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OpenStudy (anonymous):

Hello, Determine if the subset of C(-infi,infin) is a subspace of C(-inif,infin) The set of all even functions: f(-x)=f(x)

OpenStudy (anonymous):

i said yes,

OpenStudy (anonymous):

but i am unclear as to why exactly

OpenStudy (turingtest):

since we know that C(-infi,infin) is a vector space, we have to prove two theorems in order for this to be a subspace

OpenStudy (turingtest):

1)is the subspace closed under addition? 2)is the subspace closed under scalar multiplication

OpenStudy (anonymous):

i dont think its closed under multiplication because if we mucltiply by 0 we get 0, not C

OpenStudy (amistre64):

i think it rests on 3 thrms; closure closure and zero

OpenStudy (turingtest):

the existence of the 0 element is a requirement for all vector spaces, we take it for granted here because we know it exists in C

OpenStudy (anonymous):

okay,

OpenStudy (amistre64):

im not to sure I understand the notation C(...)

OpenStudy (turingtest):

so that just leaves us with two and it should be clear that if f(-x)=f(x), then cf(-x)=cf(x) so that's proven (I think, maybe not rigorously enough)

OpenStudy (anonymous):

i believe it is continous functions

OpenStudy (turingtest):

@amistre64 yes it is the set of all continuous functions

OpenStudy (turingtest):

I'm not sure how to demonstrate what I think is a well-known theorem, which is that the sum of two even functions is also even, but that is what is required to prove the first part

OpenStudy (anonymous):

so then for subspaces, we also need to prove the 10 axioms

OpenStudy (amistre64):

x^2 + x^4 + x^6 + ... + x^(2n) is an even function

OpenStudy (turingtest):

given that the subspace we are talking about is the subspace of a known vector space, we only need to prove the two theorems I stated the other 8 come free with the fact that C(...) is a vector space

OpenStudy (turingtest):

f(-x)+g(-x)=f(x)+g(x) ....maybe that proves that this is closed under addition, not sure

OpenStudy (anonymous):

i guess i am not understanding the difference between a subset and a subspace, i know that a vector spaces had to fulfill those ten axioms

OpenStudy (turingtest):

here are the ten you refer to http://tutorial.math.lamar.edu/Classes/LinAlg/VectorSpaces.aspx if we want to know that some space W is a subspace of a vector space V, all we need to do is prove that addition and scalar multiplication are /closed/ in W since eight of the axioms that are fulfilled by V carry over to W, we only need to check the closure properties, as I said (i.e. do we stay 'in bounds' if we perform vector operations in that space) they describe how the other 8 axioms come for free in this page http://tutorial.math.lamar.edu/Classes/LinAlg/Subspaces.aspx

OpenStudy (anonymous):

so for closure under addition, in thos case we can say that f(-x)+f(-x)=f(x)

OpenStudy (turingtest):

well you need two different vectors, so call one f and the other g

OpenStudy (anonymous):

f(-x)+g(-x)=w(x)

OpenStudy (turingtest):

eh, that doesn't seem to prove anything, since we have no indication that w should be even...

OpenStudy (anonymous):

but we know what f(-x)=f(x), so both function since they are even would yeid even function, for any -x

OpenStudy (turingtest):

yes, but I think we have to try to prove that the sum of two even functions is also even

OpenStudy (anonymous):

but it is always even isnt

OpenStudy (turingtest):

I do not understand what you just said...

OpenStudy (turingtest):

here's what I'm thinking f(-x)+g(-x)=w(-x)=f(x)+g(x)=w(x) hence f+g=w where w is even this is the proof I would give

OpenStudy (anonymous):

okay,i think i have the idea, i just need to be able to formalize it into a proof, like you so kindly did

OpenStudy (turingtest):

yeah, I have trouble with this kind of things at times too I've learned a lot by practicing on this site though

OpenStudy (turingtest):

but I'm pretty sure that one works

OpenStudy (anonymous):

so for closure under mulitplication: f(-x)=f(x) c(f(-x))=c(x)---->c(f(x))=cf(x)

OpenStudy (turingtest):

I think so, but you wrote it a little strangely f(-x)=f(x) c[f(-x)]=c[f(x)]=cf(x) I think this is sufficient

OpenStudy (anonymous):

For this proof that you gave, for closure under addition, i think the w(-x) should be w(x) f(-x)+g(-x)=w(-x)=f(x)+g(x)=w(x)

OpenStudy (turingtest):

f(-x)+g(-x)=w(-x)=f(x)+g(x)=w(x) ^ you mean here?

OpenStudy (anonymous):

yes

OpenStudy (turingtest):

no, because I cannot assume that f(-x)+g(-x)=w(x) I have to basis for that assumption

OpenStudy (preetha):

Wow, what a really cool and intense discussion, with everyone working together! Made my day!

OpenStudy (anonymous):

This guy turingtest is really helping me out, i would not have been able to do my homework if it wasnt for him

OpenStudy (turingtest):

@Preetha thanks, you're always so sweet :)

OpenStudy (turingtest):

let me do it more slowly 1) let us define a function f(x)+g(x)=w(x)

OpenStudy (preetha):

Thank you Turing, see how you have helped Ryoback! I can do sweet, but not Math at your level!

OpenStudy (anonymous):

TurningTest for president

OpenStudy (turingtest):

haha, much love to all :D 2) by assumption (1) we have that f(-x)+g(-x)=w(-x)

OpenStudy (turingtest):

however since f and g are even we have that 3) f(-x)+g(-x)=f(x)+g(x)=w(x)

OpenStudy (turingtest):

hence, putting it all together we have that 4) f(-x)+g(-x)=f(x)+g(x) -> w(-x)=w(x) therefor w(x) is even, and so the set of even functions is closed under addition QED :D

OpenStudy (anonymous):

yay, nice jobTuringTest. For closure under multiplcation, i can say: c[f(-x)]=cf(-x)=cf(x)

OpenStudy (turingtest):

yeah, I believe so not too much to much explanation required for that one I don't think

OpenStudy (anonymous):

can you help me with this other one f(x)=c

OpenStudy (anonymous):

the set of all constant functions

OpenStudy (turingtest):

if it's a subspace? try to prove closure under addition: what is the sum of two constants?

OpenStudy (anonymous):

yes prove its a subspace of (-inif,inif), sum of two constants is a constnat

OpenStudy (turingtest):

right, so does that mean it's closed under addition?

OpenStudy (anonymous):

yes

OpenStudy (turingtest):

great now try to write that out mathematically as a proof (it shouldn't be very complicated)

OpenStudy (anonymous):

f(x)+g(x)=w(x)=c

OpenStudy (turingtest):

because we know the functions are constants I think it is more clear to just write f(x)=a g(x)=b let a+b=c=h(x) (where a, b, and c are elements of R) f(x)+g(x)=h(x) -> a+b=c since c is an element of R, the closure has been shown

OpenStudy (anonymous):

and for multiplication i got a[f(x))=af(x)=c*a

OpenStudy (anonymous):

were a and c are constants

OpenStudy (turingtest):

oh my first proof was better anyway, your other one is fine \(c[f(x)]=c(a)=ca=constant\) some teachers may want you to add more steps, but this is pretty clear in my opinion

OpenStudy (anonymous):

youve been a great help, i might need you for this last problem, but let me try and work it out first okay

OpenStudy (turingtest):

sure

OpenStudy (turingtest):

actually I messed up the addition proof a bit above it should be because we know the functions are constants I think it is more clear to just write f(x)=a g(x)=b let a+b=c=h(x) (where a, b, and c are constants) f(x)+g(x)=h(x) -> a+b=c since c is a constant, the closure has been shown ( I mentioned R earlier, but that doesn't help because we need to prove that we are closed not just in R, but specifically the subspace of R that is the set of all constants. basic idea works though... )

OpenStudy (anonymous):

are you still here

OpenStudy (anonymous):

So, the question asks: Let A be a fixed 2by3 matrix. Prove that the set: w={x is in R^3:Ax=[1] [2] is not a subspace of R^3

OpenStudy (turingtest):

we want to prove that\[A\vec x=\left(\begin{matrix}1 \\ 2\end{matrix}\right)\]is not a subspace of \(\mathbb R^3\) ?

OpenStudy (anonymous):

but how do i expant that Ax matrix if it is a 2x3

OpenStudy (turingtest):

expand oh...

OpenStudy (anonymous):

basically they are saying that the solutions oof the matrix have to satify teh 2 conidtions for addition and mulitplication right

OpenStudy (turingtest):

right I have to think about the multiplication part, but the scalar part is easy

OpenStudy (turingtest):

\[c(A\vec x)=c\left(\begin{matrix}1 \\ 2\end{matrix}\right)=\left(\begin{matrix}c \\ 2c\end{matrix}\right)\]which is not in W

OpenStudy (anonymous):

Cant we say that its not a subspace becauase the zero vector is not cotainied

OpenStudy (turingtest):

no, look... we are working in a supposed subspace of \(\mathbb R^3\) which we know has the zero vector, so we do not need to check that axiom that fact that\[0 \left(\begin{matrix}1 \\ 2\end{matrix}\right)=\vec 0\]just illustrates the same thing I did above with c=0 you just chose a specific c trying this trick with any c except c=1 will show that the result is not contained in W

OpenStudy (anonymous):

okay, so what we have is this: we have two equations with three variable. So we need to find x values (x1,x2,x3), we are trying to prove that the vector(x1,x2,x3) is not a subspace of R3.

OpenStudy (turingtest):

this one is a little weird for me, I have to think about it... I think I have shown that this is not a vector space simply by multiplying by a constant \(c\neq1\) So really, no more needs to be done here, but if you \(must\) show that this is nbot closed under addition as well, I'll have to think about that part I will call for some backup on this one: @phi linear algebra subspace help?

OpenStudy (anonymous):

can you explain why or how you determiend it no to be closed under multiplication

OpenStudy (turingtest):

because\[c(A\vec x)=c\left(\begin{matrix}1 \\ 2\end{matrix}\right)=\left(\begin{matrix}c \\ 2c\end{matrix}\right)\]and if \(c\neq1\) then the resultant vector on the RHS is not\[\left(\begin{matrix}1 \\ 2\end{matrix}\right)\]as is required by our vector space hence we do not have closure

OpenStudy (turingtest):

what if we had to test if f(x)=1 is a vector space? let c=3 3f(x)=3(1)=3 which is not in W similar argument above

OpenStudy (phi):

How about this for an argument: u+v closed under addition? say u in V, and v in V is (u+v) in V? A(u+v) =? [1 2] Au + Av = [1 2]+[1 2] = [2 4] ≠ [1 2] so (u+v) is not in V

OpenStudy (turingtest):

nice and simple, just how I like it :)

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