Basis of a Vector Space

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In the previous post, we saw that every element of the \(\mathbb{K}\)-vector space \(\mathbb{K}[\x]\) can be expressed as a linear combination of elements of the set \(S=\{1,\x,\x^2,\ldots\}\). If such a set \(S\) is given, we can understand the properties of \(\mathbb{K}[\x]\) by examining only the elements of \(S\) instead of all elements of the vector space.

Spanning Sets

We define this situation as follows.

Definition 1 For a \(\mathbb{K}\)-vector space \(V\), a subset \(S\) of \(V\) is called a spanning set of \(V\) if, for every \(v\in V\), there exist suitable \(\alpha_1,\ldots,\alpha_n\in \mathbb{K}\) and \(x_1,\ldots, x_n\in S\) such that

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

In the previous post, we also defined linear combinations of infinitely many elements, and using this, the above definition is equivalent to saying that any element of \(V\) can be written as a linear combination of elements of \(S\).

Any \(\mathbb{K}\)-vector space \(V\) always has a spanning set. This is trivial because \(V\) spans itself. However, in this situation, it is also clear that \(S\) is too large to serve the original purpose of examining \(V\) more conveniently. We will revisit the condition for choosing a sufficiently small \(S\) shortly. First, rather than obtaining a spanning set \(S\) from a given \(V\), let us conversely find a vector space that has a given set \(S\) as its 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 \(S\) consists of a single element \(x\), then all elements of \(\span\{x\}\) are of the form

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

so \(\span \{x\}\) is also written as \(\mathbb{K}x\).

The subspace \(\span S\) is unique if it exists. If \(W,W'\) are two subspaces satisfying the condition of Definition 2, then by the minimality of \(W'\) we have \(W'\leq W\), and by the minimality of \(W\) we have \(W\leq W'\). Therefore, to justify the above definition, it suffices to show that such a subspace exists.

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

Proof

For arbitrary \(w_1, w_2\in W\), we show that \(w_1+w_2\in W\). Since \(w_1\) and \(w_2\) are each elements of \(W\), for all \(i\in I\), \(w_1,w_2\) are elements of \(W_i\), and therefore \(w_1+w_2\in W_i\) holds for all \(i\). Now by the definition of \(W\) again, \(w_1+w_2\in W\) holds.

Similarly, for arbitrary \(w\in W\) 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 also closed under scalar multiplication, and \(0\in W\) is trivial, so \(W\) is a subspace.

Now, for arbitrary \(S\), let \((W_i)_{i\in I}\) be the family of subspaces of \(V\) containing \(S\). This family is nonempty because it contains \(V\). Therefore, the intersection \(\bigcap_{i\in I} W_i\) is well-defined, and it can be seen that the space defined in this way satisfies the condition of Definition 2.

The above argument is good for showing the existence and uniqueness of \(\span S\), but it gives no information at all about what \(\span S\) actually looks like. In the following lemma, the way of defining \(\span S\) is less elegant, but it gives perfect information about the elements of \(\span S\).

Lemma 4 \(\span S\) is equal to the set of all linear combinations of elements of \(S\).

Proof

Let us denote the set of all linear combinations of elements of \(S\) by \(\langle S\rangle\). That is, all elements of \(\langle S\rangle\) can be expressed in the form of a linear combination \(\alpha_1x_1+\cdots+\alpha_nx_n\) for suitable \(x_1,\ldots, x_n\in S\), and conversely all vectors of this form are elements of \(\langle S\rangle\).

Let two elements \(v,w\) of \(\langle S\rangle\) be given by the following expressions

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

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 an arbitrary scalar \(\gamma\),

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

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

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

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

Linear Independence

Now we answer the question we postponed. That is, for a given vector space \(V\), we want to find a suitable subset \(S\subseteq V\) such that \(V=\langle S\rangle\), but with \(S\) having a certain minimality.

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

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

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

Definition 5 For a \(\mathbb{K}\)-vector space \(V\), a subset \(S\) of \(V\) is linearly independent if, whenever for some finitely supported family \((\alpha_x)_{x\in S}\)

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

holds, then necessarily \(\alpha_x=0\) for all \(x\).

Then

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

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

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

Proof

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

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

so if \(\sum_{x\in S}\alpha_xx=0\), then by uniqueness \(\alpha_x=0\) always holds. Therefore \(S\) is linearly independent.

To show the converse, assume for contradiction that \(S\) is linearly independent but that some element can be expressed as two distinct linear combinations of elements 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 all \(x\). This contradicts the assumption that \(\sum\alpha_xx\) and \(\sum\beta_xx\) are different expressions, so the proof is complete.

Basis of a Vector Space

Definition 7 For a \(\mathbb{K}\)-vector space \(V\), a set \(\mathcal{B}\subset V\) is 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 exactly equivalent to:

Whenever any element \(v\) of \(V\) is given, there exist suitable \(x_1,\ldots, x_n\in\mathcal{B}\), \(\alpha_1,\ldots, \alpha_n\in \mathbb{K}\) uniquely such that

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

In §Product of Sets, we know that the family \((\alpha_x^v)_{x\in\mathcal{B}}\) is a function from \(\mathcal{B}\) to \(\mathbb{K}\). This is the function that takes \(x\in\mathcal{B}\) and outputs \(\alpha_x^v\), and is determined solely by \(v\). Therefore, we briefly write the function value at \(x\) as \(v_x\) instead of \(\alpha_x^v\).

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

Let us look at some examples.

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

Example 9 Now, if we consider \(\mathbb{K}\) itself as a \(\mathbb{K}\)-vector space, then \(\{1\}\) is a basis of \(\mathbb{K}\). For arbitrary \(\alpha\in\mathbb{K}\), since \(\alpha=\alpha\cdot 1\), this set spans \(\mathbb{K}\), and also the only \(\alpha\in\mathbb{K}\) satisfying \(0=\alpha\cdot 1\) is \(\alpha=0\), so the linear independence condition is also satisfied.

More generally, if we 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 therefore for arbitrary \(\alpha\in\mathbb{K}\)

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

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

Given two fields \(\mathbb{K}'\supset \mathbb{K}\), then \(\mathbb{K}'\) naturally has 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 the set \(\{1,i\}\) as its basis. On the other hand, if \(\mathbb{K}'=\mathbb{R}\) and \(\mathbb{K}=\mathbb{Q}\), finding a basis of \(\mathbb{K}'\) as a \(\mathbb{K}\)-vector space is not easy. For example, the following 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 Any \(\mathbb{K}\)-vector space \(V\) has a basis.

The proof of this is accessible at the undergraduate level, but requires some set-theoretic knowledge, so we leave it for a separate post. 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 its basis. We call these the standard basis of \(\mathbb{K}^n\), and denote them by \(e_1,\ldots,e_n\). Of course, the fact that there is a standard basis means that there are non-standard bases, 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

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

can also be verified to be a basis of \(\mathbb{K}^n\) from the definition.

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