Mathematics 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=  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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