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.
Subspaces
Looking at §Vector Spaces, ⁋Example 6, we can see that a subset of a vector space often itself forms a vector space. Let us define this as follows.
Definition 1 For a \(\mathbb{K}\)-vector space \(V\), a subset \(W\) of \(V\) is called a subspace of \(V\) if the operations obtained by restricting the addition and scalar multiplication defined on \(V\) to \(W\) again define a \(\mathbb{K}\)-vector space on \(W\). We denote this by \(W\leq V\).
By definition, \(C^k(I)\) is a subspace of \(C(I)\), and \(C(I)\) is a subspace of \(\Fun(I,\mathbb{R})\).
To check whether an arbitrary subset \(W\) of \(V\) is a subspace using the definition directly, we would have to verify whether addition forms an abelian group, whether scalar multiplication satisfies all the conditions of §Vector Spaces, ⁋Definition 1, and so on. However, since the addition and scalar multiplication to be defined on \(W\) are inherited from \(V\), there are several properties that we do not need to check separately.
For example, for arbitrary \(w_1,w_2\in W\), we do not need to verify whether
\[w_1+w_2=w_2+w_1\]holds. The two elements \(w_1,w_2\) are elements of \(V\) before they are elements of \(W\), and the addition \(+\) in \(W\) is just the addition in \(V\) restricted to \(W\). Based on this, the properties we need to check are as follows.
- We must separately check whether \(W\) is closed under addition.
- Similarly, we must check whether \(W\) has an identity element and inverses for addition. Of course \(V\) contains \(0\) and \(-w\), but there is no guarantee that these lie in \(W\).
- Also, for any scalar \(\alpha\in\mathbb{K}\) and \(w\in W\), we must check whether \(\alpha w\in W\).
But we can simplify the conditions a little further. If \(W\) is closed under scalar multiplication, then by §Vector Spaces, ⁋Proposition 2 and Vector Spaces, ⁋Corollary 3, the second condition can be omitted entirely. Since \(W\) is closed under scalar multiplication, \(0w\in W\) and \((-1)w\in W\) must hold, and these are \(0\) and \(-w\) respectively. Thus we have just proved the following proposition.
Proposition 2 For a \(\mathbb{K}\)-vector space \(V\), a nonempty subset \(W\) of \(V\) is a subspace of \(V\) if and only if \(W\) is closed under addition and scalar multiplication.
However, to show that \(W\) is not empty, showing that \(0\in W\) is the easiest, so in practice there is little difference in utility between the three conditions presented earlier and the preceding proposition.
Linear Combinations
Consider a \(\mathbb{K}\)-vector space \(V\) and its subspace \(W\). Since the sum of any two elements of \(W\) is again an element of \(W\), by induction finite sums are again elements of \(W\). More generally, the following holds.
Proposition 3 Let \(V\) be a \(\mathbb{K}\)-vector space and \(W\) be a subspace of \(V\). For elements \(w_1,\ldots, w_n\) of \(W\) and scalars \(\alpha_1,\ldots,\alpha_n\), the following finite sum
\[\sum_{i=1}^n\alpha_i w_i=\alpha_1w_1+\alpha_2w_2+\cdots+\alpha_nw_n\tag{1}\]is an element of \(W\).
Proof
We proceed by induction. The case \(n=1\) is trivial, so we begin with \(n=2\). In this case, by Proposition 2, \(\alpha_1w_1\) and \(\alpha_2w_2\) are each elements of \(W\), and therefore their sum \(\alpha_1w_1+\alpha_2w_2\) is also an element of \(W\).
For general \(n\), since addition in \(W\) is associative,
\[\alpha_1w_1+\alpha_2w_2+\cdots+\alpha_nw_n=(\alpha_1w_1+\cdots\alpha_{n-1}w_{n-1})+\alpha_nw_n\]holds. Now by the inductive hypothesis, \(\alpha_1w_1+\cdots\alpha_{n-1}w_{n-1}\) and \(\alpha_nw_n\) are each elements of \(W\), and therefore their sum \(\sum_{i=1}^n\alpha_iw_i\) is also an element of \(W\).
Vectors of the form (1) in the above proposition are generally given the following name.
Definition 4 For a \(\mathbb{K}\)-vector space \(V\) and its elements \(v_1,\ldots, v_n\), a linear combination of these elements is a vector of the form
\[\alpha_1v_1+\cdots+\alpha_nv_n\]More generally, when infinitely many elements \((v_i)_{i\in I}\) of \(V\) are given, their linear combination is defined as
\[\sum_{i\in I}\alpha_iv_i\qquad\text{$\alpha_i=0$ for all but finitely many $i$}\]For example, if we view \(\mathbb{R}\) as a \(\mathbb{Q}\)-vector space as in §Vector Spaces, ⁋Example 4, then \(0.111\ldots\) is not a linear combination of the vectors
\[0.1,\quad 0.01,\quad0.001,\quad\cdots\]Indeed, if we were to represent \(0.111\ldots\) as the infinite sum above, all the coefficients would be nonzero. In a similar vein, let us consider the following example.
Example 5 Define the set \(\mathbb{K}[\x]\) as
the set of polynomials in \(\x\) with coefficients in \(\mathbb{K}\)
That is, each element of \(\mathbb{K}[\x]\) is of the form
\[p(\x)=\alpha_n\x^n+\alpha_{n-1}\x^{n-1}+\cdots+\alpha_1\x+\alpha_0\]for some natural number \(n\) and \(\alpha_i\in\mathbb{K}\). In this case, the natural number \(\max\supp(\alpha_i)=n\) is called the degree of \(p(\x)\), and \(\alpha_n\x^n\) is called the leading term. A polynomial whose leading coefficient is 1 is called a monic polynomial. Now suppose another element of \(\mathbb{K}[\x]\)
\[q(\x)=\beta_m\x^m+\beta_{m-1}\x^{m-1}+\cdots+\beta_1\x+\beta_0\]is given. Then their sum is, if \(m>n\),
\[\sum_{i=0}^na_i\x^i+\sum_{i=0}^mb_i\x^i=\sum_{i=0}^m c_i\x^i,\qquad c_i=\begin{cases}a_i+b_i&\text{if $0\leq i\leq n$}\\ b_i&\text{if $n < i\leq m$}\end{cases}\]and in the opposite case
\[\sum_{i=0}^na_i\x^i+\sum_{i=0}^mb_i\x^i=\sum_{i=0}^m c_i'\x^i,\qquad c_i'=\begin{cases}a_i+\beta_i&\text{if $0\leq i\leq m$}\\ a_i&\text{if $m < i\leq n$}\end{cases}.\]Also, for any scalar \(\gamma\in\mathbb{K}\),
\[\gamma p(\x)=\gamma\alpha_n\x^n+\gamma\alpha_{n-1}\x^{n-1}+\cdots+\gamma\alpha_1\x+\alpha_0\]It is not difficult to verify that these definitions give \(\mathbb{K}[\x]\) the structure of a \(\mathbb{K}\)-vector space.
Now we can verify that the set \(\mathbb{K}[\x]_\text{degree\scriptsize$\leq n$}\) of polynomials of degree at most \(n\) is a subspace of \(\mathbb{K}[\x]\). On the other hand, the set of polynomials of exactly degree \(n\) is not a subspace because it does not contain \(0\), but if we adjoin \(0\) it becomes a subspace.
Example 6 Now let \(\mathbb{K}[[\x]]\) be the set
of formal power series in \(\x\) with coefficients in \(\mathbb{K}\)
Defining addition of vectors and scalar multiplication in the same way as in the preceding Example 7, \(\mathbb{K}[[\x]]\) becomes a \(\mathbb{K}\)-vector space as well.
By definition, \(\mathbb{K}[\x]\) is a subspace of \(\mathbb{K}[[\x]]\). Also, all elements of \(\mathbb{K}[\x]\) can be expressed as linear combinations of vectors in the set \(\{1,\x,\x^2,\ldots\}\), but elements of \(\mathbb{K}[[\x]]\) cannot.
References
[Goc] M.S. Gockenbach, Finite-dimensional linear algebra, Discrete Mathematics and its applications, Taylor&Francis, 2011.
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