# Linear Algebra, proof

1. Jun 8, 2010

### jwhite2531

Hi all,
I need some proofs for my assignment, the question is like below:

Let L be a linear map from the vector space V to the vector space W.

• ker(L) is a subset of V which consists of vectors u such that Lu = 0. Is ker(L) a
vector subspace of V ? Give a proof.
• Let S be a subset of V which consists of vectors which are not contained in ker(L).
Is S a vector subspace of V ? Give a proof.
• Let T be a subset of W which consists of vectors which are contained in Range L.
Is T a vector subspace of W? Give a proof.
• Let T be a subset of W which consists of vectors which are not contained in
Range L. Is T a vector subspace of W? Give a proof.

2. Jun 9, 2010

### hgfalling

You'll need to show some of your work; what you've tried, concepts you think might be involved, etc. Then we can help you!

3. Jun 9, 2010

### jwhite2531

you are right, but unfortunately I don't have many things to show. We have done with entire linear algebra in two weeks and now I am supposed to finish this assignment. It just does not settle this fast in my mind,I am trying though. That's why I need some help. Thanks again

4. Jun 9, 2010

### lanedance

so people are happy to help lead you through the problemif you try, but won;t just do it for you ;)

If you get lost the definitions are a good place to start... so for 1) teh question is:

Is Ker(L) a subspace of V, where L is a linear map from V to V?

so what is the definition of a subspace?

5. Jun 9, 2010

### jwhite2531

well, (it says here) a subspace of a vector space is a nonempty subset that satisfies the requirements for a vector space.

6. Jun 9, 2010

### jwhite2531

ok,from the definition ker(L) is a vector subspace of V (I guess). but is this definition enough for the proof?

7. Jun 9, 2010

### lanedance

NO - you need to show it. There are quite a few properties required to show a set, equipped with addition & scalar multiplication is a vector space (~10 if i remember correctly)

To show a subset of a vector space is a "subspace" is a little less onerous, you need to show 3 things:
- it has the zero vector
- it si "closed" under addition
- it is "closed" under scalar multiplication

see this for subspace
http://en.wikipedia.org/wiki/Linear_subspace

and for a general vector space
http://en.wikipedia.org/wiki/Vector_space

Last edited: Jun 9, 2010
8. Jun 10, 2010

### jwhite2531

Thank you very much, here is my final proof :)
- ker(L) is nonempty since Lu=0, the zero vector of V, is in ker(L)
- if u belongs ker(L) and a is a scalar, then
L(au)=aL(U)=a.0=0, therefore au belongs ker(L)
-if u1, u2 belong ker(L), then
L(u1+u2)=Lu1+Lu2=0+0=0, so u1+u2 belongs ker(L)
hence ker(L) is a subspace of V. Am I right?

Proof for the second statement:
S is a subset of V which consists of vectors which are not contained in ker(L).
therefore S does not have a zero vector, 0v belongs ker(L), and S is not a subspace of V.

The last two are similar to first two.
I guess I makes sense but do you agree?
Thanks

9. Jun 10, 2010

### lanedance

subspace proof looks good

2nd statement is valid as well

another way to do the 2nd one without using the zero vector is, pick u & v such that:
$$L(u) = 0, \ u \ \in Ker(L)$$
$$L(v) \neq 0, \ v\ \in V$$
$$L(u+v) = L(u) + L(v) = 0 + L(v) \neq 0, \ \to (u+v) \in V$$
$$(u+v) + (-v) = u \notin V$$
so its not closed

10. Jun 10, 2010

### jwhite2531

Thank you.

11. Jun 11, 2010

### HallsofIvy

Well done, lanedance and jwhite2531.