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.

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

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

is a basis for $W$. If

$$r_1\varphi (a_1)+\mathrm{\cdots }+r_n\varphi (a_n)=0,$$

with $a_i\in A$. Then

$$\varphi (r_1{a}_1+\mathrm{\cdots }+r_n{a}_n)=0$$

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

$$r_1{a}_1+\mathrm{\cdots }+r_n{a}_n=0,$$

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

$$v=r_1{a}_1+\mathrm{\cdots }+r_n{a}_n,$$

so that

$$w=\varphi (v)=r_1\varphi (a_1)+\mathrm{\cdots }+r_n\varphi (a_n).$$

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

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

($⟸$) 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 $\varphi$, as follows: $\varphi (a)=f(a)$ for any $a\in A$. For $v\in V$, write

$$v=r_1{a}_1+\mathrm{\cdots }+r_n{a}_n$$

with $a_i\in A$, set

$$\varphi (v)=r_1\varphi (a_1)+\mathrm{\cdots }+r_n\varphi (a_n).$$

$\varphi$ 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 $\varphi$ is indeed a linear transformation. To see that $\varphi$ is one-to-one, let $\varphi (v)=0$. 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 $B$:

$$w=s_1{b}_1+\mathrm{\cdots }+s_m{b}_m.$$

Each $b_i\in B$ is the image of some $a\in A$ via $f$. For simplicity, let $f(a_i)=b_i$. Then

$$w=s_{1}f(a_1)+\mathrm{\cdots }+s_{m}f(a_m)=s_1\varphi (a_1)+\mathrm{\cdots }+s_m\varphi (a_m)=\varphi (s_1{a}_1+\mathrm{\cdots }+s_m{a}_m),$$

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

Title	vector spaces are isomorphic iff their bases are equipollent
Canonical name	VectorSpacesAreIsomorphicIffTheirBasesAreEquipollent
Date of creation	2013-03-22 18:06:55
Last modified on	2013-03-22 18:06:55
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	7
Author	CWoo (3771)
Entry type	Result
Classification	msc 13C05
Classification	msc 15A03
Classification	msc 16D40