Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 20 September 2014

Lecture 8: Bases and dimension

Let $𝔽$ be a field. Let $V$ be a vector space over $𝔽\text{.}$ Let $S$ be a subset of $V\text{.}$ A linear combination of elements of $S$ is $$c_1{s}_2+c_2{s}_2+\cdots +c_k{s}_k,$$ with $k\in {\mathbb{Z}}_{>0},$ $c_1,x_2,\dots ,c_k\in 𝔽,$ $s_1,s_2,\dots ,s_k\in S\text{.}$

The span of $S$ is $$\text{span}\left(S\right)=\{c_1{s}_1+\cdots +c_k{s}_k\hspace{0.17em}|\hspace{0.17em}k\in {\mathbb{Z}}_{>0},c_1,\dots ,c_k\in 𝔽,s_1,s_2,\dots ,s_k\in S\}\text{.}$$

The set $S$ is linearly independent if $S$ satisfies: If $k\in {\mathbb{Z}}_{>0},$ $c_1,\dots ,c_k\in 𝔽,$ $s_1,s_2,\dots ,s_k\in S$ and $c_1{s}_1+c_2{s}_2+\cdots +c_k{s}_k=0$ then $c_1=0$ and $c_2=0$ and $c_3=0$ and ... and $c_k=0\text{.}$

A basis of $V$ is a subset $B\subseteq V$ such that

(a)	$\text{span}\left(B\right)=V,$
(b)	$B$ is linearly independent.

The dimension of $V$ is $\text{Card}\left(B\right),$ the number of elements in $B,$ for a basis $B$ of $V\text{.}$

$$ℝ\left[t\right]=\{c_0+c_{1}t+c_2{t}^2+\cdots \hspace{0.17em}|\hspace{0.17em}\genfrac{}{}{0ex}{}{c_i\in ℝ\hspace{0.17em}\text{and all but a}}{\text{finite number of}\hspace{0.17em}{c}_i\hspace{0.17em}\text{equal 0}}\}$$ is a vector space over $ℝ\text{:}$ $$(3+4t+t^2)+(0+7t+1t^2)=3+11t+2t^2$$ and $$8.6(1+2t^2)=8.6+17.2t^2\text{.}$$ Let $S=\{1,t,t^2,t^3,\dots \}\text{.}$ Then $\text{span}\left(S\right)=ℝ\left[t\right]$ and $S$ is linearly independent since

(a)	If $p\in ℝ\left[t\right]$ then $$p=p_0+p_{1}t+p_2{t}^2+\cdots +p_N{t}^N$$ with $N\in {\mathbb{Z}}_{>0}$ and $1,t,t^2,\dots ,t^N\in S$ and $p_0,p_1,\dots ,p_N\in ℝ,$
(b)	If $p=p_0+p_{1}t+\cdots +p_N{t}^N=0$ then $p_0=0$ and $p_1=0$ and ... and $p_N=0\text{.}$ Since $\text{Card}\left(S\right)=\infty ,$ the dimension of $ℝ\left[t\right]$ is $\infty \text{.}$

Let $V$ be a vector space over $𝔽\text{.}$ Let $f:V\to V$ be a linear transformation. Let $B=\{b_1,b_2,\dots \}$ be a basis of $V\text{.}$ The matrix of $f$ with respect to $B$ is $$B_f=\left(f_{ij}\right)$$ given by $$f\left(b_j\right)=f_{1j}{b}_1+f_{2j}{b}_2+\cdots \text{.}$$

Let $ℝ{\left[t\right]}_{\le 3}=\{c_0+c_{1}t+c_2{t}^2\hspace{0.17em}|\hspace{0.17em}{c}_0,c_1,c_2\in ℝ\}\text{.}$ Then $B=\{1,t,t^2\}$ is a basis of $ℝ{\left[t\right]}_{\le 3}\text{.}$

Let $f:ℝ{\left[t\right]}_{\le 3}\to ℝ{\left[t\right]}_{\le 3}$ be given by $$f\left(p\right)=p(t-3)\text{.}$$ For example, $$\begin{array}{ccc}{\displaystyle f(3+2t-5t^2)}& =& {\displaystyle 3+2(t-3)-5{(t-3)}^2}\\ & =& {\displaystyle 3+2t-6-5(t^2-6t+9)}\\ & =& {\displaystyle 3+2t-6-5t^2+30t-45}\\ & =& {\displaystyle -48+32t-5t^2\text{.}}\end{array}$$ To calculate the matrix of $f$ with respect to $B,$ $$\begin{array}{ccc}{\displaystyle f\left(1\right)}& =& {\displaystyle 1=1+0t+0t^2,}\\ {\displaystyle f\left(t\right)}& =& {\displaystyle t-3=-3+t+0t^2,}\\ {\displaystyle f\left(t^2\right)}& =& {\displaystyle {(t-3)}^2=t^2-6t+9=9-6t+t^2}\end{array}$$ and $$B_f=\left(\begin{array}{ccc}1& -3& 9\\ 0& 1& -6\\ 0& 0& 1\end{array}\right)\text{.}$$ Note that $$f(3+2t-5t^2)=-48+32t-5t^2$$ and $$\left(\begin{array}{ccc}1& -3& 9\\ 0& 1& -6\\ 0& 0& 1\end{array}\right)\left(\begin{array}{c}3\\ 2\\ -5\end{array}\right)=\left(\begin{array}{c}3-6-45\\ 2+30\\ -5\end{array}\right)=\left(\begin{array}{c}-48\\ 32\\ -5\end{array}\right)\text{.}$$

Let $f:V\to V$ be a linear transformation. The kernel, or nullspace, of $f$ is $$\text{ker}\hspace{0.17em}f=\{v\in V\hspace{0.17em}|\hspace{0.17em}f\left(v\right)=0\}$$ and the nullity of $f$ is $\text{dim}\left(\text{ker}\hspace{0.17em}f\right)\text{.}$

The image of $f$ is $$\text{im}\hspace{0.17em}f=\left\{f\left(v\right)\hspace{0.17em}\right|\hspace{0.17em}v\in V\}$$ and the rank of $f$ is $\text{dim}\left(\text{im}\hspace{0.17em}f\right)\text{.}$

Let $f:ℝ\left[t\right]\to ℝ\left[t\right]$ be given by $$f\left(p\right)=tp\text{.}$$ For example, $f(1+3t+2t^2)=t+3t^2+2t^3\text{.}$ Then $$\begin{array}{ccc}{\displaystyle f\left(1\right)}& =& {\displaystyle t,}\\ {\displaystyle f\left(t\right)}& =& {\displaystyle t^2,}\\ {\displaystyle f\left(t^2\right)}& =& {\displaystyle t^3,}\\ {\displaystyle }& ⋮& {\displaystyle }\end{array}$$ and the matrix of $f$ with respect to the basis $B=\{1,t,t^3,\dots \}$ is $$B_f=\left(\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1& \ddots & \\ 0& 0& 0\\ ⋮& ⋮& ⋮\end{array}\right)\text{.}$$ Also, $$\begin{array}{ccc}{\displaystyle \text{ker}\hspace{0.17em}f}& =& {\displaystyle \left\{0\right\},}\\ {\displaystyle \text{im}\hspace{0.17em}f}& =& {\displaystyle \text{span}\{t,t^2,t^3,\dots \}\text{.}}\end{array}$$ So $f$ is injective but not surjective.

Let $d:ℝ\left[t\right]\to ℝ\left[t\right]$ be given by $$d\left(p\right)=\frac{d}{dt}·p\text{.}$$ For example $d(1+3t+2t^2)=3+4t\text{.}$ Then $$\begin{array}{ccc}{\displaystyle d\left(1\right)}& =& {\displaystyle 0,}\\ {\displaystyle d\left(t\right)}& =& {\displaystyle 1,}\\ {\displaystyle d\left(t^2\right)}& =& {\displaystyle 2t,}\\ {\displaystyle d\left(t^3\right)}& =& {\displaystyle 3t^2,}\\ {\displaystyle }& ⋮& {\displaystyle }\end{array}$$ and the matrix of $d$ with respect to the basis $B=\{1,t,t^2,t^3,\dots \}$ is $$B_d=\left(\begin{array}{cccccc}0& 1& 0& 0& 0& \cdots \\ 0& 0& 2& 0& 0& \cdots \\ 0& 0& 0& 3& 0& \cdots \\ 0& 0& 0& 0& 4& \ddots \\ ⋮& ⋮& ⋮& ⋮& ⋮\end{array}\right)\text{.}$$ Also $$\begin{array}{ccc}{\displaystyle \text{ker}\hspace{0.17em}d}& =& {\displaystyle \left\{c_0\hspace{0.17em}\right|\hspace{0.17em}{c}_0\in ℝ\}=\text{span}\left\{1\right\},}\\ {\displaystyle \text{im}\hspace{0.17em}d}& =& {\displaystyle ℝ\left[t\right],}\end{array}$$ since $$d(p_{0}t+\frac{p_1{t}^2}{2}+\frac{p_2{t}^3}{3}+\cdots )=p_0+p_{1}t+p_2{t}^2+p_3{t}^3+\cdots \text{.}$$ So $d$ is not injective, $d$ is surjective, the nullity of $d$ is $1,$ and the rank of $d$ is $\infty \text{.}$

Notes and References

These are a typed copy of Lecture 8 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 10, 2011.

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