Dimension
Theorem. Let \(V\subset\mathbb{R}^{n}\) be a subspace and \(S = \{v_{1},v_{2},\ldots,v_{k}\}\) a basis for \(V\). If \(v\in V\) then there exists a unique set of scalars \(\{a_{1},a_{2},\ldots,a_{k}\}\) such that
\[v=\sum_{i=1}^{k}a_{i}v_{i}.\]
Proof. Since \(S\) is a basis, in particular, \(S\) spans \(V\), and hence there are scalars \(a_{1},a_{2},\ldots,a_{k}\) such that
\[v=\sum_{i=1}^{k}a_{i}v_{i}.\]
We must show that these scalars are unique. Suppose we have another set of scalars \(b_{1},\ldots,b_{k}\) such that
\[v=\sum_{i=1}^{k}b_{i}v_{i}.\]
Subtracting, we have \[0 = v-v = \sum_{i=1}^{k} (a_{i}-b_{i})v_{i}.\] Since \(S\) is linearly independent, we conclude that \(a_{i} -b_{i} = 0\) for each \(i\in\{1,\ldots,k\}.\ \Box\)
Theorem. Let \(V\subset\mathbb{R}^{n}\) be a subspace. If \[S = \{v_{1},v_{2},\ldots,v_{k}\}\quad \text{and} \quad R = \{w_{1},w_{2},\ldots,w_{\ell}\}\] are both bases for \(V\), then \(|S| = |R|\).
Proof. For each \(i\in \{1,2,\ldots,k\}\) there are scalars \(a_{i1},a_{i2},\ldots,a_{i\ell}\) such that \[v_{i} = a_{i1}w_{1} + a_{i2}w_{2} + \cdots + a_{i\ell}w_{\ell}.\]
Similarly, for each \(j\in\{1,2,\ldots,\ell\}\) there are scalars \(b_{j1},b_{j2},\ldots,b_{jk}\) such that \[w_{j} =b_{j1}v_{1} + b_{j2}v_{2} +\cdots +b_{jk}v_{k}.\] Combining these we have \[v_{i_{0}} = \sum_{j=1}^{\ell}a_{i_{0}j}w_{j} = \sum_{j=1}^{\ell}a_{i_{0}j}\left(\sum_{i=1}^{k}b_{ji}v_{i}\right) = \sum_{i=1}^{k}\left(\sum_{j=1}^{\ell}a_{i_{0}j}b_{ji}\right)v_{i}.\] By the previous theorem \[\sum_{j=1}^{\ell}a_{i_{0}j}b_{ji} = \begin{cases} 1 & i=i_{0},\\ 0 & i\neq i_{0}.\end{cases}\]
\[v_{i_{0}} = a_{i_{0}1}w_{1} + a_{i_{0}2}w_{2} + \cdots + a_{i_{0}\ell}w_{\ell}\]
\[ = a_{i_{0}1}\left(b_{11}v_{1} + b_{12}v_{2} + \cdots + b_{1k}v_{k}\right) + a_{i_{0}2}\left(b_{21}v_{1} + b_{22}v_{2} + \cdots + b_{2k}v_{k}\right) \]
\[+ \cdots + a_{i_{0}\ell}\left(b_{\ell 1}v_{1} + b_{\ell 2}v_{2} + \cdots + b_{\ell k}v_{k}\right)\]
\[= \left(a_{i_{0}1}b_{11} + a_{i_{0}2}b_{21} + \cdots a_{i_{0}\ell}b_{\ell 1}\right) v_{1} + \left(a_{i_{0}1}b_{12} + a_{i_{0}2}b_{22} + \cdots a_{i_{0}\ell}b_{\ell 2}\right) v_{2}\]
\[+ \cdots + \left(a_{i_{0}2}b_{2k} + \cdots a_{i_{0}\ell}b_{\ell k}\right) v_{k} \]
\[=\sum_{i=1}^{k}\left(\sum_{j=1}^{\ell}a_{i_{0}j}b_{ji}\right)v_{i}\]
If we decompose \(w_{j_{0}}\) similarly, then we find \[\sum_{i=1}^{k}b_{j_{0}i} a_{ij}= \begin{cases} 1 & j=j_{0},\\ 0 & j\neq j_{0}.\end{cases}\]
Hence, \[\ell = \sum_{j=1}^{\ell}\sum_{i=1}^{k}b_{ji}a_{ij} = \sum_{j=1}^{\ell}\sum_{i=1}^{k}a_{ij}b_{ji} = \sum_{i=1}^{k}\sum_{j=1}^{\ell}a_{ij}b_{ji} = k.\ \Box\]
Definition. Let \(U\) be a subspace of \(\R^{n}\). The dimension of \(U\), denoted \(\text{dim}\, U\), is the number of vectors in a basis for \(U\).
Example. Let \(A = \begin{bmatrix} 1 & 1 & 0 & -1\\ 1 & 1 & 0 & -2\\ 0 & 0 & 0 & 1\end{bmatrix}\).
Find the dimension of \(C(A)\) and \(N(A)\).
\[\text{rref}(A) = \begin{bmatrix} 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\end{bmatrix}\]
\(\left\{\begin{bmatrix} 1\\ 1\\ 0\end{bmatrix},\begin{bmatrix} -1\\ -2\\ 1\end{bmatrix}\right\}\) is a basis for \(C(A)\), so \(\text{dim}\,C(A) = 2\), and
\(\left\{\begin{bmatrix} -1\\ 1\\ 0\\ 0\end{bmatrix},\begin{bmatrix} 0\\ 0\\ 1\\ 0\end{bmatrix}\right\}\) is a basis for \(N(A)\) \(\text{dim}\,N(A) = 2\).
Corollary 1. If \(V\subset \R^{n}\) is a subspace, and \(S\subset V\) is a spanning set for \(V\), then \(|S|\geq \operatorname{dim}V.\)
Proof. We have already seen that we can remove elements from a spanning set until the remaining elements are still spanning, but also independent. Hence, for a spanning set \(S\) there is a basis \(B\subset S\). Thus,
\[\operatorname{dim}V = |B|\leq |S|.\ \Box\]
Corollary 2. If \(V\subset \R^{n}\) is a subspace, and \(S\subset V\) is linearly independent, then \(|S|\leq \operatorname{dim}V.\)
Proof. We will show that there is a superset \(B\supset S\) such that \(B\) is a basis for \(V\). If \(S\) does not span \(V\), then there is a vector \(v\in V\) such that \(v\notin\operatorname{span}S\). Add this vector to the set \(S\). Repeat until there are \(\operatorname{dim}V\) elements in the set. This must be a basis for \(V\). \(\Box\).