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 zero vector space arising from \(0_V.\) Then, by Definition 3.21 and the properties of Definition 3.1, it follows that \(U\) is a vector subspace of \(V.\)

Taking \(I=\mathbb{R}\) and \(\mathbb{K}=\mathbb{R}\) in Example 3.5, 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 3.21, it follows that \(U\) is a vector subspace of \(\mathsf{F}(\mathbb{R},\mathbb{R}).\)

The kernel of the linear map \(\frac{\mathrm{d}}{\mathrm{d}x} : \mathsf{P}_n(\mathbb{R}) \to \mathsf{P}_{n-1}(\mathbb{R})\) consists of the constant polynomials satisfying \(f(x)=c\) for all \(x \in \mathbb{R}\) and where \(c \in \mathbb{R}\) is some constant.

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 \(a,b,c \in \mathbb{R}.\)

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 \(\mathcal{S}.\)

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 \[\tag{3.9} s_1p_1+s_2p_2+s_3p_3=0_{\mathsf{P}(\mathbb{R})}=o\] where \(o : \mathbb{R}\to \mathbb{R}\) denotes the zero polynomial. Since (3.9) 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.

Thinking of a field \(\mathbb{K}\) as a \(\mathbb{K}\)-vector space, the set \(\{1_{\mathbb{K}}\}\) is linearly independent, since \(1_{\mathbb{K}}\neq 0_{\mathbb{K}}.\) Example 3.42 implies that \(\{1_{\mathbb{K}}\}\) is a basis of \(\mathbb{K}.\)

Clearly, the standard basis \(\{\vec{e}_1,\ldots,\vec{e}_n\}\) of \(\mathbb{K}^n\) is linearly independent since \[s_1\vec{e}_1+\cdots+s_n\vec{e}_n=\begin{pmatrix} s_1 \\ \vdots \\ s_n\end{pmatrix}=0_{\mathbb{K}^{n}}=\begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix} \qquad \iff \qquad s_1=\dots=s_n=0.\] It follows together with Example 3.43 that the standard basis of \(\mathbb{K}^n\) is indeed a basis in the sense of Definition 3.51.

The matrices \(\mathbf{E}_{k,l} \in M_{m,n}(\mathbb{K})\) for \(1\leqslant k \leqslant m\) and \(1\leqslant l \leqslant n\) are linearly independent. Suppose we have scalars \(s_{kl} \in \mathbb{K}\) such that \[\sum_{k=1}^m\sum_{l=1}^ns_{kl}\mathbf{E}_{k,l}=\mathbf{0}_{m,n}=\begin{pmatrix}s_{11} & \cdots & s_{1n} \\ \vdots & \ddots & \vdots\\ s_{m1} & \cdots & s_{mn}\end{pmatrix}=\begin{pmatrix}0 & \cdots & 0 \\ \vdots & \ddots & \vdots\\ 0 & \cdots & 0\end{pmatrix}\] so \(s_{kl}=0\) for all \(1\leqslant k \leqslant m\) and all \(1\leqslant l \leqslant n.\) It follows together with Example 3.44 that \(\{\mathbf{E}_{k,l}\}_{1\leqslant k\leqslant m, 1\leqslant l \leqslant n}\) is a basis of \(M_{m,n}(\mathbb{K}).\) We refer to \(\{\mathbf{E}_{k,l}\}_{1\leqslant k\leqslant m, 1\leqslant l \leqslant n}\) as the standard basis of \(M_{m,n}(\mathbb{K})\).

Combining Remark 3.40 and Remark 3.50 we conclude that the empty set is a basis for the zero vector space \(\{0\}.\)