Is V a Vector Space with These Operations?

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skoomafiend
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Homework Statement



V is the set of functions R -> R; pointwise addition and (a.f)(x) = f(ax) for all x.

is V a vector space given the operations?

Homework Equations



nil.

The Attempt at a Solution



i think it is not closed under multiplication.
if r is an element of R, then
r*a(x) . r*f(x) = (r^2)*(a.f)(x)
which is not equal to
r*f(ax)

im not really sure if i even have the correct approach.
any help would be greatly appreciated.

thanks!
 
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skoomafiend said:

Homework Statement



V is the set of functions R -> R; pointwise addition and (a.f)(x) = f(ax) for all x.

is V a vector space given the operations?


Homework Equations



nil.

There are lots of relevant equations -- they comprise the definition of a vector space.

The Attempt at a Solution



i think it is not closed under multiplication.
if r is an element of R, then
r*a(x) . r*f(x) = (r^2)*(a.f)(x)
which is not equal to
r*f(ax)

im not really sure if i even have the correct approach.
any help would be greatly appreciated.

thanks!

I don't follow your argument. There is no "multiplication" of vectors in the definition of a vector space, only addition. All you need to do is pick one of the properties of a vector space that doesn't work and give a counter-example. Which property are you working with above? It might be useful to list them.
 
suppose that f(x) = x2.

is it the case that ((a+b).f)(x) = (a.f)(x) + (b.f)(x)?

(this is the distributivity of field addition over scalar mutliplication axiom).
 
skoomafiend said:

Homework Statement



V is the set of functions R -> R; pointwise addition and (a.f)(x) = f(ax) for all x.

Deveno said:
suppose that f(x) = x2.

is it the case that ((a+b).f)(x) = (a.f)(x) + (b.f)(x)?

(this is the distributivity of field addition over scalar mutliplication axiom).
If I'm understanding the problem correctly, f(x) = x2 is not a member of set V, since af(x) [itex]\neq[/itex] f(ax).
 
Mark44 said:
If I'm understanding the problem correctly, f(x) = x2 is not a member of set V, since af(x) [itex]\neq[/itex] f(ax).

You aren't. The vector space is the set of [all] functions from R to R. It's just that scalar multiplication is defined in an unusual way.