Basis of a Vector Space

Every element of the \(\mathbb{K}\)-vector space \(\mathbb{K}[\x]\) that we saw in the previous post can be written as a linear combination of elements of the set \(S=\{1,\x,\x^2,\ldots\}\). Given such a set \(S\), we can understand the properties of \(\mathbb{K}[\x]\) by looking only at the elements of \(S\) instead of all elements of the vector space.

Generating Sets

We define this situation as follows.

Definition 1 Let \(V\) be a \(\mathbb{K}\)-vector space. A subset \(S\) of \(V\) is called a spanning set of \(V\) if, whenever an arbitrary \(v\in V\) is given, there exist suitable \(\alpha_1,\ldots,\alpha_n\in \mathbb{K}\) and \(x_1,\ldots, x_n\in S\) such that we can write

\[v=\alpha_1x_1+\cdots+\alpha_nx_n\]

Here, the sum in the case \(n=0\) is understood to be \(0\).

In a previous post, we also defined linear combinations of infinitely many elements; using that, the definition above says exactly that every element of \(V\) can be written as a linear combination of elements of \(S\).

Every \(\mathbb{K}\)-vector space \(V\) has a spanning set. This is trivial, since \(V\) spans \(V\) itself. However, it is also clear that in this situation \(S\) is too large, so it does not serve our original purpose of studying \(V\) more conveniently. We will revisit the condition for choosing a sufficiently small \(S\) shortly. First, rather than extracting a spanning set \(S\) from a given \(V\), let us do the opposite: given a set \(S\), find a vector space that has \(S\) as a spanning set.

Definition 2 For a \(\mathbb{K}\)-vector space \(V\), the smallest subspace of \(V\) containing a subset \(S\) of \(V\) is called the subspace spanned by \(S\), and is denoted by \(\span S\).

In particular, if \(S\) is a set consisting of a single element \(x\), then every element of \(\span\{x\}\) is of the form

\[\alpha x,\qquad\alpha\in \mathbb{K}\]

so we also write \(\span \{x\}\) as \(\mathbb{K}x\).

The subspace \(\span S\) is unique if it exists. Indeed, if \(W,W'\) are two subspaces satisfying the conditions of Definition 2, then the minimality of \(W'\) gives \(W'\leq W\), and the minimality of \(W\) gives \(W\leq W'\). Therefore, to justify the above definition, it suffices to show that one such subspace exists.

Lemma 3 Let \(V\) be a \(\mathbb{K}\)-vector space, and let \((W_i)_{i\in I}\) be a family of subspaces of \(V\) indexed by a nonempty index set \(I\). Then \(W=\bigcap_{i\in I} W_i\) is a subspace of \(V\).

Proof

Let us show that \(w_1+w_2\in W\) for any \(w_1, w_2\in W\). Since each of \(w_1\) and \(w_2\) is an element of \(W\), both \(w_1,w_2\) are elements of \(W_i\) for every \(i\in I\), and hence \(w_1+w_2\in W_i\) holds for every \(i\). Now, by the definition of \(W\) again, we have \(w_1+w_2\in W\).

Similarly, for any \(w\in W\) and \(\alpha\in\mathbb{K}\),

\[w\in W\implies w\in W_i\text{ for all $i$}\implies \alpha w\in W_i\text{ for all $i$}\implies \alpha w\in W\]

so \(W\) is closed under scalar multiplication as well, and since \(0\in W\) is trivial, \(W\) is a subspace.

Now, for an arbitrary \(S\), let \((W_i)_{i\in I}\) be the family of subspaces of \(V\) containing \(S\). This family is nonempty since it contains \(V\). Hence the intersection \(\bigcap_{i\in I} W_i\) is well-defined, and one can see that the space defined in this way is a subspace satisfying the conditions of Definition 2.

The argument above is good for proving the existence and uniqueness of \(\span S\), but it gives no information at all about what \(\span S\) actually looks like. The way we define \(\span S\) in the next lemma is less elegant, but it also gives us complete information about the elements of \(\span S\).

Lemma 4 \(\span S\) equals the set of all linear combinations of elements of \(S\).

Proof

Let us denote by \(\langle S\rangle\) the set of all linear combinations of elements of \(S\). In other words, every element of \(\langle S\rangle\) can be written as a linear combination \(\alpha_1x_1+\cdots+\alpha_nx_n\) for some \(x_1,\ldots, x_n\in S\), and conversely every vector of this form is an element of \(\langle S\rangle\).

Suppose two elements \(v,w\) of \(\langle S\rangle\) are given by

\[v=\alpha_1x_1+\cdots+\alpha_nx_n,\quad w=\beta_1y_1+\cdots+\beta_my_m\]

respectively. Then

\[v+w=\alpha_1x_1+\cdots+\alpha_nx_n+\beta_1y_1+\cdots+\beta_my_m\]

so \(v+w\in \langle S\rangle\). Similarly, for any scalar \(\gamma\),

\[\gamma v=\gamma\alpha_1x_1+\cdots+\gamma\alpha_nx_n\]

so \(\gamma v\) also belongs to \(\langle S\rangle\). Therefore \(\langle S\rangle\) is a subspace of \(V\) containing \(S\), so by definition \(\span S\leq \langle S\rangle\).

On the other hand, by §Subspaces, ⁋Proposition 3, any subspace containing \(S\) must also contain the linear combinations of elements of \(S\), so \(\langle S\rangle\leq\span S\) holds.

In linear algebra, \(\span S\) is the more commonly used notation, but since \(\langle S\rangle\) is the more common notation in most areas of mathematics, from now on we will write the subspace generated by \(S\) as \(\langle S\rangle\).

Linear Independence

We now answer the question we had set aside. That is, given a vector space \(V\), we want to find a subset \(S\subseteq V\) such that \(V=\langle S\rangle\) holds, with \(S\) satisfying a certain minimality condition.

This notion of minimality cannot be explained simply by the size of the set. For example, consider the subset \(S=\{1,\x,\x^2,\ldots\}\) of \(\mathbb{K}[\x]\): this set satisfies \(\mathbb{K}[\x]=\langle S\rangle\), but the set

\[S'=S\cup\{1+\x\}\]

also satisfies \(\mathbb{K}[\x]=\langle S'\rangle\), while \(S\) and \(S'\) have the same cardinality as sets.

Definition 5 For a \(\mathbb{K}\)-vector space \(V\), a subset \(S\) of \(V\) is said to be linearly independent if, whenever

\[\sum_{x\in S} \alpha_xx=0\]

holds for some finitely supported family \((\alpha_x)_{x\in S}\), then necessarily \(\alpha_x=0\) for every \(x\).

Then we have

\[0=1\cdot1+1\cdot\x-1\cdot(1+\x)\]

so we see that the set \(S'\) is not linearly independent. We describe such a situation by saying that the set is linearly dependent. More generally, the following holds.

Proposition 6 For a \(\mathbb{K}\)-vector space \(V\) and an arbitrary subset \(S\) of \(V\), \(S\) is linearly independent if and only if every element of \(V\) can be expressed as a linear combination of \(S\) in at most one way.

Proof

First, if the latter condition holds, then there is at most one way to express \(0\in V\) as a linear combination of \(S\). But since

\[0=\sum_{x\in S}0x\]

if \(\sum_{x\in S}\alpha_xx=0\), then by uniqueness we always have \(\alpha_x=0\). Hence \(S\) is linearly independent.

For the converse, suppose contrary to the conclusion that \(S\) is linearly independent but some vector is expressed as two distinct linear combinations of \(S\),

\[v=\sum_{x\in S}\alpha_xx=\sum_{x\in S}\beta_xx\]

Then since \(0=v-v\),

\[0=v-v=\sum_{x\in S}\alpha_xx-\sum_{x\in S}\beta_xx=\sum_{x\in S}(\alpha_x-\beta_x)x\]

and since \(S\) is linearly independent, by definition \(\alpha_x-\beta_x=0\) holds for every \(x\). This contradicts the assumption that \(\sum\alpha_xx\) and \(\sum\beta_xx\) are distinct expressions.

Bases of Vector Spaces

Definition 7 For a \(\mathbb{K}\)-vector space \(V\), a set \(\mathcal{B}\subseteq V\) is called a basis of \(V\) if \(\mathcal{B}\) is linearly independent and \(\langle\mathcal{B}\rangle=V\).

By Lemma 4 and Proposition 6 above, this is precisely the statement that

whenever an arbitrary element \(v\) of \(V\) is given, there exists a unique finitely supported family \((\alpha_x)_{x\in\mathcal{B}}\) such that

\(v=\sum_{x\in\mathcal{B}}\alpha_xx.\)

Since this family is determined solely by \(v\), let us write it as \((\alpha_x^v)_{x\in\mathcal{B}}\). In [Set Theory] §Product of Sets, we saw that such a family is a function from \(\mathcal{B}\) to \(\mathbb{K}\), that is, a function taking \(x\in\mathcal{B}\) and returning \(\alpha_x^v\). Accordingly, we denote the value of this function at \(x\) simply by \(v_x\) instead of \(\alpha_x^v\).

We now call the (finitely supported) family \((v_x)_{x\in \mathcal{B}}\) of coefficients the coordinate representation in basis \(\mathcal{B}\) of the vector \(v\), and denote it by \([v]_\mathcal{B}\).

Let us look at a few examples.

Example 8 A basis of the \(\mathbb{K}\)-vector space \(\{0\}\) is \(\emptyset\).

Example 9 This time, regard \(\mathbb{K}\) itself as a \(\mathbb{K}\)-vector space; then \(\{1\}\) is a basis of \(\mathbb{K}\). For any \(\alpha\in\mathbb{K}\), we have \(\alpha=\alpha\cdot 1\), so this set spans \(\mathbb{K}\), and the only \(\alpha\in\mathbb{K}\) satisfying \(0=\alpha\cdot 1\) is \(\alpha=0\), so the linear independence condition also holds.

More generally, take any nonzero element of \(\mathbb{K}\) and call it \(x\); then \(\{x\}\) is a basis of \(\mathbb{K}\). Since \(x\neq 0\), the multiplicative inverse \(x^{-1}\) of \(x\) exists, and hence for any \(\alpha\in\mathbb{K}\) we have

\[\alpha=(\alpha x^{-1})\cdot x\]

and the only \(\alpha\) satisfying \(0=\alpha\cdot x\) is also \(\alpha=0\).

Given two fields \(\mathbb{K}'\supset \mathbb{K}\), the field \(\mathbb{K}'\) naturally carries the structure of a \(\mathbb{K}\)-vector space.

If \(\mathbb{K}'=\mathbb{C}\) and \(\mathbb{K}=\mathbb{R}\), then \(\mathbb{K}'\) is an \(\mathbb{R}\)-vector space with basis \(\{1,i\}\). On the other hand, when \(\mathbb{K}'=\mathbb{R}\) and \(\mathbb{K}=\mathbb{Q}\), it is not easy to find a basis of \(\mathbb{K}'\) as a \(\mathbb{K}\)-vector space. For instance, the set

\[\{\ldots,100,10,1,0.1,0.01,\ldots\}\]

is not a basis of \(\mathbb{R}\) as a \(\mathbb{Q}\)-vector space.

Nevertheless, the following theorem holds.

Theorem 10 Every \(\mathbb{K}\)-vector space \(V\) has a basis.

A proof of this is well within reach at the (mathematics) undergraduate level, but since it requires [Set Theory] §Axiom of Choice, ⁋Theorem 4 (Zorn’s lemma), we will not cover it in this category. Instead, let us look at a few more examples.

Example 11 For any field \(\mathbb{K}\), the Euclidean \(n\)-space \(\mathbb{K}^n\) has the following \(n\) vectors

\[\begin{pmatrix}1\\0\\\vdots\\ 0\end{pmatrix},\begin{pmatrix}0\\1\\\vdots\\ 0\end{pmatrix},\ldots,\begin{pmatrix}0\\0\\\vdots\\ 1\end{pmatrix}\]

as a basis. We call these the standard basis of \(\mathbb{K}^n\), and denote them by \(e_1,\ldots,e_n\), respectively. Of course, saying that a standard basis exists means that there are non-standard bases as well, and indeed

\[\begin{pmatrix}-1\\0\\\vdots\\ 0\end{pmatrix},\begin{pmatrix}0\\-1\\\vdots\\ 0\end{pmatrix},\ldots,\begin{pmatrix}0\\0\\\vdots\\ -1\end{pmatrix}\]

is also a basis of \(\mathbb{K}^n\), and one can check directly from the definition that

\[\begin{pmatrix}0\\1\\\vdots\\ 1\end{pmatrix},\begin{pmatrix}1\\0\\\vdots\\ 1\end{pmatrix},\ldots,\begin{pmatrix}1\\1\\\vdots\\ 0\end{pmatrix}\]

is also a basis of \(\mathbb{K}^n\), provided that \(\operatorname{char}\mathbb{K}\) does not divide \(n-1\).

Example 12 \(\mathbb{K}[\x]\) has the set \(\{1,\x,\x^2,\ldots\}\) as a basis. On the other hand, \(\mathbb{K}[[\x]]\) does not have this set as a basis.

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References

[Goc] M.S. Gockenbach, Finite-dimensional linear algebra, Discrete Mathematics and its applications, Taylor&Francis, 2011.

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