Another question about transforms and subspaces

[BiGyElLoWhAt]

Homework Statement

A:
Let ##T## be the linear function ##T####:\mathbb{R}^3→\mathbb{R}^1## defined as ##T####(x,y,z) = x-3y+z##.

The nullspace of T is a 2 dimensional subspace of ##\mathbb{R}^3## (a plane through the origin). Give an example of the basis of this subspace ##\{ \vec{u},\vec{v} \}##

B:
Give an example of a vector ##\vec{w} \in \mathbb{R}^3## so that ##\{ \vec{u},\vec{v},\vec{w} \} ## is a basis for ##\mathbb{R}^3##

u and v are from part a.

Homework Equations

Not sure what to put here in this case...

The Attempt at a Solution

So, the nullspace of T is all vectors so that ##T(\vec{x})=\vec{0}##.

So ##\left [ \begin{array}{ccc}
1 & -3 & 1 \\ \end{array} \right ]## ## \left [ \begin{array}{c} x \\ y \\ z \\ \end{array} \right ] = \left [ \begin{array}{c}
0 \\ \end{array} \right ]##

which, it's pretty obvious without even working through anything that my plane is simply ## x - 3y +z = 0##

So I guess my question here, is this: when I need to setup a basis, can I treat this similar to a calc 3 problem? Both for part a and part b. So for part a, can I say that a basis is (upon solving for z)

##\{ \left [ \begin{array}{c} -1 \\ 0 \\ \end{array} \right ] \ \text{,}\ \left [ \begin{array}{c} 0 \\ 3 \\ \end{array} \right ] \}##

But through a few scalar multiplications, this can be:

##\{ \left [ \begin{array}{c} 1 \\ 0 \\ \end{array} \right ] \ \text{,}\ \left [ \begin{array}{c} 0 \\ 1 \\ \end{array} \right ] \}##

Or also, should this be in 3 dimension?

##\{ \left [ \begin{array}{c} 0 \\ -1 \\ 0 \\ \end{array} \right ] \ \text{,}\ \left [ \begin{array}{c} 0 \\ 0 \\ 3 \\ \end{array} \right ] \}##


And then for b, I probably shouldn't do it this way, as we never covered the cross product, but could I just cross my 2 bases to get a 3rd orthogonal (I guess it only needs to be linearly independant of the other 2) vector as a basis for my ##\mathbb{R}^3## ? Noted that this only works because we're working in R^3 and subspaces thereof, but work smart not hard, right?

 

[LCKurtz]

Homework Statement

A:
Let ##T## be the linear function ##T####:\mathbb{R}^3→\mathbb{R}^1## defined as ##T####(x,y,z) = x-3y+z##.

The nullspace of T is a 2 dimensional subspace of ##\mathbb{R}^3## (a plane through the origin). Give an example of the basis of this subspace ##\{ \vec{u},\vec{v} \}##

B:
Give an example of a vector ##\vec{w} \in \mathbb{R}^3## so that ##\{ \vec{u},\vec{v},\vec{w} \} ## is a basis for ##\mathbb{R}^3##

u and v are from part a.

Homework Equations

Not sure what to put here in this case...

The Attempt at a Solution

So, the nullspace of T is all vectors so that ##T(\vec{x})=\vec{0}##.

So ##\left [ \begin{array}{ccc}
1 & -3 & 1 \\ \end{array} \right ]## ## \left [ \begin{array}{c} x \\ y \\ z \\ \end{array} \right ] = \left [ \begin{array}{c}
0 \\ \end{array} \right ]##

which, it's pretty obvious without even working through anything that my plane is simply ## x - 3y +z = 0##

So I guess my question here, is this: when I need to setup a basis, can I treat this similar to a calc 3 problem? Both for part a and part b. So for part a, can I say that a basis is (upon solving for z)

##\{ \left [ \begin{array}{c} -1 \\ 0 \\ \end{array} \right ] \ \text{,}\ \left [ \begin{array}{c} 0 \\ 3 \\ \end{array} \right ] \}##

But through a few scalar multiplications, this can be:

##\{ \left [ \begin{array}{c} 1 \\ 0 \\ \end{array} \right ] \ \text{,}\ \left [ \begin{array}{c} 0 \\ 1 \\ \end{array} \right ] \}##

Or also, should this be in 3 dimension?

##\{ \left [ \begin{array}{c} 0 \\ -1 \\ 0 \\ \end{array} \right ] \ \text{,}\ \left [ \begin{array}{c} 0 \\ 0 \\ 3 \\ \end{array} \right ] \}##

And then for b, I probably shouldn't do it this way, as we never covered the cross product, but could I just cross my 2 bases to get a 3rd orthogonal (I guess it only needs to be linearly independant of the other 2) vector as a basis for my ##\mathbb{R}^3## ? Noted that this only works because we're working in R^3 and subspaces thereof, but work smart not hard, right?

You haven't shown much work, only wrong answers. Your final answer has the x coordinate always zero which would be the yz plane. Obviously your transformation doesn't map the yz plane to zero.

 

[BiGyElLoWhAt]

But I want the null space, which is the yz plane, no?

 

[BiGyElLoWhAt]

Ok I see what you're saying, but then I guess I'm misinterpreting what needs to be done, so if I go from a vector space, R^3, down to R^1, that's equivalent to saying {x,y,z} -> {x,0,0} is it not?

 

[LCKurtz]

Ok I see what you're saying, but then I guess I'm misinterpreting what needs to be done, so if I go from a vector space, R^3, down to R^1, that's equivalent to saying {x,y,z} -> {x,0,0} is it not?

Your transformation takes ##(x,y,z)\rightarrow x-3y+z##. Show us your work solving that equals 0 and writing the solution as$$
\begin{bmatrix}x\\y\\z\end{bmatrix} = ~?$$

 

[BiGyElLoWhAt]

ok so :

## \left [ \begin{array}{c} 3y-z \\ \frac{x+z}{3} \\ 3y - x \\ \end{array} \right ]##
is this what you're asking for? that's the 3 ways you can write the nullspace-plane, which all come from me solving the transform = 0.

 

[LCKurtz]

Your transformation takes ##(x,y,z)\rightarrow x-3y+z##. Show us your work solving that equals 0 and writing the solution as$$
\begin{bmatrix}x\\y\\z\end{bmatrix} = ~?$$

ok so :

## \left [ \begin{array}{c} 3y-z \\ \frac{x+z}{3} \\ 3y - x \\ \end{array} \right ]##
is this what you're asking for?

No. You can let two variables be anything. Say you let ##x=a,~y=b##. What do you get for
$$
\begin{bmatrix}x\\y\\z\end{bmatrix} = ~?$$

 

[STEMucator]

There has to be at least two free variables if you're going from ##\mathbb{R^3} → \mathbb{R}##.

So one variable can be expressed in terms of the others.

This is evident from computing:

##T(e_1) = 1##
##T(e_2) = -3##
##T(e_3) = 1##

Which form the 1x3 matrix of the transformation: ##A = [T(e_1), T(e_2), T(e_3)]##. Multiplying this to any vector ##x \in \mathbb{R^3}## is the purpose of ##T##.

 

[BiGyElLoWhAt]

No. You can let two variables be anything. Say you let ##x=a,~y=b##. What do you get for
$$
\begin{bmatrix}x\\y\\z\end{bmatrix} = ~?$$

Ok, just to help keep things straight, can we call the values put into the transform _i values (##x_{i}##) and the values out of the transform _o values (##x_{o}##)? so with that being said, we have some vector ##<x_i , y_i , z_i > ## in V, and we map them linearly onto W via T, giving us ##<x_o , y_o , z _o> = <x_i , -3y_i , z_i >##

Good so far?

We want the null space, so we want to solve for all vectors ##<x_o , y_o , z _o> = <x_i , -3y_i , z_i > = <0,0,0>##

So if we have ## \left [ \begin{array}{ccc} 1 & -3 & 1 \\ \end{array} \right ] \left [ \begin{array}{c} x_i \\ y_i \\ z_i \\ \end{array} \right ] ## and we set ##x_o=a## and ##y_o=b## then we end up with ## a + b + z_i = 0## or if we set ##x_i = a## and ##y_i = b## then we get ##a -3b + z_i##

I'm not sure which one of these you mean, though. Which x are we letting = a?

I'm going to guess that we want the first, as ##x_o## , not ##x_i##, is our solution.

So then we have ##\left [ \begin{array}{c} x \\ y \\ z \\ \end{array} \right ] = \left [ \begin{array}{c} a \\ b \\ -(a + b)\\ \end{array} \right ] ##

But in all honesty, I think I'm missing the point here.

 

[BiGyElLoWhAt]

There has to be at least two free variables if you're going from ##\mathbb{R^3} → \mathbb{R}##.

So one variable can be expressed in terms of the others.

This is evident from computing:

##T(e_1) = 1##
##T(e_2) = -3##
##T(e_3) = 1##

Which form the 1x3 matrix of the transformation: ##A = [T(e_1), T(e_2), T(e_3)]##. Multiplying this to any vector ##x \in \mathbb{R^3}## is the purpose of ##T##.

Ok,

and this is equal to ## \left [ \begin{array}{ccc} T(e_1) & T(e_2) & T(e_3) \end{array} \right ] \left [ \begin{array}{c} x \\ 0 \\ 0 \\ \end{array} \right ] + \left [ \begin{array}{ccc} T(e_1) & T(e_2) & T(e_3) \end{array} \right ] \left [ \begin{array}{c} 0 \\ y \\ 0 \\ \end{array} \right ] + \left [ \begin{array}{ccc} T(e_1) & T(e_2) & T(e_3) \end{array} \right ] \left [ \begin{array}{c} 0 \\ 0 \\ z \\ \end{array} \right ] ## which I thought was basically what I did.

So when I'm describing the nullspace of ##T \in W## am I supposed to be describing it in terms of ##V## ? because that's not what I did, and that could very well be where my confusion is coming from. So would the fore-mentioned sum (T(e_1)... 0,0,z) be my basis for the null space? I mean, from those 3 vectors, I could encompass all vectors NULL(T) in terms of the vector space V.

 

[LCKurtz]

Ok, just to help keep things straight, can we call the values put into the transform _i values (##x_{i}##) and the values out of the transform _o values (##x_{o}##)? so with that being said, we have some vector ##<x_i , y_i , z_i > ## in V, and we map them linearly onto W via T, giving us ##<x_o , y_o , z _o> = <x_i , -3y_i , z_i >##

Good so far?

We want the null space, so we want to solve for all vectors ##<x_o , y_o , z _o> = <x_i , -3y_i , z_i > = <0,0,0>##

So if we have ## \left [ \begin{array}{ccc} 1 & -3 & 1 \\ \end{array} \right ] \left [ \begin{array}{c} x_i \\ y_i \\ z_i \\ \end{array} \right ] ## and we set ##x_o=a## and ##y_o=b## then we end up with ## a + b + z_i = 0## or if we set ##x_i = a## and ##y_i = b## then we get ##a -3b + z_i##

I'm not sure which one of these you mean, though. Which x are we letting = a?

I'm going to guess that we want the first, as ##x_o## , not ##x_i##, is our solution.

So then we have ##\left [ \begin{array}{c} x \\ y \\ z \\ \end{array} \right ] = \left [ \begin{array}{c} a \\ b \\ -(a + b)\\ \end{array} \right ] ##

But in all honesty, I think I'm missing the point here.

You are making this way too complicated. Why are you introducing all those subscripts? Look at your last equation. If ##x=a## and ##y=b## and ##x-3y+z=0## it isn't rocket science to figure out what ##z## is. And it isn't ##-(a+b)##.

 

[BiGyElLoWhAt]

Ok so a \\ b \\ 3b -a \\

 

[verty]

But I want the null space, which is the yz plane, no?

You don't want the null space, this is very inaccurate. You want a basis for the null space. If you're not sure what that means, look it up, confirm that you know what you are being asked, and hopefully you'll then have an idea of how to do that.

 

[BiGyElLoWhAt]

You don't want the null space, this is very inaccurate. You want a basis for the null space. If you're not sure what that means, look it up, confirm that you know what you are being asked, and hopefully you'll then have an idea of how to do that.

Ok that's true. I do want the basis for the null space. The nullspace is everywhere that T(v) = 0, which in this case is a plane x - 3y + z =0, and I want a basis, so 3(?) linearly independant vectors from which I can perform linear combinations with to get any vectors contained within the null space.

Honestly, I don't even know anymore. I was going through a final review that was used by the chair when he taught the class (our teacher sent it out 2 days ago), and he made his own and just sent that out today, and there's literally nothing about transforms on it. I don't really remember covering transforms very well. (it's a summer class, so if we did it was probably for 1 day)

 

[HallsofIvy]

I thought I had responded here but can't find it now. Yes, x- 3y+ z=0 is a two dimensional plane. The simplest way to find a basis is to solve for z: z=3y - x. So any vector in that space is of the form
[tex]\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix}x \\ y \\ 3y- x\end{pmatrix}= \begin{pmatrix}x \\ 0 \\ -x\end{pmatrix}+ \begin{pmatrix}0 \\ y \\ 3y\end{pmatrix}= x\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}+ y\begin{pmatrix}0 \\ 1 \\ 3\end{pmatrix}[/tex].

Do you see the two vector basis now?

 

[BiGyElLoWhAt]

I thought I had responded here but can't find it now. Yes, x- 3y+ z=0 is a two dimensional plane. The simplest way to find a basis is to solve for z: z=3y - x. So any vector in that space is of the form
[tex]\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix}x \\ y \\ 3y- x\end{pmatrix}= \begin{pmatrix}x \\ 0 \\ -x\end{pmatrix}+ \begin{pmatrix}0 \\ y \\ 3y\end{pmatrix}= x\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}+ y\begin{pmatrix}0 \\ 1 \\ 3\end{pmatrix}[/tex].

Do you see the two vector basis now?

Ahhhh! awesome. I thought that was what I was doing (or at least trying to do) Thanks a million.