MHB Vector Space - Proving Associativity

[Dethrone]

Let $V$ be a vector space, and define $V^n$ to be the set of all n-tuples $(v_1, v_2,...,v_n)$ of n vectors $v_i$, each belonging to $V$. Define addition and scalar multiplcation in $V^n$ as follows:
$(u_1,u_2,...,u_n)+(v_1,v_2,...,v_n)=(u_1+v_1, u_2+v_2,...,u_n+v_n)$
$a(v_1,v_2,...,v_n)=(av_1,av_2,...,av_n)$, $a \in \Bbb{R}$

Proving this is quite trivial, but I'm quite confused about something. In proving that it is associative, then $(u+v)+w=u+(v+w)$. Let $u=(u_1,u_2,...,u_n), v=(v_1,v_2,...,v_n), w=(w_1,w_2,...,w_n)$, where $u, v,w \in V$. $\left[\left((u_1,u_2,...,u_n)+(v_1,v_2,...,v_n)\right)+(w_1,w_2,...,w_n)\right]$
$=[(u_1+v_1, u_2+v_2,...,u_n+v_n)+(w_1,w_2,...,w_n)]$
$=[(u_1+v_1)+w_1, (u_2+v_2)+w_2,...,(u_n+v_n)+w_n]$
$=[u_1+(v_1+w_1), u_2+(v_2+w_2),...,u_n+(v_n+w_n)]$

Now, at this step, my TA justifies this step of switching the brackets by saying since $u_1$, $v_1$, and $w_1$ are in the vector space, then by the associative axiom $(u_1+v_1)+w_1=u_1+(v_1+w_1)$. I'm not sure if I agree with that...aren't we trying to prove that it satisfies the associative axiom, so why are we using that in our proof? This is what I think it should be: since the components of the tuples are real numbers, then they are equivalent because the addition of real numbers is associative. Am I right?

 

[Opalg]

Let $V$ be a vector space, and define $V^n$ to be the set of all n-tuples $(v_1, v_2,...,v_n)$ of n vectors $v_i$, each belonging to $V$. Define addition and scalar multiplcation in $V^n$ as follows:
$(u_1,u_2,...,u_n)+(v_1,v_2,...,v_n)=(u_1+v_1, u_2+v_2,...,u_n+v_n)$
$a(v_1,v_2,...,v_n)=(av_1,av_2,...,av_n)$, $a \in \Bbb{R}$

Proving this is quite trivial, but I'm quite confused about something. In proving that it is associative, then $(u+v)+w=u+(v+w)$. Let $u=(u_1,u_2,...,u_n), v=(v_1,v_2,...,v_n), w=(w_1,w_2,...,w_n)$, where $u, v,w \in V$. $\left[\left((u_1,u_2,...,u_n)+(v_1,v_2,...,v_n)\right)+(w_1,w_2,...,w_n)\right]$
$=[(u_1+v_1, u_2+v_2,...,u_n+v_n)+(w_1,w_2,...,w_n)]$
$=[(u_1+v_1)+w_1, (u_2+v_2)+w_2,...,(u_n+v_n)+w_n]$
$=[u_1+(v_1+w_1), u_2+(v_2+w_2),...,u_n+(v_n+w_n)]$

Now, at this step, my TA justifies this step of switching the brackets by saying since $u_1$, $v_1$, and $w_1$ are in the vector space, then by the associative axiom $(u_1+v_1)+w_1=u_1+(v_1+w_1)$. I'm not sure if I agree with that...aren't we trying to prove that it satisfies the associative axiom, so why are we using that in our proof? This is what I think it should be: since the components of the tuples are real numbers, then they are equivalent because the addition of real numbers is associative. Am I right?

The components of the tuples are not real numbers, but elements of the space $V$. You are told that $V$ is a vector space, so you can assume that addition in $V$ is associative. If you replace "real numbers" by "elements of $V$", then your reasoning is correct. Addition in $V$ is associative, and it follows that addition in $V^n$ is associative.

 

[Dethrone]

I get it! But what if we weren't told that $V$ was a vector space? For example: The set $V$ of all ordered pairs $(x,y)$ with the addition and scalar multiplication of $\Bbb{R^2}$. Prove whether or not it is a vector space.
$(u+v)+w=((x_1,x_2)+(y_1,y_2))+(z_1,z_2)=((x_1+y_1)+z_1,(x_2+y_2)+z_1)$
Now that we don't know that $V$ is a vector space, then can I say that since $x_1,y_1,z_1$ are all real numbers (assuming that the vector field is of real numbers), and we know that the addition of real numbers are associative, then $(x_1+y_1)+z_1=x_1+(y_1+x_1)$?

 

[I like Serena]

I get it! But what if we weren't told that $V$ was a vector space? For example: The set $V$ of all ordered pairs $(x,y)$ with the addition and scalar multiplication of $\Bbb{R^2}$. Prove whether or not it is a vector space.
$(u+v)+w=((x_1,x_2)+(y_1,y_2))+(z_1,z_2)=((x_1+y_1)+z_1,(x_2+y_2)+z_1)$
Now that we don't know that $V$ is a vector space, then can I say that since $x_1,y_1,z_1$ are all real numbers (assuming that the vector field is of real numbers), and we know that the addition of real numbers are associative, then $(x_1+y_1)+z_1=x_1+(y_1+x_1)$?

Hey Rido!

Yep. That is correct. ;)

So from this you can deduce that addition of the elements in V is associative.

To prove V is a vector space, you have to go through the whole lot though (see Definition of a Vector Space). (Nerd)

 

[Dethrone]

Thanks for the help, Opalg and ILS! :D

To prove V is a vector space, you have to go through the whole lot though (see Definition of a Vector Space). (Nerd)

Yes...(Crying), the simple ones like these are quite tedious, but at least I find the more difficult ones fun to prove :D