# Linear Algebra - Field Subspace

1. Sep 25, 2013

### 1LastTry

1. The problem statement, all variables and given/known data
1. Let X be a set and F a Field, and consider the vector space F(X; F) of functions from X to F. For
a subset Y$\subseteq$ X, show that the set U = {f $\in$ F(X; F) : f |Y = 0 } is a subspace of F(X; F). NB: the
expression \f |Y = 0" means that f(y) = 0 whenever y $\in$ Y .

2. Relevant equations

3. The attempt at a solution
is this similar to proving a subspace for f(x) =0 for real numbers of x ?

zero vector
for all y in Y 0(y)=0

let g also be in U, and c in Field
multiplication (cf)(y) = c(f(y)) = c(g(y)) + (cg)(y)

Last edited: Sep 25, 2013
2. Sep 25, 2013

### brmath

You are asked to show that U is a subspace of F(X;F) where F(X:F) is a vector space. That means you have to run down the list of requirements for a vector space and see if U fits, and I can see you are thinking about the right kind of thing.

I find the problem statement a little unclear, in that they say F(X;F) is a vector space, but don't say over what field. You have assumed that field is F itself. As that is the only field around, it is a natural assumption. But the scalars would be in whatever field is connected to the vector space F(X;F), and they do not say what that is. It does not have to be F.

Note that the set X is not necessarily contained in the field F. It could be anything.

The first thing you say is (f+g)(y) = f(y) + g(y). This is one of the conditions you must prove to claim that U is a vector space. How are you going to show that?

Then in dealing with the scalars (whatever they are) you have to show that cf(y) is in U.

Finally, you have to show that U contains a zero function. Now it is true that f(y) = 0 for every y in Y, but this zero function must still be in F(X;F), and so must be zero for every x not just in Y. And then you have to show that U contains an additive inverse so that for every $f \in U$ there is a $g \in U$ such that f(y) + g(y) = 0 for every $y \in Y$. Again, your potential g(x) must be in (F(X;F), and that narrows it down.

3. Sep 25, 2013

### Staff: Mentor

Since it is given that the set of functions from X to the field F is already a vector space, and U is a subset, then you DON'T need to go through all 10 or so vector space axioms. You need to show only three things:
0 (in whatever suitable form) $\in$ U
If u1 and u2 are in U, then so it u1 + u2
If u $\in$ U and k is in the field F, then ku $\in$ U

As a side note, the notation F(X; F) bothers me, as F seems to be used to denote a set of functions as well as the field F. Having two roles for one letter makes things ripe for confusion.

4. Sep 25, 2013

### 1LastTry

Uhmmm I am kinda stuck how do I prove f(y)=0 like what do I need for the 0v and addition and multiplication?

like i can come up with f(y)+f(g) = (f+g)(y)=0 how to proceed from here?
and similar with multiplication.

5. Sep 26, 2013

### brmath

1LastTry -- you have to depend on the fact that F(X;F) is a vector space. Use the vector space axioms that apply to F(X;F) which you know are true, and see how much carries over to U.

Mark44 -- I don't like the notation F(X;F) either.

6. Sep 26, 2013

### 1LastTry

maybe this will clear things up:

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7. Sep 26, 2013

### Staff: Mentor

Let's try this as a problem statement (with changes in bold):
Let X be a set and F a Field, and consider the vector space V(X; F) of functions from X to F. For
a subset Y ⊆ X, show that the set U = {f ∈ V(X; F) : f |Y = 0 } is a subspace of V(X; F). NB: the
expression "f |Y = 0" means that f(y) = 0 whenever y ∈ Y .​

For each y $\in$ Y, f(y) = 0, so it looks like U has what passes for a zero element (an additive identity).

Now take two elements of U, say u1 and u2. These are functions. Can you show that u1 + u2 is also in U? What's the criterion for a function belonging to set U?

Now take an element of U, say u, and a scalar k from the field F. Is ku $\in$ U?

If you can show the two things I've asked, then you have shown that U is a subspace of V.

All of the functions in set U belong to the vector space V, so the notions of vector addition and scalar multiplication are defined and behave according to the vector space axioms.

8. Sep 26, 2013

### Staff: Mentor

Yes, I thought that the original problem statement must have made some distinction between the field and the vector space.

9. Sep 26, 2013

### 1LastTry

criteria is that f(y) = 0

so let g(y) also be in U

f(y) + g(y) = (f+g)(y) = 0

for multiplication

c(f(y)) = (cf)(y) = 0

This is what I can come up with.

so is 0v just passes by the additive identity where f(y) + g(y) = 0 + 0 = 0?

10. Sep 26, 2013

### Staff: Mentor

I think you more-or-less have the idea, but have a hard time saying it.

So if f and g are in U, what do you conclude about f + g?

If c is in F, what do you conclude about cf?

What is the zero vector in U? (The answer is NOT 0v, whatever that means.)

11. Sep 26, 2013

### 1LastTry

f+g = 0 and cf =0?
and they are both in U?

Zero vector 0(y) = 0 ?

12. Sep 26, 2013

### Staff: Mentor

No, those aren't what I was looking for. If you have a set W that is a subset of a vector space V, what three things do you need to show to convince someone that W is a subspace of V?

13. Sep 26, 2013

### 1LastTry

that W is closed under addition and multiplication and it contains the zero vector.

But I don't understand how to show this, i mean like cant you just say f(y) + g(y) = 0 + 0 = 0 , where both f and g are in U?

14. Sep 26, 2013

### brmath

What is the zero vector in V? Does that help with U?

to Mark44: thanks for fixing the notation.

15. Sep 26, 2013

### Staff: Mentor

I would make a small change: That W is closed under vector addition and scalar multiplication. Otherwise, yes.
For the addition part, assuming that f and g are in the set U (as defined in post #7), and y $\in$ set X, then
(f + g)(y) = <you finish> , so we conclude that U is closed under vector addition.

If f is in set U, and k is a scalar in the field F, then
(kf)(y) = <you finish>, so we conclude that U is closed under scalar multiplication.

If you can finish what I started above, and can figure out what the zero "vector" is and whether it's actually in set U, you'll be done. You will have shown that U is actually a subspace of the vector space I'm calling V.

Clear?

16. Sep 26, 2013

### 1LastTry

isn't f(y) itself is a zero vector since it add itself = 0?

17. Sep 26, 2013

### Staff: Mentor

This part "since it add itself = 0" makes no sense.

The answer to the first part - "isn't f(y) itself is a zero vector" is no. Every "vector" (really, every function) in U is a zero vector due to the way U is defined.
U = {f ∈ V(X; F) : f |Y = 0 }​

For each y ∈ Y, f(y) = 0, so each f in U acts as a zero element. Suppose h is a function in V that's not also in U, and f is a function in U. Then h + f = h. So f is an additive identity (a zero element) in the same way that 5 + 0 = 5 works with numbers.

18. Sep 26, 2013

### vela

Staff Emeritus
I think you've overlooked the fact that the functions f and h are defined from X into F, not Y into F. If $x\in X-Y$, then you can't conclude that h(x)+f(x)=h(x).

19. Sep 26, 2013

### Staff: Mentor

That's a good point, and one I actually gave some thought to. In what I said, I assumed (without saying so), that any input x was an element of set Y.