Isomorphism of Vector Spaces

Algebra Level 2

Let $V$ and $W$ be two vector spaces over $\mathbb Q$ and there is a bijection $T:V\rightarrow W$ which preserves addition, that is, for all vectors $u$ and $v$ in $V$ ,

$T(u+v)=T(u)+T(v).$

Must $V$ and $W$ be isomorphic?

Bonus: What if $\mathbb Q$ is replaced by $\mathbb R$ ?

No Yes

1 solution

Brian Lie
Oct 5, 2018

It suffices to prove that $T$ preserves scalar multiplication. $T(0)=T(0+0)=T(0)+T(0)\implies T(0)=0$ $T(v)+T(-v)=T(v+(-v))=T(0)=0\implies T(-v)=-T(v)$ where $0$ denotes zero vector and $-v$ denotes additive inverse of $v$ .

Using mathematical induction, we have $T(nv)=nT(v), \text{for all }n\in\mathbb Z,v\in V.$ For all $r\in\mathbb Q$ , there exist two integers $p$ and $q$ such that $r=\frac pq.$ According to $qT(\frac pq v)=T(pv)=pT(v),$ we get $T(rv)=rT(v), \text{for all }r\in\mathbb Q,v\in V.$ Therefore, $V$ and $W$ are isomorphic.

For the Bonus question: See the comment below or the discussion .

The answer to the bonus is "no".

Here's the post I put in the discussion on this topic:

Using the axiom of choice, we can see that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as $\mathbb{Q}$ -vector spaces, but obviously they are not isomorphic as $\mathbb{R}$ -vector spaces.

That is, here's a counterexample: set $V = \mathbb{R}, W = \mathbb{R}^2$ and set $T : V\to W$ equal to some isomorphism of $\mathbb{Q}$ -vector spaces.

Brian Moehring - 2 years, 8 months ago

Isomorphism of Vector Spaces

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