0
$\begingroup$

I am trying to prove that a real vector space V not equal to the zero vector has infinitely many bases, using only the axioms of vector spaces. My strategy is assuming a valid basis $B_1$ = {$v_1$, ..., $v_n$}, multiplying every member of that basis by some number a $\subset \Re$ such that $B_2$ = {$av_1$, ... , $av_n$}, and showing that in general for every a $\subset \Re$, $av$ is a distinct vector.

My method is as follows:

  1. Pick a vector V in v that is not the zero vector
  2. Pick two numbers $a_1$ and $a_2$ in $\Re$ such that $a_1 \neq a_2$. Assume $a_1v = a_2v$.
  3. $a_1v - a_2v = a_2v - a_2v = 0$, where $0$ is the zero vector.
  4. $(a_1 - a_2)v = 0$, but $v \neq 0$, so
  5. $a_1 - a_2 = 0 \Rightarrow a_1 = a_2$, which is a contradiction.

My problem is the jump from 4 to 5. I don't think any of the 8 basic axioms of a vector space imply that multiplication of a non-zero vector by $0$ results in the $0$ vector, so what property of real vector spaces allows for this? Or is my method completely wrong?

$\endgroup$

1 Answer 1

1
$\begingroup$

You're overcomplicating things. You can just multiply one of the vectors by a nonzero number $a$ (and one vector in the basis exists).

The set $\{av_1,v_2,\dots,v_n\}$ is easily seen to be linearly independent, hence a basis.

For every nonzero vector $v\in V$, the map $f\colon\mathbb{R}\to V$, $f(a)=av$ is linear and injective, so indeed the vectors you get are distinct.

Why does $av=0$, with $v\ne0$, implies $a=0$? Because if $av=0$ and $a\ne0$, we can do $$ v=1v=(a^{-1}a)v=a^{-1}(av)=a^{-1}0=0 $$ There would be another check to do, namely that for $a\ne b$ (both nonzero), the sets $\{av_1,v_2,\dots,v_n\}$ and $\{bv_1,v_2,\dots,v_n\}$ are distinct, but their equality means that either $av_1=bv_1$ or $av_1=v_i$ and $bv_1=v_j$, for some $i,j>1$. The latter possibility is excluded by linear independence, the former by the already established fact that the map $f$ is injective

$\endgroup$
1
  • $\begingroup$ Great point about the overcomplication, thank you. $\endgroup$
    – Chidi
    Commented Oct 16, 2020 at 23:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .