Question

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)

Answer 1

i said yes,

Answer 2

but i am unclear as to why exactly

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

okay,

Answer 9

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

Answer 10

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)

Answer 11

i believe it is continous functions

Answer 12

@amistre64 yes it is the set of all continuous functions

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

but it is always even isnt

Answer 27

I do not understand what you just said...

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

but I'm pretty sure that one works

Answer 32

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

Answer 33

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

Answer 34

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)

Answer 35

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

Answer 36

yes

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

TurningTest for president

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

yay, nice jobTuringTest. For closure under multiplcation, i can say:

c[f(-x)]=cf(-x)=cf(x)

Answer 48

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

Answer 49

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

Answer 50

the set of all constant functions

Answer 51

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

Answer 52

yes prove its a subspace of (-inif,inif),

sum of two constants is a constnat

Answer 53

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

Answer 54

yes

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

were a and c are constants

Answer 60

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

Answer 61

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

Answer 62

sure

Answer 63

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... )

Answer 64

are you still here

Answer 65

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

Answer 66

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

Answer 67

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

Answer 68

expand oh...

Answer 69

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

Answer 70

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

Answer 71

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

Answer 72

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

Answer 73

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

Answer 74

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.

Answer 75

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?

Answer 76

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

Answer 77

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

Answer 78

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

Answer 79

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

Answer 80

nice and simple, just how I like it :)