Basis and Dimension

Let \(V\) be a vector space over a field \(\mathbb{F}\). A basis for \(V\) is a list of vectors in \(V\) which \(V\).

A list \(v_1, \dots, v_m\) of vectors in \(V\) is linearly independent if \(\sum_{i=1}^m a_i v_i = 0 \) implies all the \(a_i =0\).

The span of a list of vectors is the set of all linear combinations of vectors.

A list \(v_1, \dots, v_m\) is a basis iff every \(v \in V\) has a unique representation \(v = \sum_{i=1}^m a_i v_i\); here by uniqueness we mean there is only one choice of field elements \(a_1, \dots, a_m\) such that \(v = \sum_{i=1}^m a_i v_i\) (Proposition Axler 2.29). Here, the fact that every vector can be written this way corresponds to the fact that the list is a spanning list, while the uniqueness of anything expressable as a linear combination of the list elements corresponds to the list being linearly independent.

Note how close this "has a unique representation" definition is to the definition of direct sum. In fact we can say that a list of vectors \(v_1, \dots, v_m\) is a basis for \(V\) iff \(V\) is the of the one-dimensional subspaces \(U_i := \text{span}(v_i)\) spanned by the vectors in the list.

A sum of subspaces \(W:=U_1 + \cdots + U_m\) of a vector space \(V\) is a direct sum if every vector \(v\) in the sum has a unique representation as a sum of elements \(v = u_1 + \cdots + u_m\), one in each space. In this case we write the sum as \(W = U_1 \oplus \cdots \oplus U_m\).

Every spanning list can be reduced to a basis (Axler 2.31) and every linearly independent list can be extended to a basis (Axler 2.33). You will need to make use of both of these facts frequently in proofs. Since a by definition, has a finite spanning list, by reducing to a basis we can see that every finite-dimensional vector space has a basis.

If some finite list of vectors \( v_1, \dots, v_m \) spans \(V\), we say \(V\) is finite-dimensional. If it is impossible for any finite list of vectors to span \(V\), we say \(V\) is infinite dimensional.

We can also develop our first notion of a complementary vector space (Axler Proposition 2.34). If we have a finite-dimensional vector space \(V\) with a subspace \(U\), then there exists another subspace \(W\) such that \(V = U \oplus W\).

To prove this, just consider the subspace \(U\) (which you should convince yourself is also finite dimensional). Being finite dimensional , \(U\) has a basis, and we can extend this basis to a basis for all of \(V\). The new vectors we added in this extension then span a new space, which intersects \(U\) only at the zero vector, yet \(V = U + W\), so \(V = U \oplus W\) by the two-subspace version of the direct sum criterion.

Dimension

One way to think about dimension is that it is a measure of the size of a vector space. It should satisfy laws like the cardinality of finite sets: let \(A,B\) be finite sets, then \begin{equation} |A \cup B| = |A| + |B| - |A \cap B|. \label{eq:cardinality} \end{equation} Here \(|A|\) means the number of elements in \(A\). Similarly if \(A,B\) are subsets of \(\mathbb{R}^2\), we have a similar rule for area: \begin{equation} \text{area}(A \cup B) = \text{area}(A) + \text{area}(B) - \text{area}(A \cap B). \label{eq:area} \end{equation}

Now this is not literally true for vector spaces, because whenever we discuss vector spaces we don't want to think about sets and subsets; everything should be a vector space. We've already seen that if \(U,W\) are subspaces of a vector space \(V\), then their intersection \(U \cap W\) is also a subspace. However their union is not; instead we should use the smallest subspace containing the union, namely \(U+W\) in the version of equations \( \eqref{eq:cardinality} \) and \(\eqref{eq:area}\) applicable to dimension.

Before we show that there is a definition of the dimension of a vector space that has this property, let's consider a proposition (Axler 2.23) that follows from the :

The Linear Dependence Lemma states that if \(v_1, \dots, v_m\) is a linearly dependent list in \(V\), then there exists \(j \in \{1,\dots, m\} \) such that \(v_j \in \text{span}(v_1, \dots, v_{j-1})\) and removing \(v_j\) from the list doesn't change its span.

The length of any linearly independent list in a finite-dimensional vector space is at most the length of any spanning list for that space.

Now a basis is both a spanning list and a linearly independent list. So, in a finite dimensional vector space, every basis is at most as long as any other. Thus every basis of a finite dimensional vector space has the same length; we call this length the dimension.

The dimension \(\text{dim}(V)\) of a finite-dimensional vector space \(V\) is the length of any (and every) basis for that space.

This definition satisfies a formula like \( \eqref{eq:cardinality} \) and \(\eqref{eq:area}\), except that we use the sum instead of the union (Axler Proposition 2.43).

Let \(U,W\) be subspaces of a finite dimensional vector space. Then \begin{equation*} \text{dim}(U+W) = \text{dim}(U) + \text{dim}(W) - \text{dim}(U \cap W). \end{equation*}