Proving a set to be a vector space,

u5022494
Messages
2
Reaction score
0
I know that a set (let's call it V) of all functions which map (R -> R) is a vector space under the usual multiplication and addition of real numbers, but i am having trouble proving it, i understand that the zero vector is f(x)=0, do i just have to prove that each element of V remains in V under additon and scalar multiplication? If that's the case then i can't work out how to go about proving it, if it isn't the case, then what exactly am i proving?
 
Physics news on Phys.org
You have to notice that when you want to see how a set of functions forms a vector space, you are *not * taking the addition operation of real numbers.
f ( c ) may be a real number for any c, but the images of a function are not the function "object" itself. You need to think about how you can take two functions, add them, and have a resulting function that is still from R to R
 
u5022494 said:
I know that a set (let's call it V) of all functions which map (R -> R) is a vector space under the usual multiplication and addition of real numbers

No, that is not true. You don't have the correct definition of addition and scalar multiplication of functions. The usual multiplication of real numbers doesn't apply, since we are dealing with a space of functions.

The correct definitions are that V is a vector space under pointwise addition and scalar multiplication. What does that mean? If f and g are elements of V, then we define f+g to be the function defined by

(f+g)(x) = f(x) + g(x).

Now we have to verify that f+g, as defined, is in fact an element of V. You should prove that.

Scalar multiplication of a function f by a scalar a is defined as

(af)(x) = a * f(x).

Again, we need to show that if f is in V, then so is af for any scalar a.

Having shown that V is closed under addition and scalar multiplication as defined, you then need to verify each of the vector space axioms.

u5022494 said:
i understand that the zero vector is f(x)=0, do i just have to prove that each element of V remains in V under additon and scalar multiplication?

Yes, you need to prove that; then you need to prove that V satisfies the vector space axioms. I think you'll find it easier once you understand how addition and scalar multiplication of functions are defined. Did you understand the definitions I gave?
 
@ SteveL27, How would i go about proving those? perhaps an example would help considerably,
What i was thinking was that i could set V := {v1, v2, ... , vn} where each V is a function which maps R -> R and thus an element of V, but i can't quite get a rigorous proof using this notation..
 
u5022494 said:
What i was thinking was that i could set V := {v1, v2, ... , vn} where each V is a function which maps R -> R and thus an element of V, but i can't quite get a rigorous proof using this notation..

I hope you can clear something up for me.

I'm not quite sure what you mean here with V=\{v_{1} ,v_{2} ,\dots ,v_{n} \}. This would imply that V has n elements, which is not true if V is a vector space. This can be seen that if V is a vector space over the field \mathbb R and f\in V, then af\in V,\,\forall a\in \mathbb R. However, there are uncountably many a\in \mathbb R, so V has much more than only n elements.

Does that make sense? I hope it helps.

Do you mean for \{v_{1} ,v_{2} ,\dots ,v_{n} \} to be a basis of V? Because if so, then I want you to ask yourself if we know that V is finite-dimensional.
 
u5022494 said:
@ SteveL27, How would i go about proving those? perhaps an example would help considerably,
What i was thinking was that i could set V := {v1, v2, ... , vn} where each V is a function which maps R -> R and thus an element of V, but i can't quite get a rigorous proof using this notation..

We need to get the definition of addition and scalar multiplication clear first. Do you understand what the definitions of f+g and af for a scalar a? Your idea is unfortunately not in the right direction, but before discussing that, it's important to get the definition of V nailed down.

Are you taking a class or working from a book? Have you worked through other examples yet of being given a set with operations, and showing that it is or isn't a vector space?
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
Back
Top