Existence of Linear Operators with Matching Subspaces in Vector Spaces

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Homework Statement



Prove for every subspace B of vector space C, there is at least 1 linear operator L: C→C with ker (L) = B and there's at least 1 linear operator L':C→C with L'(C) = B.

Homework Equations





The Attempt at a Solution



The first operator with Ker(L) = B would be anything mapping something to 0. So this would map to 0. So by the definition of a vector space, there exists a 0 element. And since the definition of a subspace is something that has the same conditions as the vector space. Therefore, the kernel exists in both?

And for the second operator, this would be the identity element. So would the same argument as above work?
 
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LosTacos said:

Homework Statement



Prove for every subspace B of vector space C, there is at least 1 linear operator L: C→C with ker (L) = B and there's at least 1 linear operator L':C→C with L'(C) = B.

Homework Equations


The Attempt at a Solution



The first operator with Ker(L) = B would be anything mapping something to 0. So this would map to 0. So by the definition of a vector space, there exists a 0 element. And since the definition of a subspace is something that has the same conditions as the vector space. Therefore, the kernel exists in both?

And for the second operator, this would be the identity element. So would the same argument as above work?

You define what a linear operator is by defining what it does to a basis. So pick a basis for B, say ##v_1,v_2,...,v_k## and extend it to a basis for C, ##{v_1,v_2,...,v_k,v_{k+1},...,v_n}##. Now define your operators by saying what they do to each of those basis vectors.
 
Okay so B = (v1, v2, ..., vK) is a basis and C = (v1, v2, ..., vk, vk+1, ..., vn) is also a basis. I need an operator where the kernel of the linear operator is equal to the subspace B of C.
 
LosTacos said:
Okay so B = (v1, v2, ..., vK) is a basis and C = (v1, v2, ..., vk, vk+1, ..., vn) is also a basis. I need an operator where the kernel of the linear operator is equal to the subspace B of C.

Let f be your operator. What should f(v1) be?
 
LosTacos said:
Okay so B = (v1, v2, ..., vK) is a basis and C = (v1, v2, ..., vk, vk+1, ..., vn) is also a basis. I need an operator where the kernel of the linear operator is equal to the subspace B of C.
You're going to confuse yourself by using the same names for subspaces as you do for basis sets. I don't think you need to give names to the basis sets.

Let {v1, v2, ... , vk} be a basis for B, and
let {v1, v2, ... , vk, vk+1, ... , vn} be a basis for C.
 
f(v1) = 0, because the kernel is all the vectors that map to 0
 
LosTacos said:
f(v1) = 0, because the kernel is all the vectors that map to 0

Good so far. Now take a crack at defining the rest. You have to make sure all of the vectors in B map to 0 and none of the vectors in C that aren't in B map to 0.
 
Last edited:
f(v1)=0
f(v2)=0
...
f(vk)=0
f(vk+1) = a1
...
f(vn) = an

where a1,...,an are natural numbers
 
LosTacos said:
f(v1)=0
f(v2)=0
...
f(vk)=0
f(vk+1) = a1
...
f(vn) = an

where a1,...,an are natural numbers

f is supposed to map C to C. Not C to the natural numbers.
 
  • #10
f(v1)=0
f(v2)=0
...
f(vk)=0
f(vk+1) = c1
...
f(vn) = cn

where c is contained in C?
 
  • #11
LosTacos said:
f(v1)=0
f(v2)=0
...
f(vk)=0
f(vk+1) = c1
...
f(vn) = cn

where c is contained in C?

You need to be a little more definite than that. You have to be able to show that if an element of C, call it v, is not an element of B, then f(v) is not zero. If v=a1*v1+...+ak*vk+...+an*vn, where the v's are your basis vectors and the a's are scalars, how can you tell if v is in B?
 
  • #12
Let v be an element of C. Need to show that if v is not an element of B, then f(v) is not zero.
Let v = a1*v1+...+ak*vk+...+an*vn. v is an element of B, if every vector maps to 0.
 
  • #13
LosTacos said:
Let v be an element of C. Need to show that if v is not an element of B, then f(v) is not zero.
Let v = a1*v1+...+ak*vk+...+an*vn. v is an element of B, if every vector maps to 0.

No, how would you tell v is an element of C that's not in B by looking at the a's?
 
  • #14
If the a is not equal to zero. For instance L[(a1, a2, ..., an)] = [a1, a2,0,...0]. Here, a1=a2=0, but a3,...,an can have any values
 
  • #15
LosTacos said:
If the a is not equal to zero. For instance L[(a1, a2, ..., an)] = [a1, a2,0,...0]. Here, a1=a2=0, but a3,...,an can have any values

I don't see what that answer has to do with my question.
 
  • #16
B/c if the a is not 0, then it would not map to 0. Therefore, it would not be apart of the kernel and it would be in C not B.
 
  • #17
LosTacos said:
B/c if the a is not 0, then it would not map to 0.
Not true. A vector doesn't have to be the zero vector to be in the kernel.
LosTacos said:
Therefore, it would not be apart of the kernel and it would be in C not B.
 
  • #18
LosTacos said:
B/c if the a is not 0, then it would not map to 0. Therefore, it would not be apart of the kernel and it would be in C not B.

The idea here is to pick values for the ##f(v_i)## so that you can prove that to be true. Suppose ##a_{k+1}## is not zero. Is v in B? Why or why not?
 
  • #19
v is in B if the sum of a1v1 + ... + akvk+ ... + anvn = 0
So, if ak+1 is not zero, v is not in B.
 
  • #20
LosTacos said:
v is in B if the sum of a1v1 + ... + akvk+ ... + anvn = 0
So, if ak+1 is not zero, v is not in B.

Right answer. Completely wrong reason. v1,...,vk is supposed to be a basis for B. What does that mean? Define 'basis'.
 
  • #21
B is a basis iff B spans V and B is linearly independent. So, B is the set of all possible linear combinations. So, if v is not in the spanning set, it will not be in B
 
  • #22
LosTacos said:
B is a basis iff B spans V and B is linearly independent.
Yes, adding "B is a basis for V, ..."
LosTacos said:
So, B is the set of all possible linear combinations.
No. B is a set of vectors as you described above. The span of B (or span(B) is the set of all linear combinations of vectors in B.
LosTacos said:
So, if v is not in the spanning set, it will not be in B
A basis is a spanning set (the minimal spanning set), so this is trivially true.
 
  • #23
LosTacos said:
B is a basis iff B spans V and B is linearly independent. So, B is the set of all possible linear combinations. So, if v is not in the spanning set, it will not be in B

Ok, so if v is not in B then it's not in span(v1...vk). That means some of the ai's with i>k must be nonzero. It's for those vectors you have to show f(v) is nonzero. Can you write an expression for f(v)?
 
  • #24
Okay well B=(v1,v2, ..., vk) is basis and C=(v1,v2,...,vk,vk+1,..,vn) is other basis. The basis of B spans all of B and is linearly independent. The basis of C spans all of C and is also linearly independent. Pick an element of C, call it v. If v is not an element of B, then f(v) does not equal zero. Since v is not in B, v is not in span (v1,v2,...,vk). So, there must exist an ai such that i>k and does not equal zero. f(v) = aivi + ... + anvn where i>k.
 
  • #25
LosTacos said:
Okay well B=(v1,v2, ..., vk) is basis and C=(v1,v2,...,vk,vk+1,..,vn) is other basis.
C is your vector space and B is a subspace of C.
{v1, v2, ... , vk} is a basis for B, and {v1, v2, ... , vk, vk+1, vn} is a basis for C.
LosTacos said:
The basis of B spans all of B and is linearly independent.
You don't need to say this - this is what it means for a set of vectors to be a basis.
LosTacos said:
The basis of C spans all of C and is also linearly independent.
Same as above.
LosTacos said:
Pick an element of C, call it v. If v is not an element of B, then f(v) does not equal zero. Since v is not in B, v is not in span (v1,v2,...,vk). So, there must exist an ai such that i>k and does not equal zero. f(v) = aivi + ... + anvn where i>k.
 
  • #26
LosTacos said:
Okay well B=(v1,v2, ..., vk) is basis and C=(v1,v2,...,vk,vk+1,..,vn) is other basis. The basis of B spans all of B and is linearly independent. The basis of C spans all of C and is also linearly independent. Pick an element of C, call it v. If v is not an element of B, then f(v) does not equal zero. Since v is not in B, v is not in span (v1,v2,...,vk). So, there must exist an ai such that i>k and does not equal zero. f(v) = aivi + ... + anvn where i>k.

That works, but I'm not sure you are totally clear on it. f(v)=a1*f(v1)+...ak*f(vk)+...an*f(vn). That's always true. You've already figured out that f(v1)...f(vk) must be 0. And yes, the others must be nonzero. Now you need to pick f(vi) for i>k. There are lots of choices. You've chosen f(vi)=vi, which is a good one. Can you say why? Why is ##a_{k+1} v_{k+1}+...a_n v_n## never 0 if any of the a's are nonzero? What property does the set of vectors ##\{ f(v_{k+1}),...,f(v_n) \}## have to have?
 
  • #27
You choose f(vi)=vi because we know that only k elements span the ker(L). So we needed an element that was greater than those, that would not span B. We know there exists an element since B is a subset of C.

(ak+1vk+1)+ ... + anvn is never 0 if any of the a's are nonzero because from the definition of a basis, it must be linearly independent, which means that ai does not equal zero. The set of {f(vk+1),...,f(vn)} must be linearly independent and it would represent the range.
 
  • #28
For the second part, I need to find a linear operator L':C→C with L'(C) = B.

For this linear operator, must it be the identity element. Since, the linear transformation is unique, it must be the identity element?
 
  • #29
LosTacos said:
You choose f(vi)=vi because we know that only k elements span the ker(L). So we needed an element that was greater than those, that would not span B. We know there exists an element since B is a subset of C.

(ak+1vk+1)+ ... + anvn is never 0 if any of the a's are nonzero because from the definition of a basis, it must be linearly independent, which means that ai does not equal zero. The set of {f(vk+1),...,f(vn)} must be linearly independent and it would represent the range.

The first paragraph is way off. I don't know what you mean by a vector being 'greater than' another vector.

The second paragraph is pretty correct. ##\{ f(v_{k+1}),...,f(v_n) \}## only needs to linearly independent. You don't really need to pick ##f(v_i)=v_i##. You could also have chosen ##f(v_i)=v_{k-i}##. Or any other linearly independent set of vectors. If you understand that, I think you've got it.
 
  • #30
LosTacos said:
For the second part, I need to find a linear operator L':C→C with L'(C) = B.

For this linear operator, must it be the identity element. Since, the linear transformation is unique, it must be the identity element?

No, it doesn't. And it's hardly unique. And the identity doesn't work. For the identity, I, I(C)=C, not B. Define ##L'(v_i)## again in a way that makes L'(C)=B.
 
  • #31
I'm confused as to how to think about this. I need a vi where L'(C)=B. So I would need a basis for C that would map an element to B?
 
  • #32
LosTacos said:
I'm confused as to how to think about this. I need a vi where L'(C)=B. So I would need a basis for C that would map an element to B?

No! You just need to define a mapping that takes all of the vi's into B. Use the same basis as before. Some of those vi's are in B, aren't they? Just define f(vi) for each i.
 
  • #33
Okay so then {v1, v2, ... , vk} is a basis for B, and {v1, v2, ... , vk, vk+1, vn} is a basis for C.
Could I use the map where for all i>k L(vi) = L(v1) which is in B.
 
  • #34
LosTacos said:
Okay so then {v1, v2, ... , vk} is a basis for B, and {v1, v2, ... , vk, vk+1, vn} is a basis for C.
Could I use the map where for all i>k L(vi) = L(v1) which is in B.

Yes, you could. Now what to do you propose for i<=k?
 
  • #35
L(v1) = 1
L(v2) = 2
...
For all i<=k: L(vi) = i
 
  • #36
LosTacos said:
L(v1) = 1
L(v2) = 2
...
For all i<=k: L(vi) = i

I think you mean the right thing. But you certainly aren't saying the right thing. 1 and 2 aren't vectors, are they?
 
  • #37
Sorry. It would be L(v1) = v1, L(v2) = v2,... L(vi)=vi
 
  • #38
LosTacos said:
Sorry. It would be L(v1) = v1, L(v2) = v2,... L(vi)=vi

Ok, so L(vi)=vi for i<=k and L(vi)=v1 for i>k? Does that work? Is L(C)=B? Explain why? For extra points tell me some other ways you could have defined L such that L(C)=B.
 
  • #39
Yes. Since {v1, v2, ... , vk} is a basis for B, and {v1, v2, ... , vk, vk+1, vn} is a basis for C,
for i<=k, L(vi) = vi which is apart of both Basis of B and Basis of C because v1 < vi < vk. Then, for any i > k, L(vi) = L(v1) = v1 which is apart of Basis B and Basis C.
 
  • #40
I could of defined L(C)=B such that if i>k, then L(vi) = ker(L) = B = 0
 
  • #41
LosTacos said:
I could of defined L(C)=B such that if i>k, then L(vi) = ker(L) = B = 0

Yes, I THINK you've essentially got it, but you don't really express yourself very well, so I'm guessing. The point is that both ways you've defined L, L(C)=span(v1,v2,...,vk). Which defines B. There are many other ways to define L as well, agree?
 

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