Definition. A basis of a nontrivial subspace \(S\) of a vector space \(V\) is a subset \(B\) of \(S\) such that

1. \(\operatorname{Span}(B)=S\) and


2. \(B\) is linearly independent set.

We define the basis of the trivial subspace \(\{\overrightarrow{0_V}\}\) to be \(B=\varnothing\). The number of vectors in a basis \(B\) is the dimension of \(S\) denoted by \(\operatorname{dim}(S)\) or \(\operatorname{dim} S\).

Remark. If a basis of \(V\) consists of \(n\) vectors, then each basis of \(V\) has exactly \(n\) vectors and \(\operatorname{dim}(V)=n\). If \(\operatorname{dim}(V)\) is a nonnegative integer, \(V\) is called a finite-dimensional vector space. Otherwise \(V\) is called an infinite-dimensional vector space. If \(H\) is a subspace of a finite-dimensional vector space \(V\), then \(\dim H\leq \dim V\) (See Extension Theorem below).

Example.

1. \(\{\overrightarrow{e_1},\overrightarrow{e_2},\ldots,\overrightarrow{e_n}\}\) is a basis of \(\mathbb R^n\). So \(\operatorname{dim}(\mathbb R^n)=n\).


2. \(\{\overrightarrow{1},\overrightarrow{t},\overrightarrow{t^2},\ldots,\overrightarrow{t^n}\}\) is a basis of \(P_n\). So \(\operatorname{dim}(P_n)=n+1\).


3. \(B=\{\overrightarrow{E_{i,j}}: 1 \leq i \leq m,1 \leq j \leq n\}\) is a basis of \(M_{m, n}(\mathbb R)\) where \(\overrightarrow{E_{i,j}}\) is an \(m\times n\) matrix with \((i,j)\)-entry 1 and \(0\) elsewhere. So \(\operatorname{dim}(M_{m, n}(\mathbb R))=mn\).


4. \(\{\overrightarrow{e_1},\overrightarrow{e_2},\ldots,\overrightarrow{e_n},\ldots\}\) is a basis of \(\mathbb{R}^{\infty}\) where \(\overrightarrow{e_i}\) is the infinite sequence with \(1\) in the \(i\)th place and \(0\) elsewhere. So \(\mathbb{R}^{\infty}\) is an infinite-dimensional vector space.

Now we present some important theorems regarding bases of a subspace of a vector space. Proofs will be similar to that in Basis and Dimension in \(\mathbb R^n\).

Theorem.(Unique Representation Theorem) Let \(S\) be a subspace of a vector space \(V\). Then \(B=\{\overrightarrow{b_1},\overrightarrow{b_2},\ldots,\overrightarrow{b_k}\}\) is a basis of \(S\) if and only if each vector \(\overrightarrow{v}\) of \(S\) is a unique linear combination of \(\overrightarrow{b_1},\overrightarrow{b_2},\ldots,\overrightarrow{b_k}\), i.e., \(\overrightarrow{v}=c_1\overrightarrow{b_1}+c_2\overrightarrow{b_2}+\cdots+c_k\overrightarrow{b_k}\) for unique scalars \(c_1,c_2,\ldots,c_k\).

Theorem.(Reduction Theorem) Let \(S\) be a subspace of a vector space \(V\). If a set \(B=\{\overrightarrow{b_1},\overrightarrow{b_2},\ldots,\overrightarrow{b_k}\}\) of vectors of \(S\) spans \(S\), then either \(B\) is a basis of \(S\) or a subset of \(B\) is a basis of \(S\).

Theorem.(Extension Theorem) Let \(S\) be a subspace of a vector space \(V\). If a set \(B=\{\overrightarrow{b_1},\overrightarrow{b_2},\ldots,\overrightarrow{b_k}\}\) of vectors of \(S\) is linearly independent, then either \(B\) is a basis of \(S\) or a superset of \(B\) is a basis of \(S\).

Example. For \(\overrightarrow{p_1}(t)=t+2t^2\), \(\overrightarrow{p_2}(t)=2+2t^2\), and \(\overrightarrow{p_3}(t)=1-t-t^2\) in \(P_2\), \(\overrightarrow{p_2}=2\overrightarrow{p_1}+2\overrightarrow{p_3}\). Then \(\operatorname{Span} \{\overrightarrow{p_1},\;\overrightarrow{p_2},\;\overrightarrow{p_3}\}=\operatorname{Span}\{\overrightarrow{p_1},\;\overrightarrow{p_3}\}\) and \(\{\overrightarrow{p_1},\;\overrightarrow{p_3}\}\) is a basis of the subspace \(\operatorname{Span}\{\overrightarrow{p_1},\;\overrightarrow{p_2},\;\overrightarrow{p_3}\}\) of \(P_2\).

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