# 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.

($\u27f9$) 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

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

($\u27f8$) 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 |
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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 |