What is the proof for the dimensions of null and range in linear algebra?

[evilpostingmong]

Yes. range(S)=(3x,x) for all values of x. That's x*(3,1). The subspace basis is {(3,1)}. It's a one dimensional subspace of R^2.

The kernel also makes sense. I feel so stupid in not seeing that the reason why
T maps (0 0 z) to (0 0) is simply because it should be treated just like any other
vector. T(x y z)=(3x, 2y). T( x y z)=T(0 0 z) by letting y=0 and x=0. Then
T(x y z)=(3x 2y)=(3*0 2*0)=(0 0)

 

[Dick]

The kernel also makes sense. I feel so stupid in not seeing that the reason why
T maps (0 0 z) to (0 0) is simply because it should be treated just like any other
vector. T(x y z)=(3x, 2y). T( x y z)=T(0 0 z) by letting y=0 and x=0. Then
T(x y z)=(3x 2y)=(3*0 2*0)=(0 0)

Dead right. You seem to be addicted to thinking of things in complicated and confusing ways when in fact they are neither. Now to go back somewhat in the direction of where we started, T* is a map from R^2->R^3. I.e. T*(x,y) is a 3-vector. What's the formula for the 3-vector? I ask this question in spite of my fear of the 'wall of text'.

 

[evilpostingmong]

Dead right. You seem to be addicted to thinking of things in complicated and confusing ways when in fact they are neither. Now to go back somewhat in the direction of where we started, T* is a map from R^2->R^3. I.e. T*(x,y) is a 3-vector. What's the formula for the 3-vector? I ask this question in spite of my fear of the 'wall of text'.

(x y 0)

 

[Dick]

(x y 0)

Dead wrong. Are you just pushing 'submit reply' while you are thinking about things? I'll give you big hint. I told you before that the matrix representations of T and T* were transposes of each other. Does that help you solve the question of what is T*(x,y)? Can you prove that statement from the definition of adjoint?

 

[evilpostingmong]

Dead wrong. Are you just pushing 'submit reply' while you are thinking about things? I'll give you big hint. I told you before that the matrix representations of T and T* were transposes of each other. Does that help you solve the question of what is T*(x,y)? Can you prove that statement from the definition of adjoint?

oh that's right. I'd say that a 2x3 matrix would do the trick on (x y).
row space of this matrix applied to (x y) is three dimensional
representing the map to R^3.

 

[Dick]

oh that's right. I'd say that a 2x3 matrix would do the trick on (x y).
row space of this matrix applied to (x y) is three dimensional
representing the map to R^3.

Vague, as always, whatever that means. T*(x,y)=(3x,2y,0). That's what I would have said.

 

[evilpostingmong]

Vague, as always, whatever that means. T*(x,y)=(3x,2y,0). That's what I would have said.

Oh, crap, so let's say T(x y z)=(5x) then T*(x)=(5x 0 0). Sorry about that.

 

[Dick]

Let's escalate. The other problem was too easy. Suppose T:R^3->R^2. T(x,y,z)=(x+2y,y-3z). What's null(T), range(T), and what's the transformation T*:R^2->R^3? What's null(T*) and range(T*)? If you can answer these coherently, I think you've got the basics.

 

[evilpostingmong]

Let's escalate. The other problem was too easy. Suppose T:R^3->R^2. T(x,y,z)=(x+2y,y-3z). What's null(T), range(T), and what's the transformation T*:R^2->R^3? What's null(T*) and range(T*)? If you can answer these coherently, I think you've got the basics.

I have an idea for the basis for range(T) but its not that pretty. But then again, it isn't
terrible either.

 

[Dick]

I have an idea for the basis for range(T) but its not that pretty. But then again, it isn't
terrible either.

How can it be 'not pretty'? The range(T) is a subspace of R^2, it's either {0}, spanned by a single vector or it's all of R^2. Your 'not pretty' idea had better correspond to one of those.

 

[evilpostingmong]

How can it be 'not pretty'? The range(T) is a subspace of R^2, it's either {0}, spanned by a single vector or it's all of R^2. Your 'not pretty' idea had better correspond to one of those.

in that case, it is {(1 0) (0 1)}. why? x+2y is a scalar by itself. y-3z is a scalar by itself.
x+2y and y-3z are different if y=/=0 and/or x=/=y. In that case, x+2y can be anything
and y-3z can be any scalar so letting x=0 and y=1/2 or
letting x=1 and y=0 we have (1 0) for (x+2y, 0) and letting y=1/2 and letting z=-1/6
gives (0, 1) for (0, y-3z).

 

[Dick]

in that case, it is {(1 0) (0 1)}. why? x+2y is a scalar by itself. y-3z is a scalar by itself.
x+2y and y-3z are different if y=/=0 and/or x=/=y. In that case, x+2y can be anything
and y-3z can be any scalar so letting x=0 and y=1/2 or
letting x=1 and y=0 we have (1 0) for (x+2y, 0) and letting y=1/2 and letting z=-1/6
gives (0, 1) for (0, y-3z).

I don't know what being a 'scalar by itself' means, so I skipped the first part. The second part where you put in values is fine.

 

[evilpostingmong]

I don't know what being a 'scalar by itself' means, so I skipped the first part. The second part where you put in values is fine.

(y-3z) is a scalar and (x+2y) is a scalar. Sorry for the confusion.

 

[Dick]

(y-3z) is a scalar and (x+2y) is a scalar. Sorry for the confusion.

Is it really necessary to mention that in your solution? I'm trying to get you to leave out the unnecessary parts.

 

[evilpostingmong]

Let's escalate. The other problem was too easy. Suppose T:R^3->R^2. T(x,y,z)=(x+2y,y-3z). What's null(T), range(T), and what's the transformation T*:R^2->R^3? What's null(T*) and range(T*)? If you can answer these coherently, I think you've got the basics.

basis for nullT={(0 0 1)}. Now the transformation T*:R^2->R^3 is the map from
R^2 to R^3's 2d plane so T*(x y)=(x+2y, y-3z, 0). Basis for range(T*)={(1 0 0), (0 1 0)}
its logical in that if {(1 0 ) (0 1)} is a basis for range(T) then T*(x 0)+T*(0 y)=
(x+2y 0 0)+(0 y-3z 0) to the 2d plane of R^3. Basis for null(T*) is {(0 0)} since
T*(0)+T*(0)=(0+2*0,0-3*0, 0)=(0 0 0).

 

[Dick]

basis for nullT={(0 0 1)}. Now the transformation T*:R^2->R^3 is the map from
R^2 to R^3's 2d plane so T*(x y)=(x+2y, y-3z, 0). Basis for range(T*)={(1 0 0), (0 1 0)}
its logical in that if {(1 0 ) (0 1)} is a basis for range(T) then T*(x 0)+T*(0 y)=
(x+2y 0 0)+(0 y-3z 0) to the 2d plane of R^3. Basis for null(T*) is {(0 0)} since
T*(0)+T*(0)=(0+2*0,0-3*0, 0)=(0 0 0).

You've got T* completely wrong. Write T as a matrix and take it's transpose. Carefully.

 

[evilpostingmong]

You've got T* completely wrong. Write T as a matrix and take it's transpose. Carefully.

top row[1 2 0] bottom row[0 1 -3] then the transpose is
[1 0] [2 1] [0 -3] top to bottom.

 

[Dick]

top row[1 2 0] bottom row[0 1 -3] then the transpose is
[1 0] [2 1] [0 -3] top to bottom.

Ok, so T*(x,y)=?

 

[evilpostingmong]

Ok, so T*(x,y)=?

T*(x,y)= (x, 2x+y, -3y). I'll be back in an hour or so.

 

[Dick]

T*(x,y)= (x, 2x+y, -3y). I'll be back in an hour or so.

Right. I probably won't be back until tomorrow, but do you see why you were having such a problem with original proof? Your concepts of what things actually were are all vague and fuzzy. Hopefully, they are getting better.

 

[evilpostingmong]

Right. I probably won't be back until tomorrow, but do you see why you were having such a problem with original proof? Your concepts of what things actually were are all vague and fuzzy. Hopefully, they are getting better.

You bet. Much improved!

 

[evilpostingmong]

T*(x,y)= (x, 2x+y, -3y). I'll be back in an hour or so.

What about the basis for nullT*? Would that be {0}? apply 0 to the matrix I used
and you will get (0 0 0). And I can see how dim rangeT=dim rangeT*
the basis for rangeT* is {(1 2 0) (0 1 -3)} and for range T its {(1 0) (0 1)}.
At least for this case.

 

[Dick]

What about the basis for nullT*? Would that be {0}? apply 0 to the matrix I used
and you will get (0 0 0). And I can see how dim rangeT=dim rangeT*
the basis for rangeT* is {(1 2 0) (0 1 -3)} and for range T its {(1 0) (0 1)}.
At least for this case.

Ok, now how do you see it in general?

 

[evilpostingmong]

Yes. range(S)=(3x,x) for all values of x. That's x*(3,1). The subspace basis is {(3,1)}. It's a one dimensional subspace of R^2. Who cares what the basis for R^3 is? You keep changing your posts.

I have the proof in store for the case where dimrangeT=dimW. But what about when dimrangeT<dimW? The range with the basis {(3,1)} is an example of where my confusion is.
I know that one member of nullS=(0 0 z). But since dim R^3=3, dim rangeS=1 and
dim nullS=2, there must be another vector in the basis for nullS. But I just can't
think of another member of this basis. But then again, I am half asleep.

 

[Dick]

I have the proof in store for the case where dimrangeT=dimW. But what about when dimrangeT<dimW? The range with the basis {(3,1)} is an example of where my confusion is.
I know that one member of nullS=(0 0 z). But since dim R^3=3, dim rangeS=1 and
dim nullS=2, there must be another vector in the basis for nullS. But I just can't
think of another member of this basis. But then again, I am half asleep.

Sleep on it. Think about how row rank of a matrix equals the column rank might help.

 

[HallsofIvy]

What about the basis for nullT*? Would that be {0}? apply 0 to the matrix I used
and you will get (0 0 0). And I can see how dim rangeT=dim rangeT*
the basis for rangeT* is {(1 2 0) (0 1 -3)} and for range T its {(1 0) (0 1)}.
At least for this case.

Any matrix times 0 is 0. Any linear transformation of 0 gives 0. 0 is in any null space. The question is what other vectors are in that space.

You find the null space of T*(x,y)= (x, 2x+y, -3y), not by observing that T*(0,0)= (0, 0, 0), but by solving x= 0, 2x+ y= 0, -3y= 0. Yes, the first and last equation easily give x= 0, y= 0 so (0, 0) is the only vector in the null space.

 

[evilpostingmong]

Any matrix times 0 is 0. Any linear transformation of 0 gives 0. 0 is in any null space. The question is what other vectors are in that space.

You find the null space of T*(x,y)= (x, 2x+y, -3y), not by observing that T*(0,0)= (0, 0, 0), but by solving x= 0, 2x+ y= 0, -3y= 0. Yes, the first and last equation easily give x= 0, y= 0 so (0, 0) is the only vector in the null space.

ok so it involves using algebra I learned in high school. I will consider that when looking
for null space vectors. Thank you! Oh btw yes I know that any transformation of 0 gives 0 and that
any matrix*0=0 but I was talking about this specific case. But that tip (solving for the variables) should
help. I hit a real wall when it comes for determining null space members for rangespaces like
the one associated with the basis {(3, 1)} . Besides (0 0 1) that is.

 

[evilpostingmong]

Sleep on it. Think about how row rank of a matrix equals the column rank might help.

I guess I should consider top[3 0 0] bottom[0 1 0] being the transformation S
and apply this to the vector (1 1 0). Row rank=2 column rank=2. Theres two nonzero columns
and two nonzero rows. I think we are only mapping from a subspace of R^3 and not R^3 entirely because
I can only come up with one vector in nullS and that's (0 0 1).

 

[Dick]

I guess I should consider top[3 0 0] bottom[0 1 0] being the transformation S
and apply this to the vector (1 1 0). Row rank=2 column rank=2. Theres two nonzero columns
and two nonzero rows. I think we are only mapping from a subspace of R^3 and not R^3 entirely because
I can only come up with one vector in nullS and that's (0 0 1).

I can't tell what matrix you are working with or what you are trying to say about it.

 

[evilpostingmong]

I can't tell what matrix you are working with or what you are trying to say about it.

[3 0 0] is the top row [0 1 0] is the bottom row. I cannot find a vector in nullS's basis besides
(0 0 1)that would map to 0 after applying this matrix to it.This matrix is the "S"
matrix so taking x*(1 1 0) and applying this matrix to it gives x*(3 1) in the span
{(3,1)}.