# all bases for a vector space have the same cardinality

In this entry, we want to show the following property of bases for a vector space^{}:

###### Theorem 1.

All bases for a vector space $V$ have the same cardinality.

Let $B$ be a basis for $V$ ($B$ exists, see this link (http://planetmath.org/ZornsLemmaAndBasesForVectorSpaces)). If $B$ is infinite^{}, then all bases for $V$ have the same cardinality as that of $B$ (proof (http://planetmath.org/CardinalitiesOfBasesForModules)). So all we really need to show is where $V$ has a finite basis.

Before proving this important property, we want to prove something that is almost as important:

###### Lemma 1.

If $A$ and $B$ are subsets of a vector space $V$ such that $A$ is linearly independent^{} and $B$ spans $V$, then $\mathrm{|}A\mathrm{|}\mathrm{\le}\mathrm{|}B\mathrm{|}$.

###### Proof.

If $A$ is finite and $B$ is infinite, then we are done. Suppose now that $A$ is infinite. Since $A$ is linearly independent, there is a superset^{} $C$ of $A$ that is a basis for $V$. Since $A$ is infinite, so is $C$, and therefore all bases for $V$ are infinite, and have the same cardinality as that of $C$. Since $B$ spans $V$, there is a subset $D$ of $B$ that is a basis for $V$. As a result, we have $|A|\le |C|=|D|\le |B|$.

Now, we suppose that $A$ and $B$ are both finite. The case where $A=\mathrm{\varnothing}$ is clear. So assume $A\ne \mathrm{\varnothing}$. As $B$ spans $V$, $B\ne \mathrm{\varnothing}$. Let $A=\{{a}_{1},\mathrm{\dots},{a}_{n}\}$ and

$$B=\{{b}_{1},\mathrm{\dots},{b}_{m}\}$$ |

and assume $$. So ${a}_{i}\ne 0$ for all $i=1,\mathrm{\dots},n$. Since $B$ spans $V$, ${a}_{1}$ can be expressed as a linear combination^{} of elements of $B$. In this expression, at least one of the coefficients (in the field $k$) can not be $0$ (or else ${a}_{1}=0$). Rename the elements if possible, so that ${b}_{1}$ has a non-zero coefficient in the expression of ${a}_{1}$. This means that ${b}_{1}$ can be written as a linear combination of $a$ and the remaining $b$’s. Set

$${B}_{1}=\{{a}_{1},{b}_{2},\mathrm{\dots},{b}_{m}\}.$$ |

As every element in $V$ is a linear combination of elements of $B$, it is therefore a linear combination of elements of ${B}_{1}$. Thus, ${B}_{1}$ spans $V$. Next, express ${a}_{2}$ as a linear combination of elements in ${B}_{1}$. In this expression, if the only non-zero coefficient is in front of ${a}_{1}$, then ${a}_{1}$ and ${a}_{2}$ would be linearly dependent, a contradiction^{}! Therefore, there must be a non-zero coefficient in front of one of the $b$’s, and after some renaming once more, we have that ${b}_{2}$ is the one with a non-zero coefficient. Therefore, ${b}_{2}$, likewise, can be expressed as a linear combination of ${a}_{1},{a}_{2}$ and the remaining $b$’s. It is easy to see that

$${B}_{2}=\{{a}_{1},{a}_{2},{b}_{3},\mathrm{\dots},{b}_{m}\}$$ |

spans $V$ as well. Continue this process until all of the $b$’s have been replaced, which is possible since $$. We have finally arrived at the set

$${B}_{m}=\{{a}_{1},\mathrm{\dots},{a}_{m}\}$$ |

which is a proper subset^{} of $A$. In addition, ${B}_{m}$ spans $V$. But this would imply that $A$ is linearly dependent, a contradiction.
∎

###### Proof.

Suppose $A$ and $B$ are bases for $V$. We apply the lemma. Then $|A|\le |B|$ since $A$ is linearly independent and $B$ spans $V$. Similarly, $|B|\le |A|$ since $B$ is linearly independent and $A$ spans $V$. An application of Schroeder-Bernstein theorem completes the proof. ∎

Title | all bases for a vector space have the same cardinality |
---|---|

Canonical name | AllBasesForAVectorSpaceHaveTheSameCardinality |

Date of creation | 2013-03-22 18:06:51 |

Last modified on | 2013-03-22 18:06:51 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 6 |

Author | CWoo (3771) |

Entry type | Result |

Classification | msc 16D40 |

Classification | msc 13C05 |

Classification | msc 15A03 |