Subspaces

Subspaces

As seen in §Vector Spaces, ⁋Example 6, a subset of a vector space often forms a vector space in its own right. Let us define this notion 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 directly from the definition whether an arbitrary subset \(W\) of \(V\) is a subspace, we would have to verify that its addition forms an abelian group, that scalar multiplication satisfies all the conditions of §Vector Spaces, ⁋Definition 1, and so on. However, since the addition and scalar multiplication on \(W\) are inherited from \(V\), some properties need not be checked.

For example, for arbitrary \(w_1,w_2\in W\), there is no need to check whether

\[w_1+w_2=w_2+w_1\]

holds. This is because the two elements \(w_1,w_2\) are elements of \(V\) before they are elements of \(W\), and the addition \(+\) on \(W\) is the restriction of the addition on \(V\) to \(W\). Taking this into account, the properties we do need to check are as follows.

1. We must check separately whether \(W\) is closed under addition.
2. Similarly, we must check whether \(W\) contains the additive identity and additive inverses. Of course \(V\) contains \(0\) and \(-w\), but there is no guarantee that these belong to \(W\).
3. Also, for an arbitrary scalar \(\alpha\in\mathbb{K}\) and \(w\in W\), we must check whether \(\alpha w\in W\).

But the conditions can be trimmed down a bit further here. If \(W\) is closed under scalar multiplication alone, then by §Vector Spaces, ⁋Proposition 2 and §Vector Spaces, ⁋Corollary 3, the second condition can be omitted entirely. Indeed, since \(W\) is closed under scalar multiplication, we must have \(0w\in W\) and \((-1)w\in W\), and these are precisely \(0\) 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, since the easiest way to show that \(W\) is nonempty is to show that \(0\in W\), in practice there is little difference in usefulness between the three conditions presented earlier and the preceding proposition.

Linear Combinations

Consider a \(\mathbb{K}\)-vector space \(V\) and a subspace \(W\) of it. Since the sum of any two elements of \(W\) is again an element of \(W\), by induction any finite sum is again an element of \(W\). More generally, the following holds.

Proposition 3 Let \(V\) be a \(\mathbb{K}\)-vector space and let \(W\) be a subspace of \(V\). For elements \(w_1,\ldots, w_n\) of \(W\) and scalars \(\alpha_1,\ldots,\alpha_n\), the 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 \(n=1\) gives \(\alpha_1w_1\in W\) by Proposition 2. Now consider the case \(n=2\). In this case, by Proposition 2, each of \(\alpha_1w_1,\alpha_2w_2\) is an element of \(W\), and hence their sum \(\alpha_1w_1+\alpha_2w_2\) is also an element of \(W\).

For a general \(n\), since addition on \(W\) satisfies the associative law,

\[\alpha_1w_1+\alpha_2w_2+\cdots+\alpha_nw_n=(\alpha_1w_1+\cdots+\alpha_{n-1}w_{n-1})+\alpha_nw_n\]

holds. Now by the inductive hypothesis, each of \(\alpha_1w_1+\cdots+\alpha_{n-1}w_{n-1}\) and \(\alpha_nw_n\) is an element of \(W\), and therefore their sum \(\sum_{i=1}^n\alpha_iw_i\) is also an element of \(W\).

In general, a vector of the form (1) in the above proposition is 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 them 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, a linear combination of them 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 regard \(\mathbb{R}\) as a \(\mathbb{Q}\)-vector space as in §Vector Spaces, ⁋Example 4, then \(\sqrt{2}-1=0.4142\ldots\) is not a linear combination of the vectors

\[0.1,\quad 0.01,\quad0.001,\quad\cdots\]

. Following the decimal expansion, we can write \(\sqrt{2}-1=4\cdot0.1+1\cdot0.01+4\cdot0.001+\cdots\) as an infinite sum, but since this infinite sum has infinitely many nonzero coefficients, it is not a linear combination in the sense defined above. Moreover, since all the listed vectors are rational, any finite linear combination of them is always rational, and the irrational number \(\sqrt{2}-1\) cannot be expressed as a linear combination of them in any way. In a similar vein, let us look at 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 scalars \(\alpha_i\in\mathbb{K}\). If \(\alpha_n\neq0\), we call \(n\) the degree of \(p(\x)\), and in this case \(\alpha_n\x^n\) is called the leading term. The degree of the zero polynomial \(0\) is either left undefined or formally regarded as \(-\infty\); accordingly, when we speak of polynomials of degree at most \(n\), we agree to include \(0\) as well. 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 defined, if \(m>n\), by

\[\sum_{i=0}^n\alpha_i\x^i+\sum_{i=0}^m\beta_i\x^i=\sum_{i=0}^m \gamma_i\x^i,\qquad \gamma_i=\begin{cases}\alpha_i+\beta_i&\text{if $0\leq i\leq n$}\\ \beta_i&\text{if $n < i\leq m$}\end{cases}\]

and in the opposite case by

\[\sum_{i=0}^n\alpha_i\x^i+\sum_{i=0}^m\beta_i\x^i=\sum_{i=0}^n \gamma_i'\x^i,\qquad \gamma_i'=\begin{cases}\alpha_i+\beta_i&\text{if $0\leq i\leq m$}\\ \alpha_i&\text{if $m < i\leq n$}\end{cases}\]

, and for an arbitrary scalar \(\gamma\in\mathbb{K}\), scalar multiplication is defined by

\[\gamma p(\x)=\gamma\alpha_n\x^n+\gamma\alpha_{n-1}\x^{n-1}+\cdots+\gamma\alpha_1\x+\gamma\alpha_0\]

. It is not difficult to verify that these definitions endow \(\mathbb{K}[\x]\) with the structure of a \(\mathbb{K}\)-vector space.

Now we can check 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 since it does not contain \(0\), and when \(n\geq1\), even after adjoining \(0\) it is still not a subspace because it is not closed under addition, as the sum of \(\x^n\) and \(1-\x^n\) shows.

Example 6 This time, let the set \(\mathbb{K}[[\x]]\) be

the set of formal power series in \(\x\) with coefficients in \(\mathbb{K}\)

. If we define addition and scalar multiplication in exactly the same way as in the preceding Example 5, then \(\mathbb{K}[[\x]]\) likewise becomes a \(\mathbb{K}\)-vector space.

By definition, \(\mathbb{K}[\x]\) is a subspace of \(\mathbb{K}[[\x]]\). Also, every element of \(\mathbb{K}[\x]\) can be expressed as a linear combination of vectors from the set \(\{1,\x,\x^2,\ldots\}\), but \(\mathbb{K}[[\x]]\) has elements that cannot be expressed in that way.

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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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