# Determining subspaces for all functions in a Vector space

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1. Oct 14, 2015

### Andrew Pierce

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

First, I'd like to say that this question is from an Introductory Linear Algebra course so my knowledge of vector space and subspace is limited. Now onto the question.

Q: Which of the following are subspaces of F(-∞,∞)?
(a) All functions f in F(-∞,∞) for which f(0) = 0
(b) All functions f in F(-∞,∞) for which f(0) = 1
(c) All functions f in F(-∞,∞) for which f(-x) = f(x)
(d) All polynomials of degree 2

2. Relevant equations

Definitions: A subset W of a vector space V is called a subspace of V if W is itself a vector space under the addition and scalar multiplication deﬁned on V.

Theorems: If W is a set of one or more vectors in a vector space V, then W is a subspace of V if and only if the following conditions are satisﬁed.
( a ) If u and v are vectors in W, then u + v is in W.
( b ) If k is a scalar and u is a vector in W, then ku is in W.

Subspaces:

Pn < C^∞ (-∞,∞) < C^m (-∞,∞) < C^1 (-∞,∞), < C(-∞,∞) < F(-∞,∞)
(as defined by a diagram in my book)

3. The attempt at a solution

I am absolutely stuck. I have read through the chapter but it is general solutions with vague wording that tells you to prove it yourself. "We leave it for you to convince yourself that the vector spaces discussed in Examples 7 to 10 are“nested” one inside the other as illustrated in Figure 4.2.5. Copyright | Wiley | Elementary Linear Algebra | Edition 11". I know it's my lack of understanding of subspaces and vector spaces that is preventing me from doing this kind of problem. I understand the R3 space somewhat but I'm stuck on transferring what I know from that vector space to this function vector space.

I understand that a subspace has to be defined under the addition and scalar multiplication of that set. But I do not know how that applies to these functions.

A previous question I worked out to show my logic in thinking, (it is probably flawed so please point out my errors):

Determine Which of the following are subspaces of R3:

(a,0,0)

this is a subspace of R3 because it has the zero vector for
a=0
and can be added to itself under a= 0 to get 0
also multiplying any real number to this vector if a= 0 still gives 0

(a,b,c) where b= a + c + 1

not a subspace of R3 because it fails to have the zero vector for any a or c that influences b

test:
for a = 0
c = 0
b = (0) + (0) + 1
=1

a=-1
c=0
b= -1 + 0 +1
=0

So this is my thought process for these problems.

2. Oct 14, 2015

### andrewkirk

The elements of your vector space are functions from $\mathbb{R}$ to $\mathbb{R}$. I am not entirely sure what is meant by your notation F(-∞,∞). It may mean that for a function to be in the vector space its integral over (-∞,∞) must exist and be finite. But whether that additional requirement applies won't matter until part (d).

The sum (f+g) of two functions f and g is defined as the function that maps x to f(x)+g(x).
The result of multiplying function f by real scalar k is defined as the function that maps x to kf(x).

With that under your belt, try to prove that criteria (a) and (b) in your Theorem hold for the first case ('functions f in F(-∞,∞) for which f(0) = 0').

3. Oct 14, 2015

### Andrew Pierce

Ok so could I set g to be g(x) and g(0)=0
then g(0) + f(0) = 0

then setting k=0
kf(x) = 0
making it constrained as a subspace of F(-∞,∞)?

I don't know if I'm focusing to much on zero or if that is the best way to go about solving these problems.

4. Oct 14, 2015

### andrewkirk

Well not the best way for 'these problems' generally, but in (a) the only constraint on elements of the subspace is that they must have function value zero at x=0. So the value at x=0 is all you need to focus on in that case.

Checking for scalar multiplication by k=0 is always good, because it tests whether the subset includes the zero vector, which is necessary for it to be a subspace. But you should also check for multiplication by nonzero scalars. Does it still work in that case? Can we conclude that it works for all scalars?

5. Oct 14, 2015

### Andrew Pierce

Thanks so much for all your help! Alright, so, for checking scalar multiplication I should set f(x) = 0? then multiply by any nonzero scalar and the answer will still come out to be zero?

Then onto part (b)
All functions f in F(-∞,∞) for which f(0) = 1

could I set g(x) to be g(0) = -1
and add f(0) + g(0) = 0?

then k=0
kf(x) = 0

but, if k isn't equal to zero and
f(0) = 1
does that mean that the zero vector doesn't exist?
Thereby not being a subspace of F(-∞,∞)?

6. Oct 15, 2015

### HallsofIvy

Staff Emeritus
Yes, that is the basic idea. In (a), if f and g are both functions having f(0)= 0 and g(0)= 0, then (f+ g)(0)= 0 and kf(0)= 0 for any k so this is as subspace. In (b), if f and g are both functions having f(0)= 1 and g(0)= 1, then (f+ g)(0)= 2 so this is NOT a subspace.

"(d) All polynomials of degree 2" is just a little different. Suppose $f(x)= x^2+ 2x- 1$ and $g(x)= -x^2+ x+ 3$. Are f and g in this set? What about f+ g?

7. Oct 15, 2015

### Fredrik

Staff Emeritus
You should tell us what you mean by F(-∞,∞). This isn't a standard notation.

Your notation and language is rather sloppy. I don't understand what you're doing. I suggest that you start each problem by choosing a notation for the set that you're supposed to examine to see if it's a subspace. For example, in part (a), you can start like this: Let S be the set of all f in F(-∞,∞) such that f(0)=0.

To prove that S is a subspace of F(-∞,∞), it's sufficient to prove the following three things:

1. S is closed under addition.
2. S is closed under scalar multiplication.
3. S is not empty.

The standard way to prove 3 is to prove that S contains the identity element of the addition operation on F(-∞,∞), i.e. the "zero vector" of F(-∞,∞). To do this, you have to know what the zero vector of F(-∞,∞) is. Do you? (Most people who come here to ask how to prove that a given subset is a subspace don't know this).

I'll show you how to do part 1 of problem (a). I'm leaving parts 2 and 3 to you.

Let f,g be elements of S. Since f(0)=0 and g(0)=0, we have (f+g)(0)=f(0)+g(0)=0+0=0, so f+g is in S. Since f and g are arbitrary elements of S, this means that S is closed under addition.