Dimension theorem for vector spaces

In mathematics, the dimension theorem for vector spaces states that all bases of a vector space have equally many elements. This number of elements may be finite, or given by an infinite cardinal number, and defines the dimension of the space.

Formally, the dimension theorem for vector spaces states that

Given a vector space V, any two linearly independent generating sets (in other words, any two bases) have the same cardinality.

If V is finitely generated, then it has a finite basis, and the result says that any two bases have the same number of elements.

While the proof of the existence of a basis for any vector space in the general case requires Zorn's lemma and is in fact equivalent to the axiom of choice, the uniqueness of the cardinality of the basis requires only the ultrafilter lemma,^[1] which is strictly weaker (the proof given below, however, assumes trichotomy, i.e., that all cardinal numbers are comparable, a statement which is also equivalent to the axiom of choice). The theorem can be generalized to arbitrary R-modules for rings R having invariant basis number.

The theorem for finitely generated case can be proved with elementary arguments of linear algebra, and requires no forms of the axiom of choice.

Proof

Assume that { a_i: i ∈ I } and { b_j: j ∈ J } are both bases, with the cardinality of I bigger than the cardinality of J. From this assumption we will derive a contradiction.

Case 1

Assume that I is infinite.

Every b_j can be written as a finite sum

$b_j = \sum_{i\in E_j} \lambda_{i,j} a_i$, where $E_{j}$ is a finite subset of $I$.

Since the cardinality of I is greater than that of J and the E_j's are finite subsets of I, the cardinality of I is also bigger than the cardinality of $\bigcup_{j\in J} E_j$. (Note that this argument works only for infinite I.) So there is some $i_0\in I$ which does not appear in any $E_{j}$. The corresponding $a_{i_0}$ can be expressed as a finite linear combination of $b_{j}$'s, which in turn can be expressed as finite linear combination of $a_i$'s, not involving $a_{i_0}$. Hence $a_{i_0}$ is linearly dependent on the other $a_{i}$'s.

Case 2

Now assume that I is finite and of cardinality bigger than the cardinality of J. Write m and n for the cardinalities of I and J, respectively. Every a_i can be written as a sum

$a_i = \sum_{j\in J} \mu_{i,j} b_j$

The matrix $(\mu_{i,j}: i\in I, j\in J)$ has n columns (the j-th column is the m-tuple $(\mu_{i,j}: i\in I)$), so it has rank at most n. This means that its m rows cannot be linearly independent. Write $r_i = (\mu_{i,j}: j\in J)$ for the i-th row, then there is a nontrivial linear combination

$\sum_{i\in I} \nu_i r_i = 0$

But then also $\sum_{i\in I} \nu_i a_i = \sum_{i\in I} \nu_i \sum_{j\in J} \mu_{i,j} b_j = \sum_{j\in J} \biggl(\sum_{i\in I} \nu_i\mu_{i,j} \biggr) b_j = 0,$ so the $a_i$ are linearly dependent.

Alternative Proof

The proof above uses several non-trivial results. If these results are not carefully established in advance, the proof may give rise to circular reasoning. Here is a proof of the finite case which requires less prior development.

Theorem 1: If $A = (a_1,\dots,a_n) \subseteq V$ is a linearly independent tuple in a vector space $V$, and $B_0 = (b_1,...,b_r)$ is a tuple that spans $V$, then $n\leq r$.^[2]

The argument is as follows:

Since $B_{0}$ spans $V$, the tuple $(a_1,b_1,\dots,b_r)$ also spans. Since $a_1\neq 0$ (because $A$ is linearly independent), there is at least one $t \in \{1,\ldots,r\}$ such that $b_{t}$ can be written as a linear combination of $B_1 = (a_1,b_1,\dots,b_{t-1}, b_{t+1}, ... b_r)$. To see this, write $a_{1}$ as a linear combination of the $b_{i}$, and note that because ${\displaystyle a_{i}\neq 0}$, at least one of the coefficients of the $b_{i}$ must be non-zero. Isolating this term on one side of the equation and dividing by its coefficient - here we make critical use of the assumption that a vector space is always defined over a field - yields the result. Thus, $B_{1}$ is a spanning tuple, and its length is the same as $B_{0}$'s.

Repeat this process. Because $A$ is linearly independent, we can always remove an element from the list $B_{i}$ which is not one of the $a_{j}$'s that we prepended to the list in a prior step (because $A$ is linearly independent, and so there must be some nonzero coefficient in front of one of the $b_{i}$'s). Thus, after $n$ iterations, the result will be a tuple $B_n = (a_1, \ldots, a_n, b_{m_1}, \ldots, b_{m_k})$ (possibly with $k=0$) of length $r$. In particular, $A \subseteq B_n$, so $|A| \leq |B_n|$, i.e., $n \leq r$.

To prove the finite case of the dimension theorem from this, suppose that $V$ is a vector space and $S = \{v_1, \ldots, v_n\}$ and $T = \{w_1, \ldots, w_m\}$ are both bases of $V$. Since $S$ is linearly independent and $T$ spans, we can apply Theorem 1 to get $m \geq n$. And since $T$ is linearly independent and $S$ spans, we get $n \geq m$. From these, we get $m=n$.

Kernel extension theorem for vector spaces

This application of the dimension theorem is sometimes itself called the dimension theorem. Let

T: U → V

be a linear transformation. Then

dim(range(T)) + dim(kernel(T)) = dim(U),

that is, the dimension of U is equal to the dimension of the transformation's range plus the dimension of the kernel. See rank-nullity theorem for a fuller discussion.

References

1. Howard, P., Rubin, J.: "Consequences of the axiom of choice" - Mathematical Surveys and Monographs, vol 59 (1998) ISSN 0076-5376.
2. S. Axler, "Linear Algebra Done Right," Springer, 2000.