This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. 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 \(S=\{1,\x,\x^2,\ldots\}\). 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.
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
\[v=\alpha_1x_1+\cdots+\alpha_nx_n\]holds.
In the previous post we also defined linear combinations of infinitely many elements; using this, the above definition is equivalent to saying that any element of \(V\) can be written as a linear combination of elements of \(S\).
Every \(\mathbb{K}\)-vector space \(V\) always has a spanning set. This is obvious because \(V\) spans itself. However, in this situation \(S\) is too large, so it clearly does not serve our original purpose of studying \(V\) more conveniently. We will return shortly to the condition for choosing a
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 every element of \(\span\{x\}\) has the form
\[\alpha x,\qquad\alpha\in \mathbb{K}\]so we also write \(\span \{x\}\) as \(\mathbb{K}x\).
The subspace \(\span S\), if it exists, is unique. Indeed, 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 \(V\) be a \(\mathbb{K}\)-vector space, and let \((W_i)_{i\in I}\) be a family of subspaces of \(V\) for a nonempty index set \(I\). Then \(W=\bigcap_{i\in I} W_i\) is a subspace of \(V\).
Proof
For any \(w_1, w_2\in W\), we show that \(w_1+w_2\in W\). Since \(w_1\) and \(w_2\) are each elements of \(W\), for every \(i\in I\) we have \(w_1,w_2\in W_i\), and therefore \(w_1+w_2\in W_i\) holds for all \(i\). By the definition of \(W\), we obtain \(w_1+w_2\in W\).
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 also closed under scalar multiplication, and \(0\in W\) is obvious. Hence \(W\) is a subspace.
Now, for any \(S\), let \((W_i)_{i\in I}\) be the
This argument is good for proving the existence and uniqueness of \(\span S\), but it gives no information about what \(\span S\) actually looks like. In the next lemma, the way of defining \(\span S\) is less elegant, but it gives complete 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 by \(\langle S\rangle\) the set of all linear combinations of elements of \(S\). That is, every element of \(\langle S\rangle\) can be written 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 every vector of this form belongs to \(\langle S\rangle\).
Suppose two elements \(v,w\) of \(\langle S\rangle\) are given by
\[v=\alpha_1x_1+\cdots+\alpha_nx_n,\quad w=\beta_1y_1+\cdots+\beta_my_m.\]Then
\[v+w=\alpha_1x_1+\cdots+\alpha_nx_n+\beta_1y_1+\cdots+\beta_my_m\]so \(v+w\in S\). 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 all linear combinations of elements of \(S\), so \(\langle S\rangle\leq\span S\).
Although \(\span S\) is the more commonly used notation in linear algebra, \(\langle S\rangle\) is the more prevalent notation in most areas of mathematics. Henceforth we will write the subspace generated by \(S\) as \(\langle S\rangle\).
Linear Independence
We now address the question we postponed earlier. That is, given a vector space \(V\), we want to find a subset \(S\subseteq V\) such that \(V=\langle S\rangle\), while requiring \(S\) to have a certain minimality property.
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 having the same cardinality as \(S\).
Definition 5 For a \(\mathbb{K}\)-vector space \(V\), a subset \(S\) of \(V\) is linearly independent if whenever a finitely supported family \((\alpha_x)_{x\in S}\) satisfies
\[\sum_{x\in S} \alpha_xx=0\]then \(\alpha_x=0\) for all \(s\).
Then
\[0=1\cdot1+1\cdot\x-1\cdot(1+\x)\]so we 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 \(V\) can be expressed as a linear combination of elements of \(S\) in at most one way.
Proof
First, if the latter condition holds, then \(0\in V\) can be expressed as a linear combination of elements of \(S\) in at most one way. But
\[0=\sum_{x\in S}0x\]so if \(\sum_{x\in S}\alpha_xx=0\), then uniqueness implies that \(\alpha_x=0\) always holds. Therefore \(S\) is linearly independent.
To show the converse, assume for contradiction that \(S\) is linearly independent but an element can be written as two 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 \(\alpha_x-\beta_x=0\) for all \(x\). This contradicts the assumption that \(\sum\alpha_xx\) and \(\sum\beta_xx\) are distinct 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 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 exactly equivalent to saying that
for every element \(v\) of \(V\), there exist suitable \(x_1,\ldots, x_n\in\mathcal{B}\) and \(\alpha_1,\ldots, \alpha_n\in \mathbb{K}\),
\[v=\alpha_1x_1+\cdots+\alpha_nx_n.\]uniquely determined, such that
In [Set Theory] §Product of Sets, we saw that a family \((\alpha_x^v)_{x\in\mathcal{B}}\) is a function from \(\mathcal{B}\) to \(\mathbb{K}\). This function takes \(x\in\mathcal{B}\) and returns \(\alpha_x^v\), and is determined solely by \(v\). Therefore, we briefly write the value of this function 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 consider \(\mathbb{K}\) 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 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 \(x^{-1}\) exists, and therefore for any \(\alpha\in\mathbb{K}\),
\[\alpha=(\alpha x^{-1})\cdot x\]holds, and the only \(\alpha\) satisfying \(0=\alpha\cdot x\) is also \(\alpha=0\).
If two fields \(\mathbb{K}'\supset \mathbb{K}\) are given, then \(\mathbb{K}'\) has a natural 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 having the set \(\{1,i\}\) as a 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 instance, 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 Every \(\mathbb{K}\)-vector space \(V\) has a basis.
The proof of this is accessible at the undergraduate level, but it requires some set-theoretic background, 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 a 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 as well; 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 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 \(\mathbb{K}^n\).
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.
References
[Goc] M.S. Gockenbach, Finite-dimensional linear algebra, Discrete Mathematics and its applications, Taylor&Francis, 2011.
댓글남기기