# Day 7:

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, $\ell = \sum_{j=1}^{\ell}\sum_{i=1}^{k}b_{ji}a_{ij} = \sum_{j=1}^{\ell}\sum_{i=1}^{k}a_{ij}b_{ji} = \sum_{i=1}^{k}\sum_{j=1}^{\ell}a_{ij}b_{ji} = k.\ \Box$

### Dimension of subspaces

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 $$V$$. $$\Box$$.

By John Jasper

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