# Homework Help: The kernel exists in both?

1. May 8, 2013

### LosTacos

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

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.

2. Relevant equations

3. 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?

2. May 8, 2013

### Dick

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.

3. May 8, 2013

### LosTacos

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.

4. May 8, 2013

### Dick

Let f be your operator. What should f(v1) be?

5. May 8, 2013

### Staff: Mentor

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.

6. May 8, 2013

### LosTacos

f(v1) = 0, because the kernel is all the vectors that map to 0

7. May 8, 2013

### Dick

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: May 8, 2013
8. May 9, 2013

### LosTacos

f(v1)=0
f(v2)=0
.......
f(vk)=0
f(vk+1) = a1
.........
f(vn) = an

where a1,....,an are natural numbers

9. May 9, 2013

### Dick

f is supposed to map C to C. Not C to the natural numbers.

10. May 9, 2013

### LosTacos

f(v1)=0
f(v2)=0
.......
f(vk)=0
f(vk+1) = c1
.........
f(vn) = cn

where c is contained in C?

11. May 9, 2013

### Dick

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. May 9, 2013

### LosTacos

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. May 9, 2013

### Dick

No, how would you tell v is an element of C that's not in B by looking at the a's?

14. May 9, 2013

### LosTacos

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. May 9, 2013

### Dick

I don't see what that answer has to do with my question.

16. May 9, 2013

### LosTacos

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. May 9, 2013

### Staff: Mentor

Not true. A vector doesn't have to be the zero vector to be in the kernel.

18. May 9, 2013

### Dick

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. May 9, 2013

### LosTacos

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. May 9, 2013

### Dick

Right answer. Completely wrong reason. v1,...,vk is supposed to be a basis for B. What does that mean? Define 'basis'.

21. May 9, 2013

### LosTacos

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. May 9, 2013

### Staff: Mentor

Yes, adding "B is a basis for V, ..."
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.
A basis is a spanning set (the minimal spanning set), so this is trivially true.

23. May 9, 2013

### Dick

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. May 9, 2013

### LosTacos

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. May 9, 2013

### Staff: Mentor

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.
You don't need to say this - this is what it means for a set of vectors to be a basis.
Same as above.