# vector spaces are isomorphic iff their bases are equipollent

###### Theorem 1.

Vector spaces $V$ and $W$ are isomorphic iff their bases are equipollent (have the same cardinality).

###### Proof.

($\Longrightarrow$) Let $\phi:V\to W$ be a linear isomorphism. Let $A$ and $B$ be bases for $V$ and $W$ respectively. The set

 $\phi(A):=\{\phi(a)\mid a\in A\}$

is a basis for $W$. If

 $r_{1}\phi(a_{1})+\cdots+r_{n}\phi(a_{n})=0,$

with $a_{i}\in A$. Then

 $\phi(r_{1}a_{1}+\cdots+r_{n}a_{n})=0$

since $\phi$ is linear. Furthermore, since $\phi$ is one-to-one, we have

 $r_{1}a_{1}+\cdots+r_{n}a_{n}=0,$

hence $r_{i}=0$ for $i=1,\ldots,n$, since $A$ is linearly independent. This shows that $\phi(A)$ is linearly independent. Next, pick any $w\in W$, then there is $v\in V$ such that $\phi(v)=w$ since $\phi$ is onto. Since $A$ spans $V$, we can write

 $v=r_{1}a_{1}+\cdots+r_{n}a_{n},$

so that

 $w=\phi(v)=r_{1}\phi(a_{1})+\cdots+r_{n}\phi(a_{n}).$

This shows that $\phi(A)$ spans $W$. As a result, $\phi(A)$ is a basis for $W$. $A$ and $\phi(A)$ are equipollent because $\phi$ is one-to-one. But since $B$ is also a basis for $W$, $\phi(A)$ and $B$ are equipollent. Therefore

 $|A|=|\phi(A)|=|B|.$

($\Longleftarrow$) Conversely, suppose $A$ is a basis for $V$, $B$ is a basis for $W$, and $|A|=|B|$. Let $f$ be a bijection from $A$ to $B$. We extend the domain of $f$ to all of $A$, and call this extension $\phi$, as follows: $\phi(a)=f(a)$ for any $a\in A$. For $v\in V$, write

 $v=r_{1}a_{1}+\cdots+r_{n}a_{n}$

with $a_{i}\in A$, set

 $\phi(v)=r_{1}\phi(a_{1})+\cdots+r_{n}\phi(a_{n}).$

$\phi$ is a well-defined function since the expression of $v$ as a linear combination of elements of $A$ is unique. It is a routine verification to check that $\phi$ is indeed a linear transformation. To see that $\phi$ is one-to-one, let $\phi(v)=0$. But this means that $v=0$, again by the uniqueness of expression of $0$ as a linear combination of elements of $A$. If $w\in W$, write it as a linear combination of elements of $B$:

 $w=s_{1}b_{1}+\cdots+s_{m}b_{m}.$

Each $b_{i}\in B$ is the image of some $a\in A$ via $f$. For simplicity, let $f(a_{i})=b_{i}$. Then

 $w=s_{1}f(a_{1})+\cdots+s_{m}f(a_{m})=s_{1}\phi(a_{1})+\cdots+s_{m}\phi(a_{m})=% \phi(s_{1}a_{1}+\cdots+s_{m}a_{m}),$

which shows that $\phi$ is onto. Hence $\phi$ is a linear isomorphism between $V$ and $W$. ∎

Title vector spaces are isomorphic iff their bases are equipollent VectorSpacesAreIsomorphicIffTheirBasesAreEquipollent 2013-03-22 18:06:55 2013-03-22 18:06:55 CWoo (3771) CWoo (3771) 7 CWoo (3771) Result msc 13C05 msc 15A03 msc 16D40