make mistakes when you're reducing a matrix, so it's important to check your work (as I suggested in my previous post). Also, it's helpful to indicate in your work what you're doing. Off to the left of my matrices, if I'm adding -3 times row 2 to row 1 (for example), I put -3 to the left of row 2, and 1 to the left of row 1, with an arrow going from row 2 to row 1.

That way, if my answer doesn't check out, I can go back and look at my individual steps. Having an audit trail, so to speak, makes it easier to quickly check my row-reduction work.

Just to clarify, and ensure I don't get lost in the intermediate step (I like looking at big picture).
Is the correct question I ought to be asking myself at the onset is:
Given [0, 0, 2, 4; 0, 0, 2, 4], what values must the vector[x;y;z;u] be such that
[0, 0, 2, 4; 0, 0, 2, 4][x;y;z;u] = [0;0] ?
The way I see it, [x;y;z;u] is really just the scalar in the linear combination.

 

[negation]

That's more like it! Congrats!

Your solution could be written as <x, 0, 0, 0> + <0, -(10/3)u, -2u, u>, or better yet, as x<1, 0, 0, 0> + u<0, -10/3, -2, 1>, where x and u are arbitrary real numbers.

Since there are two vectors, the nullspace of your matrix is a two-dimensional subspace of R⁴. In other words, it is a plane in four dimensions. The vectors {<1, 0, 0, 0>, <0, -10/3, -2, 1>} span the nullspace, and since they are linearly independent (which I can tell by inspection), they form a basis for the nullspace.

Strange, the system rejected my answer.
It asked what the spanning set of U is.
(x,-10u/3,-2u,u) is apparently not the right one...

 

[Mark44]

Just to clarify, and ensure I don't get lost in the intermediate step (I like looking at big picture).
Is the correct question I ought to be asking myself at the onset is:
Given [0, 0, 2, 4; 0, 0, 2, 4], what values must the vector[x;y;z;u] be such that
[0, 0, 2, 4; 0, 0, 2, 4][x;y;z;u] = [0;0] ?

Yes.

The way I see it, [x;y;z;u] is really just the scalar in the linear combination.

No, <x, y, z, u> is a vector in R⁴. Its coordinates (the x, y, z, and u values) are scalars.

 

[Mark44]

Strange, the system rejected my answer.
It asked what the spanning set of U is.
(x,-10u/3,-2u,u) is apparently not the right one...

The spanning set would be a set of one or more vectors. What I wrote at the tail end of post #10 is probably what they're looking for.

 

[negation]

The spanning set would be a set of one or more vectors. What I wrote at the tail end of post #10 is probably what they're looking for.

They're looking for the vectors in format as such, e.g., (1,2,3,4), (5,6,7,8)

 

[Mark44]

They're looking for the vectors in format as such, e.g., (1,2,3,4), (5,6,7,8)

See post 10. Note that I incorrectly copied one coordinate of one vector, but it's fixed now.