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

[evilpostingmong]

Homework Statement

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

Homework Equations

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.

 

[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.

 

[evilpostingmong]

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.

I see that. But does it make sense for nullT*=0 if T and T* are not 0 transfomations? So let's 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.

 

[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?

 

[HallsofIvy]

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.

 

[evilpostingmong]

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.

PREPARE FOR A WALL OF TEXT, BUT CORRECTING ME WOULD HELP TREMENDOUSLY
AND COULD IMPACT ANY FUTURE PROOFS THAT I DO!
Oh crap, that's 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 don't 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.

 

[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.

 

[evilpostingmong]

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.

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.

 

[Dick]

Thank you! It wouldn't make sense to expect to map to R^2 from R^3 and end up with (x y z) because (x y z) is not in R^2. So the kernel holds the guy who doesn't get mapped to R^2 or
(z). Since (x y) 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.

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.

 

[Dick]

Yes. The kernel is the set of vectors such that T(v) in V (V is the space where
T maps a vector from V to W) in that T(v)=the 0 vector. So I guess it has to do with the
following. z is a vector not in R^2. x=(1 0 0) y=(0 1 0) z=(0 0 1). (x y z)=(1 0 0)+(0 1 0)
+(0 01). Now T(x y z)= T(x)+ T(y)+T(z)= (3 0 0)+(0 2 0)+(0 0 0)=(3 2 0). Now
by the definition of range, T(x y z) is in R^3 as well as R^2 since T(x y z)=(3 2 0) a vector
in both R^3 and R^2. I got this from the defintion that says the following "Let L be a linear transformation from a vector space V to a vector space W. Then the range of L is the set of all vectors w in W such that there is a v in V with L(v) = w ". So since L(v) is in V,
and L(v) is in W, for L(v)=w, L(v) is in both V and W. Since T(1 1 1) is in R^3 since T(1 0 0)
+T(1 0 0)+(0 0 0) is a linear combination of vectors in R^3 and (3 2 0) is 2 dim (so its in
R^2), (3 2 0) is in both R^2 and R^3 whereas (0 0 1) is in R^3 but not R^2.

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.

 

[cipher42]

You've got to start distinguishing between scalars and vectors.

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.

 

[evilpostingmong]

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.

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, that's 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.

 

[Dick]

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.

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?

 

[Dick]

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, that's 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.

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'.

 

[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 can't 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.

 

[Dick]

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 can't 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.

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.

 

[evilpostingmong]

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'.

Yeah, youre right, stupid mistake there.

 

[evilpostingmong]

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.

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).

 

[Dick]

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).

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.

 

[evilpostingmong]

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.

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).

 

[evilpostingmong]

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.

Sorry, but I don't really know why range(T) is not in V in this case. For 1-1 and onto, doesn't
that mean that U=V?

 

[Dick]

Sorry, but I don't really know why range(T) is not in V in this case. For 1-1 and onto, doesn't
that mean that U=V?

Just because you have a bijection between two things doesn't mean they ARE the same thing. The xy plane is a 2 dimensional subspace of R^3 which isomorphic to R^2. That doesn't mean it IS R^2. R^2 is made of 2-vectors, the xy plane is made of 3-vectors.

 

[Dick]

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).

T(x,y,z)=(3x,2y) not (x,y). You have to show (1,0) and (0,1) are in the range. T(1/3,0,0)=(1,0), T(0,1/2,0)=(0,1). It's not true that for every S:R^3->R^2 that range(S) is R^2. Try S(x,y,z)=(3x,x). What's range(S)?

 

[evilpostingmong]

Just because you have a bijection between two things doesn't mean they ARE the same thing. The xy plane is a 2 dimensional subspace of R^3 which isomorphic to R^2. That doesn't mean it IS R^2. R^2 is made of 2-vectors, the xy plane is made of 3-vectors.

Its amazing how many of these examples run right by me. Yet they make so much sense.
True, (x y 0) is a vector in the 2 dimensional subspace of R^3 and it is NOT in R^2 which is the misconception that I had before. xy plane=/=R^2. And T(x y 0)=(3x 2y) and
T(r k 0)=(3r 2k) which are clearly different if (x y 0)=/=(r k 0). And T(x)+T(y)=3(x 0)
+2(0 y) considering x and y can be any scalar in R^3, x and y can be any scalar in
R^2 (but x and y BY THEMSELVESare unchanged after the transformation from xy to R^2 besides the fact that
they get multiplied by 3 and 2), so it only makes sense since there's no such thing as a scalar not being in space X.
Since x and y can be any scalar in R^2 and R^3, (3x 2y) can be any vector in
R^2 so the transformation is onto.

 

[Dick]

Its amazing how many of these examples run right by me. Yet they make so much sense.
True, (x y 0) is a vector in the 2 dimensional subspace of R^3 and it is NOT in R^2 which is the misconception that I had before. xy plane=/=R^2. And T(x y 0)=(3x 2y) and
T(r k 0)=(3r 2k) which are clearly different if (x y 0)=/=(r k 0). And T(x)+T(y)=3(x 0)
+2(0 y) considering x and y can be any scalar in R^3, x and y can be any scalar in
R^2 (but x and y BY THEMSELVESare unchanged after the transformation from xy to R^2 besides the fact that
they get multiplied by 3 and 2), so it only makes sense since there's no such thing as a scalar not being in space X.
Since x and y can be any scalar in R^2 and R^3, (3x 2y) can be any vector in
R^2 so the transformation is onto.

Yes, T is onto, except don't talk about scalars being 'in' R^2 or R^3. They aren't. The scalars are in R. The vectors are in R^2 and R^3.

 

[evilpostingmong]

Yes, T is onto, except don't talk about scalars being 'in' R^2 or R^3. They aren't. The scalars are in R. The vectors are in R^2 and R^3.

whoops sorry, meant to say for (x y 0) x and y can be any scalar so (3x 2y) can be any
vector. of course x and y are numbers, not lines.

 

[evilpostingmong]

T(x,y,z)=(3x,2y) not (x,y). You have to show (1,0) and (0,1) are in the range. T(1/3,0,0)=(1,0), T(0,1/2,0)=(0,1). It's not true that for every S:R^3->R^2 that range(S) is R^2. Try S(x,y,z)=(3x,x). What's range(S)?

Range (S)'s basis is ={(3x, 0) (0, x)}. Consider the vector (0 ,y). Contained in R^2 is (x, y) where x=/=y. Therefore (x, y) is not in range(S) and range(S)=/=R^2 as a result.
eg this can't be a member of range(S):(3x, y) if y=/=2
But since there is a vector in R^2 where we have (3x, x) (let x=2 so (6, 2) is such vector)
which is also in Range(S), range(S) is a subspace of R^2.

 

[Dick]

Range (S)'s basis is ={(3x, 0) (0, x)}. Consider the vector (0 ,y). Contained in R^2 is (x, y) where x=/=y. Therefore (x, y) is not in range(S) and range(S)=/=R^2 as a result.
eg this can't be a member of range(S):(3x, y) if y=/=2
But since there is a vector in R^2 where we have (3x, x) (let x=2 so (6, 2) is such vector)
which is also in Range(S), range(S) is a subspace of R^2.

That's a mess. A basis for range(S) is NOT {(3x,0),(0,x)}. (3x,0) is only in range(S) if x=0. Likewise for (0,x). (3x,0) isn't even a vector, it's lots of different vectors, depending on what x is. Try again to find a basis for range(S). The vector(s) in the basis shouldn't have variables in them.

 

[evilpostingmong]

That's a mess. A basis for range(S) is NOT {(3x,0),(0,x)}. (3x,0) is only in range(S) if x=0. Likewise for (0,x). (3x,0) isn't even a vector, it's lots of different vectors, depending on what x is. Try again to find a basis for range(S). The vector(s) in the basis shouldn't have variables in them.

one question though: is x different from y? or is it that it x can or cannot be different from y?
unless you let the basis for R^3 be { x(1 1 0) y( 0 1 0) z(0 0 1)} then the basis
for range(S) is {(3, 1)} T(1 1 0)= {(3 1)}.

 

[Dick]

one question though: is x different from y? or is it that it x can or cannot be different from y?
unless you let the basis for R^3 be { x(1 1 0) y( 0 1 0) z(0 0 1)} then the basis
for range(S) is {(3, 1)}

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.

 

[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.