LINEAR_ALGEBRA: What is a subspace of V = F(R, R), the vector space of all real funcs

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  • #1
VinnyCee
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V = F(R, R), the vector space of all real valued functions f(x) of a real variable x. Which are subspaces of V?

(A) {f | f(0) = 0}

(B) {f | f(0) = 1}

(C) {f | f(0) = f(1)}

(D) [itex]C^0(R)[/itex] = {f | f is continous}

(E) [itex]C^1(R)[/itex] = {f | f is differentiable and f' is continous}

(F) P = {f | f is a polynomial}

(G) [itex]P_d\,\,\,\,=\,\,\,\,{f\,\in\,P\,|\,deg(f)\,\le\,d}[/itex]

(H) [itex]{f\,\in\,C^1(R)\,|\,f'\,=\,f}[/itex]

I have no idea what the last five of these instances mean. Case (C) is not a subspace because it only satifies the first rule that the set {0} be in the space before it can be considered a subspace, right?

Please help, I don't understand the terminology of the last five examples!
 

Answers and Replies

  • #2
AKG
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You don't know what "polynomial", "differentiable", and "continuous" mean? If f is a polynomial, you don't know what deg(f), the degree of f, is?
 
  • #3
VinnyCee
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Yes, I know those terms, but not the [itex]C^0,\,C^1[/itex] terms or the P or [itex]P_d[/itex] terms.
 
  • #4
quasar987
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Well that's why they are defined for you

For instance, [itex]C^1(R)[/itex] denoted the set of all continuously differentiable (meaning f' is continuous) functions.
 
  • #5
Daverz
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(G) would be polynomials that look like

[itex]
f(x) = \sum_{i=0...d} a_i x^i
[/itex]
 
  • #6
VinnyCee
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Both (A) and (B) are subspaces of V, but (C) is not because each element must be unique, right? f(0) = f(1) means that it is not unique and not 1-1. But, how would I prove that (A) and (B) are subspaces of V?
 
  • #7
Daverz
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If f(0) = f(1), it's still a function (think of a horizontal line), and there's no requirement for it to be 1-1.

For (B), what is the 0 element of the set?
 
  • #8
VinnyCee
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There will be no 0 element because it always equals 1. So (B) is NOT a subset. (C) could be a horizontal line. But how would I prove these items? what would an arbitrary f(x) function look like?

[tex]f(x)\,=\,a_1\,x^{n\,+\,2}\,+a_2\,x^{n\,+\,1}\,+\,...\,+a_n\,x^n[/tex]
 
  • #9
Daverz
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It's just like the matrix case. You have functions in the set [itex]f[/itex] and [itex]g[/itex] and a real number [itex]a[/itex]. Are [itex]af[/itex], [itex]f + g[/itex],and the zero function in the set?
 
  • #10
VinnyCee
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0 function would be f = 0 or [itex]f = 0 x^2 + 0 x + 0[/itex]?

How would I write it out (prove) for case (A)?

{f | f(0) = 0}

1.) 0 [itex]\in[/itex] V.

2.) a f(0) = 0 [itex]\in[/itex] V.

3.) f(0) + g(0) = 0 + 0 = 0 [itex]\in[/itex] V.
 
Last edited:
  • #11
Daverz
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VinnyCee said:
0 function would be f = 0 or [itex]f = 0 x^2 + 0 x + 0[/itex]?

How would I write it out though for case (A)?

{f | f(0) = 0}

1.) 0 [itex]\in[/itex] V.

2.) a f(0) = 0 [itex]\in[/itex] V.

3.) f(0) + g(0) = 0 + 0 = 0 [itex]\in[/itex] V.

Those are all pretty clear, right? You just need a sentence after each of those computations saying "so <function> is in the set <whatever>." You may need a sentence explaining that the sum of a C^0 functions is also C^0. The same for C^1 functions. And similarly for a real number times a function (should be able to look this up in your calculus text.) For the polynomials you show directly by combining terms that the new function is a polynomial of whatever type is required.
 
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  • #12
VinnyCee
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What about case (B)?

It says that a function with zero plugged-in as the variable is equal to one. But this does not mean that the function cannot equal zero at some other plugged-in constant, right? So rule 1 checks out.

However, rule 3 (additive) is not satisfied becuase f(0) + g(0) = (1) + (1) = 2. Which is not 1, obviously.

Does that sound right?
 
  • #13
radou
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Yes, it does. Further on, for scalar multiplication, i.e. a real scalar [tex]a \neq 1[/tex], you'd have [tex]a\cdot f(0) = a[/tex].
 
  • #14
VinnyCee
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So, is every case except (B) and (H) a subspace of V?
 
  • #15
HallsofIvy
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VinnyCee said:
What about case (B)?

It says that a function with zero plugged-in as the variable is equal to one. But this does not mean that the function cannot equal zero at some other plugged-in constant, right? So rule 1 checks out.
No, it doesn't. In function spaces with f+ g defined by (f+g)(x)= f(x)+ g(x) (the usual definition), the 0 vector is the function f(x)= 0 for all x. Is that function in this set?

However, rule 3 (additive) is not satisfied becuase f(0) + g(0) = (1) + (1) = 2. Which is not 1, obviously.
Yes, that is true.
 

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