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# 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). If $B$ is infinite, then all bases for $V$ have the same cardinality as that of $B$ (proof). 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 $|A|\leq|B|$.

###### 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 $|A|\leq|C|=|D|\leq|B|$.

Now, we suppose that $A$ and $B$ are both finite. The case where $A=\varnothing$ is clear. So assume $A\neq\varnothing$. As $B$ spans $V$, $B\neq\varnothing$. Let $A=\{a_{1},\ldots,a_{n}\}$ and

$B=\{b_{1},\ldots,b_{m}\}$ |

and assume $m<n$. So $a_{i}\neq 0$ for all $i=1,\ldots,n$. 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

$B_{1}=\{a_{1},b_{2},\ldots,b_{m}\}.$ |

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

$B_{2}=\{a_{1},a_{2},b_{3},\ldots,b_{m}\}$ |

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

$B_{m}=\{a_{1},\ldots,a_{m}\}$ |

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

Now we can complete the proof of theorem 1.

###### Proof.

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

## Mathematics Subject Classification

16D40*no label found*13C05

*no label found*15A03

*no label found*

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