Consider \(\mathsf{P}(\mathbb{R}),\) the set of polynomial functions in one real variable, which we denote by \(x,\) with real coefficients. That is, an element \(p \in \mathsf{P}(\mathbb{R})\) is a function \[p : \mathbb{R}\to \mathbb{R}, \qquad x \mapsto a_n x^n+a_{n-1}x^{n-1}+\cdots + a_1 x+a_0=\sum_{k=0}^na_kx^k,\] where \(n \in \mathbb{N}\) and the coefficients \(a_k \in \mathbb{R}\) for \(k=0,1,\ldots,n.\) The largest \(m \in \mathbb{N}\) such that \(a_m \neq 0\) is called the degree of \(p.\) Notice that we consider polynomials of arbitrary, but finite degree. A power series \(x \mapsto \sum_{k=0}^{\infty} a_k x^k,\) that you encounter in the Calculus module, is not a polynomial, unless only finitely many of its coefficients are different from zero.

Clearly, we can multiply \(p\) with a real number \(s \in \mathbb{R}\) to obtain a new polynomial \(s\cdot_{\mathsf{P}(\mathbb{R})} p\) \[\tag{4.1} s\cdot_{\mathsf{P}(\mathbb{R})} p : \mathbb{R}\to \mathbb{R}, \qquad x \mapsto s\cdot p(x)\] so that \((s \cdot_{\mathsf{P}(\mathbb{R})}p)(x)=\sum_{k=0}^n s a_k x^k\) for all \(x \in \mathbb{R}.\) Here \(s \cdot p(x)\) is the usual multiplication of the real numbers \(s\) and \(p(x).\) If we consider another polynomial \[q : \mathbb{R}\to \mathbb{R}, \qquad x\mapsto \sum_{k=0}^n b_k x^k\] with \(b_k \in \mathbb{R}\) for \(k=0,1,\ldots,n,\) the sum of the polynomials \(p\) and \(q\) is the polynomial \[\tag{4.2} p+_{\mathsf{P}(\mathbb{R})}q : \mathbb{R}\to \mathbb{R}, \qquad x \mapsto p(x)+q(x)\] so that \((p+_{\mathsf{P}(\mathbb{R})}q)(x)=\sum_{k=0}(a_k+b_k)x^k\) for all \(x \in \mathbb{R}.\) Here \(p(x)+q(x)\) is the usual addition of the real numbers \(p(x)\) and \(q(x).\) We will henceforth omit writing \(+_{\mathsf{P}(\mathbb{R})}\) and \(\cdot_{\mathsf{P}(\mathbb{R})}\) and simply write \(+\) and \(\cdot.\)

A field \(\mathbb{K}\) is a \(\mathbb{K}\)-vector space. We may take \(V=\mathbb{K},\) \(0_V=0_{\mathbb{K}}\) and equip \(V\) with addition \(+_V=+_{\mathbb{K}}\) and scalar multiplication \(\cdot_V=\cdot_{\mathbb{K}}.\) Then the properties of a field imply that \(V=\mathbb{K}\) is a \(\mathbb{K}\)-vector space.

Let \(V=M_{m,n}(\mathbb{K})\) denote the set of \(m\times n\)-matrices with entries in \(\mathbb{K}\) and \(0_V=\mathbf{0}_{m,n}\) denote the zero vector. It follows from Proposition 2.16 that \(V\) equipped with addition \(+_V : V \times V \to V\) defined by (2.2) and scalar multiplication \(\cdot_V : \mathbb{K}\times V \to V\) defined by (2.3) is a \(\mathbb{K}\)-vector space. In particular, the set of column vectors \(\mathbb{K}^n=M_{n,1}(\mathbb{K})\) is a \(\mathbb{K}\)-vector space as well.

The set \(\mathsf{P}(\mathbb{R})\) of polynomials in one real variable and with real coefficients is an \(\mathbb{R}\)-vector space, when equipped with addition and scalar multiplication as defined in (4.1) and (4.2) and when the zero vector \(0_{\mathsf{P}(\mathbb{R})}\) is defined to be the zero polynomial \(o : \mathbb{R}\to \mathbb{R},\) that is, the polynomial satisfying \(o(x)=0\) for all \(x \in \mathbb{R}.\)

We follow the convention of calling a mapping with values in \(\mathbb{K}\) a function. Let \(I\subset \mathbb{R}\) be an interval and let \(o : I \to \mathbb{K}\) denote the zero function defined by \(o(x)=0\) for all \(x \in I.\) We consider \(V=\mathsf{F}(I,\mathbb{K}),\) the set of functions from \(I\) to \(\mathbb{K}\) with zero vector \(0_V=o\) given by the zero function and define addition \(+_V : V \times V \to V\) as in (4.2) and scalar multiplication \(\cdot_V : \mathbb{K}\times V \to V\) as in (4.1). It now is a consequence of the properties of addition and multiplication of scalars that \(\mathsf{F}(I,\mathbb{K})\) is a \(\mathbb{K}\)-vector space. (The reader is invited to check this assertion!)

A mapping \(x : \mathbb{N} \to \mathbb{K},\) from the natural numbers into a field \(\mathbb{K},\) is called a sequence in \(\mathbb{K}\) (or simply a sequence, when \(\mathbb{K}\) is clear from the context). It is common to write \(x_n\) instead of \(x(n)\) for \(n \in \mathbb{N}\) and to denote a sequence by \((x_n)_{n \in \mathbb{N}}=(x_0,x_1,x_2,\ldots).\) We write \(\mathbb{K}^{\infty}\) for the set of sequences in \(\mathbb{K}.\) For instance, taking \(\mathbb{K}=\mathbb{R},\) we may consider the sequence \[\left(\frac{1}{n + 1}\right)_{n \in \mathbb{N}}=\left(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\ldots\right)\] or the sequence \[\left(\sqrt{n+1}\right)_{n \in \mathbb{N}}=\left(1,\sqrt{2},\sqrt{3},2,\sqrt{5},\ldots\right).\] If we equip \(\mathbb{K}^{\infty}\) with the zero vector given by the zero sequence \((0,0,0,0,0,\ldots),\) addition given by \((x_n)_{n \in \mathbb{N}}+(y_n)_{n\in N}=(x_n+y_n)_{n \in \mathbb{N}}\) and scalar multiplication given by \(s\cdot(x_n)_{n \in \mathbb{N}}=(s x_n)_{n \in \mathbb{N}}\) for \(s \in \mathbb{K},\) then \(\mathbb{K}^{\infty}\) is a \(\mathbb{K}\)-vector space.

Consider a set \(V\) whose only element is a formal symbol \(\star.\) We define \(0_V = \star,\) addition by \(\star +_V \star = \star\) and scalar multiplication by \(s\cdot_V \star=\star.\) Then all the properties of Definition 4.2 are satisfied. We write \(V=\{\star\},\) or simply \(V=\{0\},\) and call \(V\) the zero vector space (over \(\mathbb{K}\)).

If \(\mathbb{F}\) and \(\mathbb{K}\) are fields, and \(\iota: \mathbb{F}\to \mathbb{K}\) is a field embedding, then \(\mathbb{K}\) is an \(\mathbb{F}\)-vector space in a natural way: the addition is the native field addition of \(\mathbb{K},\) and the scalar multiplication being given by \(s \cdot x = \iota(s) \cdot_\mathbb{K}x.\) (Exercise: check that the axioms are satisfied!) In effect, we are throwing away some of the structure from \(\mathbb{K}\) – we are “forgetting” how to multiply elements of \(\mathbb{K},\) except when one of them is in the image of \(\iota.\)

In particular, \(\mathbb{R}\) is a \(\mathbb{Q}\)-vector space. This is a really strange and puzzling concept, and shows that sometimes we have to live with our definitions having unexpected consequences!

For \(n \in \mathbb{N}\) with \(n\geqslant 2\) consider \(V=\mathsf{P}_n(\mathbb{R})\) and the polynomials \(p_1,p_2,p_3 \in \mathsf{P}_n(\mathbb{R})\) defined by the rules \(p_1(x)=1,\) \(p_2(x)=x,\) \(p_3(x)=x^2\) for all \(x \in \mathbb{R}.\) A linear combination of \(\{p_1,p_2,p_3\}\) is a polynomial of the form \(p(x)=ax^2+bx+c\) where \(a,b,c \in \mathbb{R}.\)

“Is \(\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \in \mathbb{R}^3\) a linear combination of \(\left\{\left(\begin{smallmatrix} 3 \\ -1 \\ 0\end{smallmatrix}\right), \left(\begin{smallmatrix} 0 \\ 1 \\-1 \end{smallmatrix}\right)\right\}\)?”

We’re asking if there are \(s_1, s_2\) such that \(s_1 \begin{pmatrix} 3 \\ -1 \\ 0\end{pmatrix} + s_2 \begin{pmatrix} 0 \\1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix},\) which is the same as the equations \[\begin{array}{rc} 3s_1 + 0 s_2 &= 1 \\ -s_1 + s_2 &= 2 \\ 0s_1 - s_2 &= 1 \end{array}, \qquad \text{i.e.} \qquad \left( \begin{array}{cc|c} 3 & 0 & 1 \\ -1 & 1 & 2 \\ 0 & -1 & 1 \end{array}\right).\] Since the RREF of this matrix is \(\left(\begin{array}{rr|r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right),\) there are no solutions. So the answer to the original question is no

Let \(\vec{w}\neq 0_{\mathbb{R}^3},\) then the line \[U=\{s \vec{w}\, |\, s \in \mathbb{R}\} \subset \mathbb{R}^3\] is a vector subspace. Indeed, taking \(s=0\) it follows that \(0_{\mathbb{R}^3} \in U\) so that \(U\) is non-empty. Let \(\vec{u}_1,\vec{u}_2\) be vectors in \(U\) so that \(\vec{u}_1=t_1\vec{w}\) and \(\vec{u}_2=t_2\vec{w}\) for scalars \(t_1,t_2 \in \mathbb{R}.\) Let \(s_1,s_2 \in \mathbb{R},\) then \[s_1\vec{u}_1+s_2\vec{u}_2=s_1t_1\vec{w}+s_2t_2\vec{w}=\left(s_1t_1+s_2t_2\right)\vec{w} \in U\] so that \(U\subset \mathbb{R}^3\) is a subspace.

Let \(V\) be a \(\mathbb{K}\)-vector space and \(U=\{0_V\}\) the set consisting of the zero vector of \(V.\) Then, by Definition 4.17 and the properties of Definition 4.2, it follows that \(U\) is a vector subspace of \(V\): the zero subspace \(\{0_V\}.\)

If \(V\) and \(W\) are different vector spaces over \(\mathbb{K},\) then \(\{0_V\}\) and \(\{0_W\}\) may or may not be exactly the same – it depends on what \(V\) and \(W\) are – but they have the same vector-space structure (they are isomorphic, a concept we’ll see later).

Taking \(I=\mathbb{R}\) and \(\mathbb{K}=\mathbb{R}\) in Example 4.6, we see that the functions \(f : \mathbb{R}\to \mathbb{R}\) form an \(\mathbb{R}\)-vector space \(V=\mathsf{F}(\mathbb{R},\mathbb{R}).\) Consider the subset \[U=\left\{f \in \mathsf{F}(\mathbb{R},\mathbb{R})\,|\, f\;\text{is periodic with period}\; 2\pi\right\}\] consisting of \(2\pi\)-periodic functions, that is, an element \(f \in U\) satisfies \(f(x+2\pi)=f(x)\) for all \(x \in \mathbb{R}.\) Notice that \(U\) is not empty, as \(\cos : \mathbb{R}\to \mathbb{R}\) and \(\sin : \mathbb{R}\to \mathbb{R}\) are elements of \(U.\) Suppose \(f_1,f_2 \in U\) and \(s_1,s_2 \in \mathbb{R}.\) Then, we have for all \(x \in \mathbb{R}\) \[\begin{aligned} (s_1f_1+s_2f_2)(x+2\pi)&=s_1f_1(x+2\pi)+s_2f_2(x+2\pi)=s_1f_1(x)+s_2f_2(x)\\ &=(s_1f_1+s_2f_2)(x) \end{aligned}\] showing that \(s_1f_1+s_2f_2\) is periodic with period \(2\pi.\) By Definition 4.17, it follows that \(U\) is a vector subspace of \(\mathsf{F}(\mathbb{R},\mathbb{R}).\)

Thinking of a field \(\mathbb{K}\) as a \(\mathbb{K}\)-vector space, the set \(\mathcal{S}=\{1_{\mathbb{K}}\}\) consisting of the identity element of multiplication is a generating set for \(V=\mathbb{K}.\) Indeed, for every \(x \in \mathbb{K}\) we have \(x=x\cdot_{V}1_{\mathbb{K}}.\)

The standard basis \(\mathcal{S}=\{\vec{e}_1,\ldots,\vec{e}_n\}\) is a generating set for \(\mathbb{K}^n,\) since for all \(\vec{x}=(x_i)_{1\leqslant i\leqslant n} \in \mathbb{K}^n,\) we can write \(\vec{x}=x_1\vec{e}_1+\cdots+x_n\vec{e}_n\) so that \(\vec{x}\) is a linear combination of elements of \(\mathcal{S}.\) Thus \(\mathbb{K}^n\) is finite-dimensional.

Let \(\mathbf{E}_{k,l} \in M_{m,n}(\mathbb{K})\) for \(1\leqslant k \leqslant m\) and \(1\leqslant l \leqslant n\) denote the \(m\)-by-\(n\) matrix satisfying \(\mathbf{E}_{k,l}=(\delta_{ki}\delta_{lj})_{1\leqslant i\leqslant m, 1\leqslant j \leqslant n}.\) For example, for \(m=2\) and \(n=3\) we have \[\mathbf{E}_{1,1}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},\qquad \mathbf{E}_{1,2}=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix},\qquad \mathbf{E}_{1,3}=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}\] and \[\mathbf{E}_{2,1}=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},\qquad \mathbf{E}_{2,2}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},\qquad \mathbf{E}_{2,3}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}.\] Then \(\mathcal{S}=\{\mathbf{E}_{k,l}\}_{1\leqslant k\leqslant m, 1\leqslant l \leqslant n}\) is a generating set for \(M_{m,n}(\mathbb{K}),\) since a matrix \(\mathbf{A}\in M_{m,n}(\mathbb{K})\) can be written as \[\mathbf{A}=\sum_{k=1}^m\sum_{l=1}^n A_{kl}\mathbf{E}_{k,l}\] so that \(\mathbf{A}\) is a linear combination of the elements of \(\mathcal{S}.\)

The vector space \(\mathsf{P}(\mathbb{R})\) of polynomials is infinite dimensional. In order to see this, consider a finite set of polynomials \(\{p_1,\ldots,p_n\},\) \(n \in \mathbb{N}\) and let \(d_i\) denote the degree of the polynomial \(p_i\) for \(i=1,\ldots,n.\) We set \(D =\max\{d_1,\ldots,d_n\}.\) Since a linear combination of the polynomials \(\{p_1,\ldots,p_n\}\) has degree at most \(D ,\) any polynomial \(q\) whose degree is strictly larger than \(D\) will satisfy \(q \notin \operatorname{span}\{p_1,\ldots,p_n\}.\) It follows that \(\mathsf{P}(\mathbb{R})\) cannot be generated by a finite set of polynomials.

We consider the polynomials \(p_1,p_2,p_3 \in \mathsf{P}(\mathbb{R})\) defined by the rules \(p_1(x)=1,p_2(x)=x,p_3(x)=x^2\) for all \(x \in \mathbb{R}.\) Then \(\{p_1,p_2,p_3\}\) is linearly independent. In order to see this, consider the condition \[\tag{4.6} s_1p_1+s_2p_2+s_3p_3=0_{\mathsf{P}(\mathbb{R})}=o\] where \(o : \mathbb{R}\to \mathbb{R}\) denotes the zero polynomial. Since (4.6) means that \[s_1p_1(x)+s_2p_2(x)+s_3p_3(x)=o(x),\] for all \(x \in \mathbb{R},\) we can evaluate this condition for any choice of real number \(x.\) Taking \(x=0\) gives \[s_1p_1(0)+s_2p_2(0)+s_3p_3(0)=o(0)=0=s_1.\] Taking \(x=1\) and \(x=-1\) gives \[\begin{aligned} 0&=s_2p_2(1)+s_3p_3(1)=s_2+s_3,\\ 0&=s_2p_2(-1)+s_3p_3(-1)=-s_2+s_3, \end{aligned}\] so that \(s_2=s_3=0\) as well. It follows that \(\{p_1,p_2,p_3\}\) is linearly independent.

Consider the vectors in \(\mathbb{R}^3\) given by \[v_1 = \left(\begin{smallmatrix} 0 \\ 1 \\ 2\end{smallmatrix}\right), v_2 = \left(\begin{smallmatrix} -9 \\ 4 \\ 6\end{smallmatrix}\right), v_3 = \left(\begin{smallmatrix} 3 \\ 0 \\ -1\end{smallmatrix}\right), v_4 = \left(\begin{smallmatrix} 4 \\ -1 \\ 5\end{smallmatrix}\right).\] These are linearly independent if and only if the system of equations \[\left(\begin{smallmatrix} 0 & -9 & 3 & 4 \\ 1 & 4 & 0 & -1 \\ 2 & 6 & -1 & 5 \end{smallmatrix}\right) \left(\begin{smallmatrix} s_1 \\ s_2 \\ s_3 \\ s_4 \end{smallmatrix}\right) = \left(\begin{smallmatrix} 0 \\ 0 \\ 0 \end{smallmatrix}\right)\] has only the zero solution. We already calculated the RREF of this matrix in a previous example; so we know that this system is equivalent to \[\left(\begin{array}{rrrr|r} 1 & 0 & 0 & \frac{17}{3} & 0 \\ 0 & 1 & 0 & -\frac{5}{3} & 0 \\ 0 & 0 & 1 & -\frac{11}{3} & 0 \end{array}\right).\] Obviously this system is consistent (because the rightmost column is entirely zeroes), but we care about uniqueness of solutions; and since none of the rows has its leading entry in the fourth column, \(s_4\) is a free variable, and hence the solution is not unique.