Proof in linear algebra

1. Jul 16, 2009

evilpostingmong

1. The problem statement, all variables and given/known data

Prove that
dim null T∗ = dim null T + dimW − dimV
and
dim range T∗ = dim range T
for every T ∈ L(V,W).

2. Relevant equations

3. The attempt at a solution
I have my solution written down, but just to make sure...
I think that nullT*=0 since W is a subspace of V and mapping from W to V
with T* yields the same result as mapping from W to W since W is in V, so it would only
make sense that T(wi) is in V and no vector (besides 0) is in nullT*. Just want
to make sure my reasoning here is right. This isn't my solution, btw, but a crucial finding.

2. Jul 16, 2009

Dick

That's awful, really awful. Suppose T:R^3->R^3 and T is the zero transformation. I.e. T(x)=(0,0,0). T* is also the zero transformation. null(T*)=R^3. I.e. dim(null(T*))=3. Your 'crucial finding' doesn't make any sense AT ALL. At all, at all.

3. Jul 16, 2009

evilpostingmong

I see that. But does it make sense for nullT*=0 if T and T* are not 0 transfomations? So lets say that T(1 2 3)=(3 6 0) going from R^3-->R^2. Null T is 1 dimensional.
so 1+2-3=0=dim nullT*. I feel dumb not considering the 0 transformation. My proof needs a major overhaul.

4. Jul 17, 2009

Dick

How is T(1 2 3)=(3 6 0) supposed to describe a map from R^3->R^2?? Why don't you warm up by telling me why the statements are true in the case where V=W?

Last edited: Jul 17, 2009
5. Jul 17, 2009

HallsofIvy

Staff Emeritus
To expand on Dick's statement: Saying "T(1 2 3)= (3 6 0)" tells what T does to one vector. It does not define a linear transformation. Further, (3 6 0) is in R3, not R2 so any linear transformation (or non-linear function) that mapped (1, 2, 3) to (3 6 0) would be "R3->R3". A function defined by T(x y z)= (3x 2y) would be a linear transformation from R3 to R2.

6. Jul 17, 2009

evilpostingmong

PREPARE FOR A WALL OF TEXT, BUT CORRECTING ME WOULD HELP TREMENDOUSLY
AND COULD IMPACT ANY FUTURE PROOFS THAT I DO!
Oh crap, thats what it is? I always thought that all members in a kernel would
be mapped to 0 in the dimensions that don't contain them as non zero. So I always
thought that (x y 0) is the same as (x y). No wonder this proof was off.
Here's what confused me. I'll quote someone here, I don't want him to feel bad about
this, because I was the one who took him out of context....

"That's fine as far as it goes. Sure null(T) is U^perp as you've defined it. But Halls also said you should prove the span of U and U^perp is V. You didn't do that. Do you see why that's necessary?"

I read it as this: U^perp is (0 0 z) and U is (x y 0). The inner product of these is
0 since they're orthogonal. (0 0 z) is in the null space since (0 0 z) is not in the 2 dimensional U. So (0 0 z) or (0 0 j) or whatever is in the third dimension will get mapped to 0 in the third "slot" of a vector in U. Thats what I considered for membership in the
nullspace. 0 means nothing so it is not counted when considering the dimensions of (0 0 z). Also I figured that taking the inner prod. (z) by ( x y) would not give 0 but I was told that U^perp=nullT since the inner product of any member in U to any member in
U^perp is 0. That's the source of my confusion. If anyone could resolve this confusion,
I'll be extremely happy. bTW dont think I'm saying I'm right you're wrong. I'm the beginner here, and
if I have an idea that is wrong but right in my head, and I don't know why its wrong, even after reading
the textbook or other sites (which I do) I show the experts what I think and hope for correction.

Last edited: Jul 17, 2009
7. Jul 17, 2009

Dick

Ok, so you are somewhat vague on what a kernel is. Fine. Let's not switch to talking about a different proof, ok? Take the T that Halls gave as a example. T(x y z)= (3x 2y). What is the kernel and range of T? Describe them simply and clearly. No wall of text, please. If you are clear on that try to figure out what T* is explicitly.

8. Jul 17, 2009

evilpostingmong

Thank you! It wouldn't make sense to expect to map to R^2 from R^3 and end up with (3x 2y z) because (3x 2y z) is not in R^2. So the kernel holds the guy who doesn't get mapped to R^2 or
(z). Since (3x 2y) is in R^2, it is 2 dimensional, it is in the rangespace R^2. Thats what I can make of this new way (for me)
to see kernels and rangespaces. I guess since T acts on a basis, and z is not a linear comb. of x and y, we don't need
to argue whether or not z is in the first or second dimension. So we can safely say that z is in the third dimension, nullT.

Last edited: Jul 17, 2009
9. Jul 17, 2009

Dick

That's only vaguely correct. Which is part of your problem. You say the kernel is (z). What is that? A number? A 1-vector? Look up the EXACT definitions of 'kernel' and 'range'. What type of objects are they? Now answer the question again and make sure that what you say the kernal and range are match the type of object the definition says they should be. Your 'proofs' are going nowhere otherwise.

10. Jul 17, 2009

Dick

Your use of language is really distorted. (x,y,z)=x*(1,0,0)+y*(0,1,0)+z*(0,0,1). You've got to start distinguishing between scalars and vectors. T(x,y,z)=(3x,2y)=3x*(1,0)+2y*(0,1). z is a scalar, saying it's not in R^2 doesn't make any sense. Here's the correct way to phrase it. The kernel is the subspace of all (x,y,z) such that T(x,y,z)=(0,0). If T(x,y,z)=(3x,2y)=(0,0) that means x=0 and y=0. So (x,y,z)=(0,0,z). The kernel is the set of all vectors (0,0,z). It's z*(0,0,1) where z can be any number. It's the subspace spanned by {(0,0,1)}. There, no mumbo-jumbo. R^2 and R^3 are different spaces. Saying a vector in one is also in the other makes no sense either.

Last edited: Jul 17, 2009
11. Jul 17, 2009

cipher42

Dick speaks the truth. These more complicated expressions are going to be impossible to grasp (let alone prove properties about!) until you have a rock-solid understanding of the basic principles: vectors, scalars, maps, etc. When you're studying this kind of stuff, if you realize that you don't really every bit of what you're writing, then it's time to go back and review until you do.

12. Jul 17, 2009

evilpostingmong

Makes sense now. R^2 is spanned by {(1 0 ), (0 ,1)} and going back to "base building"
the kernel is 1 dimensional so direct summing R^2 with the kernel woud produce
{(1 0 0) (0 1 0) (0 01)} a basis which spans R^3. Goes back to saying dimV=dimnullT+dimrangeT.
And yes the notation threw me off, saying the scalar z (say z=2) is not in R^2 is like
saying ( 2 2 ) is not in R^2. That doesn't make sense. And T(0 0 z)=0 only because
adding 0 to T(x y z)=(3x, 2y)+0 is still in R^2. If T(0 0 z)=(0 0 2z) well, thats not
in R^2. Its not 2 dimensional. No linear combinations of vectors in R^2 will take you there.
Simply because T acts on a basis and the third dimension is not in 2d space.

Last edited: Jul 17, 2009
13. Jul 17, 2009

Dick

That's getting better. Yes, once you know the range is R^2. Can you say in clear language why range(T) is span {(1,0),(0,1)}? In other words, all of R^2?

14. Jul 17, 2009

Dick

You are slipping again. T(x,y,z)=(3x,2y). Where are you getting this T(0,0,z)=(0,0,2z) stuff? What's that got to do with anything? T(0,0,z)=(0,0) because that's the definition of T! Period. No image vectors of T are 'not in R^2'.

15. Jul 17, 2009

evilpostingmong

Because the basis for V is the direct sum of the bases for nullT and rangeT.
Goes back to dimnullT+dimrangeT=dimV. nullT and rangeT forms a basis for dimV.
Since rangeT's space is 2 dimensional (call it that for now), nullT's space needs to be 1 dimensional (since we name V's dim=3). And since rangeT's space is 2 dimensional, its basis needs to be a linearly independent span with two elements. We can name them (1 0) and (0 1) {(1 0) (0 1)} All possible linear combinations of these vectors are closed within rangeT.
nullT's space is 1 dimensional. But since dim nullT+dimrangeT=dimV, and V has
three elements in the basis and rangeT has 2, how can we have a linearly independent basis
for V when there are only two "slots" on each vector in rangeT's basis {(1 0) (0 1)}?
It cant be {(1 0 ) (0 1) (1 1)} since a lc of these would still be in rangeT. That
is why (0 0 1) is in nullT, independent from {(1 0 ) and (0 1)} not a linear combination
of any vector in rangeT.

16. Jul 17, 2009

Dick

If T:V->W then range(T) is a subspace of W. It is not a part of a basis for V. If dim(ker(T))+dim(range(T))=dim(V) gave you that impression, you should forget it NOW. You can construct a subspace U of V such that T(U)=range(T) and the mapping is 1-1 and onto. But range(T) is not in V.

17. Jul 17, 2009

evilpostingmong

Yeah, youre right, stupid mistake there.

18. Jul 17, 2009

evilpostingmong

Ok since V is 3d and range T is 2d. {(1 0 ) (0 1)} is not a basis for V. x(1 0)+y(0 1)
is not some vector where there are three slots. (0 0 z) is not some vector where
there are two slots. No linear combination of vectors in {(0 0 1)} will give
x(1 0)+y(0 1).

19. Jul 17, 2009

Dick

Right, I think, if you are trying to say you can't mix up vectors of different dimension. To get back to business, why is range(T)=span{(1,0),(0,1)}? Without mumbo-jumbo.

20. Jul 17, 2009

evilpostingmong

range T=span{(1,0), (0,1)} because range T is 2 dimensional. It is a subspace of W.
All linear combinations of vectors in range T's span are within W. And according to the
basis, rangeT is such that T(x y z)=(x y) a linear combination of (1 0) and (0 1).