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

[Dick]

[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)}.

Ok, fine, so what point are you trying to make?

 

[evilpostingmong]

Ok, fine, so what point are you trying to make?

Cases where we need to map to a subspace whose range T is a subspace
and where all nonzero variables need to be equal (in this case all variables need
to be equal to x). When it comes to these cases, do you map from the entire
space or from a subspace? To clarify, for {(3, 1)}, I could only find one element in
the basis for nullS and its obvious that there's only one in the basis for rangeS.
So if that's the case, then we are not mapping from the entire space R^3 but from
a 2 dimensional subspace of R^3. But the problem is I don't know whether I'm right or
wrong here.

 

[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).

Is the point of this that you think rank(S)=dim(null(S))? I think rank(S)=dim(range(S)).

 

[evilpostingmong]

Is the point of this that you think rank(S)=dim(null(S))? I think rank(S)=dim(range(S)).

Yes. Wait! I think I have a better idea. top=[3 0 0] bottom=[1 0 0]. Row rank and column
rank are both 1 which=the dimension for rangeS. And (0 1 0) and (0 0 1) both map to
0. So that's two vectors in the basis for nullS and one vector (3, 1) in the basis for rangeS.
So rank(s) does not equal dim(null(S)).

 

[Dick]

Yes. Wait! I think I have a better idea. top=[3 0 0] bottom=[1 0 0]. Row rank and column
rank are both 1 which=the dimension for rangeS. And (0 1 0) and (0 0 1) both map to
0. So that's two vectors in the basis for nullS and one vector (3, 1) in the basis for rangeS.
So rank(s) does not equal dim(null(S)).

Well, no. Of course not. The columns of S are a basis for range(S), not null(S).

 

[evilpostingmong]

Well, no. Of course not. The columns of S are a basis for range(S).

Oh well in that case [3 0] [0 1].

 

[Dick]

Oh well in that case [3 0] [0 1].

When you write a sentence, please use a verb. It's really a lot of work trying to guess what you mean.

 

[evilpostingmong]

When you write a sentence, please use a verb. It's really a lot of work trying to guess what you mean.

sorry. anyway I was mistaken saying [3 0] [0 1]'s columns are a basis for
range(S). Wow I am confused, its that case where rangeT<W that is bothering me.
Here, let's start clean. I am having trouble proving the case where rangeT<W.

 

[Dick]

sorry. anyway I was mistaken saying [3 0] [0 1]'s columns are a basis for
range(S). Wow I am confused, its that case where rangeT<W that is bothering me.

If S=(3 0 0) (0 1 0) then (3 0) and (0 1) ARE a basis for range(S). They are a basis for R^2. If S=(3 0 0) (1 0 0) then dim(range(T))<dim(W).

 

[evilpostingmong]

If S=(3 0 0) (0 1 0) then (3 0) and (0 1) ARE a basis for range(S). They are a basis for R^2.

wait but haven't we determined that range(S) is one dimensional {(3,1)} and not
2 dimensional {(3 0 ) (0 1)}? I'll direct you to that post.

 

[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.

 

[Dick]

wait but haven't we determined that range(S) is one dimensional {(3,1)} and not
2 dimensional {(3 0 ) (0 1)}? I'll direct you to that post.

Yes, dim(range(S))=1, and rank(S)=1. Why should it be two dimensional?

 

[evilpostingmong]

Yes, dim(range(S))=1, and rank(S)=1. Why should it be two dimensional?

Because its a basis for a subspace in R^2?

 

[evilpostingmong]

If S=(3 0 0) (0 1 0) then (3 0) and (0 1) ARE a basis for range(S). They are a basis for R^2. If S=(3 0 0) (1 0 0) then dim(range(T))<dim(W).

I'm guessing that dim(range(T))<dim(W) is a problem and should not be considered
in the proof.

 

[Dick]

I'm guessing that dim(range(T))<dim(W) is a problem and should not be considered
in the proof.

It's NOT a special problem and it NEEDS to be covered by the proof. There's no reason why dim(range(T)) couldn't be less than dim(W). Why are you so worried about it?

 

[evilpostingmong]

It's NOT a special problem and it NEEDS to be covered by the proof. There's no reason why dim(range(T)) couldn't be less than dim(W). Why are you so worried about it?

Because I don't even know what is in nullS's basis for {(3,1)} besides (0 0 1).
What else is in nullS's basis? range(S) is 1 dimensional and I can't convince myself otherwise. I know that this may
sound specific but it is cases where dim rangeT<dim W and range T's basis has vectors with more than one
nonzero component (with the requirement that there needs to be two or more x's or y's or whatever)
that make me worried about what to put in null's basis. range(S)'s basis is 1 dimensional which we found.
But if its one dimensional, there needs to be two vectors in nullS's basis one is (0 0 1) what else in nullS's basis
can map to 0? I mean the transfomraion matrix S is (3 0 0) (0 1 0) what can you apply to this matrix that is
linearly independent to (0 0 1) and maps to 0? That's what's been bothering me. Let's say the dimension
of rangeT is n less than teh dimension of W. does that mean that the basis for nullT gets n more vectors
than what it would've had if rangeT=W?

 

[Dick]

If you are talking about S=(3 0 0) (0 1 0), range(S) is spanned by (3 0) and (0,1) and is two dimensional, null(S) is spanned by (0,0,1) and is one dimensional. If you are talking about S=(3 0 0) (1 0 0) then null(S) is spanned by (0 1 0) and (0 0 1) and is two dimensional, range(S) is spanned by (3 1) and is one dimensional. In both cases dim(null(S))+dim(range(S))=3. I don't see any problem whatsoever.

 

[evilpostingmong]

Yes. Wait! I think I have a better idea. top=[3 0 0] bottom=[1 0 0]. Row rank and column
rank are both 1 which=the dimension for rangeS. And (0 1 0) and (0 0 1) both map to
0. So that's two vectors in the basis for nullS and one vector (3, 1) in the basis for rangeS.
So rank(s) does not equal dim(null(S)).

Oh, haha! Thats what I was trying to get at all along! When you said this was wrong,
you probably thought I was implying that [3 0 0] [1 0 0] represents a basis for R^2
when I was talking about the transformation that allows range(S) to be 1 dimensional.
I thought you wanted me to give just that, you thought I was trying to give
something completely different but failing to do so! We were both falling on deaf ears.
That solves my mystery and confusion about this case, which was haunting my dreams.
I'm happy that we were able to clear that obstacle!

 

[Dick]

Oh, haha! Thats what I was trying to get at all along! When you said this was wrong,
you probably thought I was implying that [3 0 0] [1 0 0] represents a basis for R^2
when I was talking about the transformation that allows range(S) to be 1 dimensional.
I thought you wanted me to give just that, you thought I was trying to give
something completely different but failing to do so! We were both falling on deaf ears.
That solves my mystery and confusion about this case, which was haunting my dreams.
I'm happy that we were able to clear that obstacle!

Ok. So why don't you tackle the original problem now?

 

[evilpostingmong]

Ok. So why don't you tackle the original problem now?

Now here is part of the proof. This is the case where rangeT=W and where I considered
the row rank to=dimW, which makes it a requirement for the columns of M to
represent a basis for W (note that M represents T).

For T*(a1...am) to=0, we need ak=0 and to show this ((b1,1*0+...+bm,1*0)_1...(b1,m*0+...+bm,b*0)_n=(0...0)=0 so dim nullT must=0.

Heres what led to this conclusion. I figured that all scalars in (a1...am) must=0
for M^T(a1...am) to=0 since M^T represents a basis. It only makes sense for all
scalars in this vector to be zero given the fact that all bases have at least
one zero in them so there is at least one row in M^T that is [0_1,k...b_m,k]
or [b_1,k...0_m,k] or [0_1,k...b_j,k...0_m,k].

 

[Dick]

You are completely missing that the actual proof is really simple and substituting a sort of a complex thrashing around for real insight, which at this point I refuse to even read. That's not good. Here are the ingredients you need. And you've seen these before. i) row rank=column rank. ii) matrix(T)=transpose(matrix(T*)). iii) dim(null(T))+dim(range(T))=dim(domain(T)). iv) dim(range(T))=rank T. Put them together in correct sequence and you are done. I'm not going to read any more half baked ideas when the real answer is so much simpler. It's making me crosseyed. I'm really serious this time.

 

[evilpostingmong]

You are completely missing that the actual proof is really simple and substituting a sort of a complex thrashing around for real insight, which at this point I refuse to even read. That's not good. Here are the ingredients you need. And you've seen these before. i) row rank=column rank. ii) matrix(T)=transpose(matrix(T*)). iii) dim(null(T))+dim(range(T))=dim(domain(T)). iv) dim(range(T))=rank T. Put them together in correct sequence and you are done. I'm not going to read any more half baked ideas when the real answer is so much simpler. It's making me crosseyed. I'm really serious this time.

I have a simpler solution using your ingredients. I'll be back later.

 

[evilpostingmong]

Now here is the proof that is a lot simpler than the one I was going to post (fully that is).

Consider the matrix M representing T. Let the transpose of M represent T*. Now
row rank M=column rankM=dimT and row rank transposeM=column rank transposeM=dim rangeT*.
Since row rankM=column rank transposeM, and row rank M=dim rangeT and column rank transposeM=dim rangeT*,
dim rangeT=dim rangeT*.
Now let dim domT*=dim nullT*+dim rangeT*. Since T* maps from rangeT to
a subspace of V dim dom T*=dim rangeT=dim rangeT*. Therefore, dim domT*=dim rangeT*. Now dim rangeT*=dim nullT*+dim range T*. Therefore dim nullT*=dim range T*-dim rangeT*=0. Thus dim nullT*=0. We know that
dim V=dim nullT+dim rangeT. Since dim null T+dim rangeT=dim nullT+ dim W, dim V=dim nullT+ dim W.
Therefore dim nullT+dim W-dimV=0=dim nullT*.

I could show that dim nullT+dim rangeT=dim nullT+dim W even if dim rangeT<dim W but its a bit long (but not a pain to read, its pretty simple)
and I don't know whether or not it is necessary.

 

[Dick]

The first part looks ok. You can pretty much guess that if one of your proofs starts getting long and complex, it's probably also wrong.The problem starts here "Since T* maps from rangeT to
a subspace of V dim dom T*=dim rangeT=dim rangeT*." That's not true at all. domain(T*)=W. The conclusion that dim(null(T*))=0 is completely false.

 

[evilpostingmong]

The first part looks ok. You can pretty much guess that if one of your proofs starts getting long and complex, it's probably also wrong.The problem starts here "Since T* maps from rangeT to
a subspace of V dim dom T*=dim rangeT=dim rangeT*." That's not true at all. domain(T*)=W. The conclusion that dim(null(T*))=0 is completely false.

Ok. Now let dim dom T*=dim nullT*+dim rangeT*. Let dim W=dim dom T*.
Now dim W=dim nullT*+dim rangeT*. Now dim rangeT*=dim W-dim nullT*. Since dim rangeT=dim rangeT* , dim range T=dimW-dim nullT*. Since dimV=dim nullT+dim rangeT,
dim V=dim nullT+dim W-dim nullT*. Rearranging gives dim nullT*+dim V=dim nullT+dim W
so dim nullT*=dim nullT+dim W-dim V.

 

[Dick]

Ok. Now let dim dom T*=dim nullT*+dim rangeT*. Let dim W=dim dom T*.
Now dim W=dim nullT*+dim rangeT*. Now dim rangeT*=dim W-dim nullT*. Since dim rangeT=dim rangeT* , dim range T=dimW-dim nullT*. Since dimV=dim nullT+dim rangeT,
dim V=dim nullT+dim W-dim nullT*. Rearranging gives dim nullT*+dim V=dim nullT+dim W
so dim nullT*=dim nullT+dim W-dim V.

Much better and simpler and correcter.

 

[evilpostingmong]

Much better and simpler and correcter.

Yes! You see, my major issue in the beginning was the fact that I thought
that an n-dimensional subspace of R^n+k =R^n. I considered T:R^3--->R^2 and thought about the
xy plane (a basis is {(1 0 0) (0 1 0 )}) and since it is obvious that there is no
z*(0 0 1) spanned by this basis, and thinking about the xyz graph ( you know how
the xy grid on this looks like the classic 2 dimensional x-y grid) I figured
that {(1 0 0) (0 1 0)} is the same as {(1 0) (0 1)}. After all, the basis is
2-dimensional. I did know that {(0 0 1)} is a basis for nullT and figured hey its 1 dimensional
so dim R^3=dim nullT+dim rangeT=3 so why complain? That's where I got confused.
Thanks for the help, I learned a lot in this thread!

 

[Dick]

Yes! You see, my major issue in the beginning was the fact that I thought
that an n-dimensional subspace of R^n+k =R^n. I considered T:R^3--->R^2 and thought about the
xy plane (a basis is {(1 0 0) (0 1 0 )}) and since it is obvious that there is no
z*(0 0 1) spanned by this basis, and thinking about the xyz graph ( you know how
the xy grid on this looks like the classic 2 dimensional x-y grid) I figured
that {(1 0 0) (0 1 0)} is the same as {(1 0) (0 1)}. After all, the basis is
2-dimensional. I did know that {(0 0 1)} is a basis for nullT and figured hey its 1 dimensional
so dim R^3=dim nullT+dim rangeT=3 so why complain? That's where I got confused.
Thanks for the help, I learned a lot in this thread!

You're welcome. Try not to let the details of specific examples confuse you when it comes to the general case.