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[Linear Alg] Determining which sets are subspaces of R[x]

  • #1
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1. Homework Statement

Which of the following sets are subspaces of ##R[x]?##

##W_1 = {f \in \mathbf R[x] : f(0) = 0}##

##W_2 = {f \in \mathbf R[x] : 2f(0) = f(1)}##

##W_3 = {f \in \mathbf R[x] : f(t) = f(1-t) \forall t \in \mathbf R}##

##W_4 = {f \in \mathbf R[x] : f = \sum_{i=0}^n a_ix^2i}##

2. Homework Equations

The criteria for a given set being a subspace are that it be closed under addition and scalar multiplication, and also that the 0 vector belongs to the set.

3. The Attempt at a Solution

##W_1## is a subspace:

##f(0)+g(0) = (f+g)(0) \Rightarrow 0 + 0 = 0 \in W_1##
##\lambda \cdot f(0) = (\lambda \cdot f(0)) = \lambda \cdot 0 = 0 \in W_1##
##0v = 0 \in W_1##

##W_2## is not a subspace:

##2f(0) = f(1) \neq (2 \cdot f(0)) \Rightarrow## The condition of closure under scalar multiplication is not met.

I also suspect the 0 vector does not belong to this set, but I am unsure of how to show this.

The notation and executing the notation correctly is my biggest problem here, however, I appreciate all conceptual insight. I have thus far not been able to correctly describe whether or not ##W_3## and ##W_4## are subspaces. My suspicion is that ##W_3## is not a subspace because ##f(0) = f(1-0) = f(1) \neq f(0)## but I am uncertain of how to improve this train of thought. Similarly, I believe ##W_4## meets all the criteria to be a subspace of ##\mathbf R[x]## but I do not know how to correctly use notation to describe how it meets those criteria.

Additionally, I would like clarification about what exactly ##\mathbf R[x]## refers to.

Thank you very much.
 

Answers and Replies

  • #2
fresh_42
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1. Homework Statement

Which of the following sets are subspaces of ##R[x]?##

##W_1 = {f \in \mathbf R[x] : f(0) = 0}##

##W_2 = {f \in \mathbf R[x] : 2f(0) = f(1)}##

##W_3 = {f \in \mathbf R[x] : f(t) = f(1-t) \forall t \in \mathbf R}##

##W_4 = {f \in \mathbf R[x] : f = \sum_{i=0}^n a_ix^2i}##

2. Homework Equations

The criteria for a given set being a subspace are that it be closed under addition and scalar multiplication, and also that the 0 vector belongs to the set.

3. The Attempt at a Solution

##W_1## is a subspace:

##f(0)+g(0) = (f+g)(0) \Rightarrow 0 + 0 = 0 \in W_1##
##\lambda \cdot f(0) = (\lambda \cdot f(0)) = \lambda \cdot 0 = 0 \in W_1##
##0v = 0 \in W_1##
That's correct.
##W_2## is not a subspace:

##2f(0) = f(1) \neq (2 \cdot f(0)) \Rightarrow## The condition of closure under scalar multiplication is not met.
I don't understand this: ##2f(0) = 2 \cdot f(0)##, what else should it be? And if we have e.g. ##f(x)=a(x+1)## then ##2f(0)=2a=f(1)## plus ##\alpha \cdot a(x+1) + \beta \cdot b(x+1) = (\alpha a +\beta b)(x+1) ## is again of this form. So why isn't ##W_2## a subspace? I don't know whether it is, but your argumentation is wrong.
I also suspect the 0 vector does not belong to this set, but I am unsure of how to show this.
You can't. ##2\cdot 0(0)= 0 = 0(1)##.
The notation and executing the notation correctly is my biggest problem here, however, I appreciate all conceptual insight. I have thus far not been able to correctly describe whether or not ##W_3## and ##W_4## are subspaces. My suspicion is that ##W_3## is not a subspace because ##f(0) = f(1-0) = f(1) \neq f(0)## but I am uncertain of how to improve this train of thought. Similarly, I believe ##W_4## meets all the criteria to be a subspace of ##\mathbf R[x]## but I do not know how to correctly use notation to describe how it meets those criteria.

Additionally, I would like clarification about what exactly ##\mathbf R[x]## refers to.

Thank you very much.
You have to show: Given any ##f\; , \;g \in W_i## and any ##\alpha\; , \;\beta \in R## - whatever ##R## is - then ##\alpha \cdot f + \beta \cdot g \in W_i##. If this is the case, then the zero vector is automatically in ##W_i##. Do you see why?
 
  • #3
Ray Vickson
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1. Homework Statement

Which of the following sets are subspaces of ##R[x]?##

##W_1 = {f \in \mathbf R[x] : f(0) = 0}##

##W_2 = {f \in \mathbf R[x] : 2f(0) = f(1)}##

##W_3 = {f \in \mathbf R[x] : f(t) = f(1-t) \forall t \in \mathbf R}##

##W_4 = {f \in \mathbf R[x] : f = \sum_{i=0}^n a_ix^2i}##

2. Homework Equations

The criteria for a given set being a subspace are that it be closed under addition and scalar multiplication, and also that the 0 vector belongs to the set.

3. The Attempt at a Solution

##W_1## is a subspace:

##f(0)+g(0) = (f+g)(0) \Rightarrow 0 + 0 = 0 \in W_1##
##\lambda \cdot f(0) = (\lambda \cdot f(0)) = \lambda \cdot 0 = 0 \in W_1##
##0v = 0 \in W_1##

##W_2## is not a subspace:

##2f(0) = f(1) \neq (2 \cdot f(0)) \Rightarrow## The condition of closure under scalar multiplication is not met.

I also suspect the 0 vector does not belong to this set, but I am unsure of how to show this.

The notation and executing the notation correctly is my biggest problem here, however, I appreciate all conceptual insight. I have thus far not been able to correctly describe whether or not ##W_3## and ##W_4## are subspaces. My suspicion is that ##W_3## is not a subspace because ##f(0) = f(1-0) = f(1) \neq f(0)## but I am uncertain of how to improve this train of thought. Similarly, I believe ##W_4## meets all the criteria to be a subspace of ##\mathbf R[x]## but I do not know how to correctly use notation to describe how it meets those criteria.

Additionally, I would like clarification about what exactly ##\mathbf R[x]## refers to.

Thank you very much.
I do not believe your argument for ##W_2##. Clearly, if ##2f(0)=f(1)## then, for any ##c \in \mathbf{R}## we have ##2 c f(0) = c f(1)##, so ##c f \in W_2##. Does the "zero" function not satisfy the condition for ##W_2?##
 
  • #4
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That's correct.
I don't understand this: ##2f(0) = 2 \cdot f(0)##, what else should it be? And if we have e.g. ##f(x)=a(x+1)## then ##2f(0)=2a=f(1)## plus ##\alpha \cdot a(x+1) + \beta \cdot b(x+1) = (\alpha a +\beta b)(x+1) ## is again of this form. So why isn't ##W_2## a subspace? I don't know whether it is, but your argumentation is wrong.
I understand. Then in this case, I don't see why ##W_2## isn't a subspace:

##f(x) + g(x) = (f+g)(x) \in W_2##
##c \cdot f(x) = (c \cdot f(x)) \in W_2##
##0(x) = 0 \in W_2##

You have to show: Given any ##f\; , \;g \in W_i## and any ##\alpha\; , \;\beta \in R## - whatever ##R## is - then ##\alpha \cdot f + \beta \cdot g \in W_i##. If this is the case, then the zero vector is automatically in ##W_i##. Do you see why?
Yes, if your statement holds, it combines the conditions for closure under addition and scalar multiplication, and if those conditions are met then unless 0 is specifically excluded from the set, you can produce the 0 vector.

Thinking more about ##W_3## :

##f(t) + g(t) = f(1-t) + g(1-t) = (f+g)(1-t) \in W_3##
##c \cdot f(t) = (c \cdot f)(t) = (c \cdot f)(1-t) \in W_3##
##0(t) = 0(1-t) = 0 \in W_3##

I am nonetheless suspicious about scalar multiplication here...

Thanks for your help, fresh_42 and Ray Vickson!
 
  • #5
fresh_42
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I understand. Then in this case, I don't see why ##W_2## isn't a subspace:
Me neither.
Thinking more about ##W_3## :

##f(t) + g(t) = f(1-t) + g(1-t) = (f+g)(1-t) \in W_3##
##c \cdot f(t) = (c \cdot f)(t) = (c \cdot f)(1-t) \in W_3##
##0(t) = 0(1-t) = 0 \in W_3##
A bit dirty, but correct. The order is as follows:
##(f+g)(t)= f(t) + g(t) = \ldots \in W_3## which you have done right, except that you left out the first term. But now it should be the same
##(c \cdot f)(t)= c\cdot f(t) = c \cdot f(1-t) = (c \cdot f)(1-t) \in W_3##
I am nonetheless suspicious about scalar multiplication here...
Why? Do you mean the definition of ##c\cdot f## or the order which you mixed a little bit or the definition of ##c\cdot 0\,##?
 
  • #6
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Me neither.
A bit dirty, but correct. The order is as follows:
##(f+g)(t)= f(t) + g(t) = \ldots \in W_3## which you have done right, except that you left out the first term. But now it should be the same
##(c \cdot f)(t)= c\cdot f(t) = c \cdot f(1-t) = (c \cdot f)(1-t) \in W_3##

Why? Do you mean the definition of ##c\cdot f## or the order which you mixed a little bit or the definition of ##c\cdot 0\,##?
Haha, I think it was just my sloppy ordering that confused me.

So I have concluded that ##W_1, W_2, W_3## are all subspaces.

To think a little more about ##W_4## then:

(note, the exponent over x is supposed to be ##2i##)

##W_4 = \{f \in \mathbf R[x] : f = \sum_{i=0}^n a_ix^{2i}\}##

##a_ix^{2i} + b_ix^{2i} = (a_i+b_i)x^{2i} \in W_4##

##c \cdot a_ix^{2i} = (c \cdot a_i)x^{2i} \in W_4##

I'm not sure about the notation for a zero vector here, would it simply be ##0 \cdot a_ix^{2i}##?

EDIT: Additionally I would like to know if there is anything to be said about the notation ##\mathbf R[x]##. I'm not really sure what that notation tells me
 
Last edited:
  • #7
fresh_42
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##W_4 = \{f \in \mathbf R[x] : f = \sum_{i=0}^n a_ix^{2i}\}##
I edited the LaTeX code a bit. Exponents of more than one sign have to be put in brackets: x^{2i} and brackets as brackets ##\{\,\,\}## are actually \{ \}.
##a_ix^{2i} + b_ix^{2i} = (a_i+b_i)x^{2i} \in W_4##

##c \cdot a_ix^{2i} = (c \cdot a_i)x^{2i} \in W_4##
What has happened to the sum? We have ##f(x)=\sum_{i=0}^n a_ix^{2i}\; , \;g(x)=\sum_{j=0}^m b_jx^{2j}## and not only one summand. This doesn't change the argument, but it has to be mentioned as e.g. ##n## and ##m## doesn't have to be the same. However, we can always sum to ##N=\max\{\,n,m\,\}## and fill up the missing terms with zero coefficients ##a_i## or ##b_j##. So it is basically correct, just sloppy.
I'm not sure about the notation for a zero vector here, would it simply be ##0 \cdot a_ix^{2i}##?
No. The zero vector is just ##0##. This is the polynomial ##0=0+0\cdot x+0 \cdot x^2 + 0\cdot x^3 +\ldots## with all ##a_i=0##. What you have written is the zero scalar: ##0\cdot f(x) = 0##. This is formally not the same as ##c\cdot 0 = 0## although the result is of course the same.
EDIT: Additionally I would like to know if there is anything to be said about the notation ##\mathbf R[x]##. I'm not really sure what that notation tells me
Firstly I assume your ##\mathbf{R}## should be the real numbers, as you speak of vector spaces. One could also take a ring as scalar domain, which is usually noted as ##\mathbf{R}##, but this wouldn't lead to a vector space. Nevertheless, the real numbers should be written as \mathbb{R} which results in ##\mathbb{R}##.

##\mathbb{R}[x]## is the vector space of all polynomials ##p(x)= a_0 + a_1x+a_2x^2+a_3x^3+\ldots +a_{n-1}x^{n-1}+a_nx^n## with real numbers ##a_0,a_1,\ldots , a_n## and some natural number ##n \in \mathbb{N}## which is the degree of the polynomial, i.e. we assume ##a_n \neq 0## for otherwise we would have stopped at a lower ##n##. Another possible description is to identify ##p(x)## by the sequence ##(a_0,a_1,\ldots , a_n)##, in which case we would write ##p=(a_i)_i## and almost all ##a_i=0##, which means, all are zero except finitely many. But this notation is barely used. Normally one writes polynomials with a variable, ##x## in our case.
 
  • #8
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Thank you very much, fresh_42! I asked about that notation for ##\mathbb{R}[x]## because it appears in a subsequent exercise (which I might post for help shortly as it's rather involved, haha) and I wasn't entirely clear on its significance.

I appreciate your corrections regarding my proof that ##W_4## is also a subspace.

That just about concludes this thread: the answer to my exercise, with the accompanying proofs, is that all four sets ##W_1, W_2, W_3, W_4 ## are subspaces of ##\mathbb{R}[x]##
 

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