## Group Theory and Linear Algebra

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 $c1s2+ c2s2+⋯+ cksk,$ with $k\in {ℤ}_{>0},$ ${c}_{1},{x}_{2},\dots ,{c}_{k}\in 𝔽,$ ${s}_{1},{s}_{2},\dots ,{s}_{k}\in S\text{.}$

The span of $S$ is $span(S)= { c1s1+⋯+cksk | k∈ℤ>0, c1,…,ck∈𝔽, s1,s2,…,sk∈S } .$

The set $S$ is linearly independent if $S$ satisfies: If $k\in {ℤ}_{>0},$ ${c}_{1},\dots ,{c}_{k}\in 𝔽,$ ${s}_{1},{s}_{2},\dots ,{s}_{k}\in S$ 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 $V\text{.}$

$ℝ[t]= { c0+c1t+c2t2+⋯ | ci∈ℝ and all but a finite number of ci equal 0 }$ is a vector space over $ℝ\text{:}$ $(3+4t+t2)+ (0+7t+1t2)= 3+11t+2t2$ and $8.6(1+2t2)= 8.6+17.2t2.$ Let $S=\left\{1,t,{t}^{2},{t}^{3},\dots \right\}\text{.}$ 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 $p=p0+p1t+ p2t2+⋯+ pNtN$ with $N\in {ℤ}_{>0}$ 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 $B=\left\{{b}_{1},{b}_{2},\dots \right\}$ be a basis of $V\text{.}$ The matrix of $f$ with respect to $B$ is $Bf=(fij)$ given by $f(bj)= f1jb1+ f2jb2+⋯.$

Let $ℝ{\left[t\right]}_{\le 3}=\left\{{c}_{0}+{c}_{1}t+{c}_{2}{t}^{2} | {c}_{0},{c}_{1},{c}_{2}\in ℝ\right\}\text{.}$ Then $B=\left\{1,t,{t}^{2}\right\}$ 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 $f(p)=p(t-3).$ For example, $f(3+2t-5t2) = 3+2(t-3)- 5(t-3)2 = 3+2t-6-5 (t2-6t+9) = 3+2t-6-5t2+ 30t-45 = -48+32t-5t2.$ To calculate the matrix of $f$ with respect to $B,$ $f(1) = 1=1+0t+0t2, f(t) = t-3=-3+t+0t2, f(t2) = (t-3)2=t2 -6t+9=9-6t+t2$ and $Bf= ( 1-39 01-6 001 ) .$ Note that $f(3+2t-5t2)= -48+32t-5t2$ and $( 1-39 01-6 001 ) (32-5)= ( 3-6-45 2+30 -5 ) = ( -48 32 -5 ) .$

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

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

Let $f:ℝ\left[t\right]\to ℝ\left[t\right]$ be given by $f(p)=tp.$ For example, $f\left(1+3t+2{t}^{2}\right)=t+3{t}^{2}+2{t}^{3}\text{.}$ Then $f(1) = t, f(t) = t2, f(t2) = t3, ⋮$ and the matrix of $f$ with respect to the basis $B=\left\{1,t,{t}^{3},\dots \right\}$ is $Bf= ( 000 100 010 001⋱ 000 ⋮⋮⋮ ) .$ Also, $ker f = {0}, im f = span{t,t2,t3,…}.$ So $f$ is injective but not surjective.

Let $d:ℝ\left[t\right]\to ℝ\left[t\right]$ be given by $d(p)=ddt·p.$ For example $d\left(1+3t+2{t}^{2}\right)=3+4t\text{.}$ Then $d(1) = 0, d(t) = 1, d(t2) = 2t, d(t3) = 3t2, ⋮$ and the matrix of $d$ with respect to the basis $B=\left\{1,t,{t}^{2},{t}^{3},\dots \right\}$ is $Bd= ( 01000⋯ 00200⋯ 00030⋯ 00004⋱ ⋮⋮⋮⋮⋮ ) .$ Also $ker d = {c0 | c0∈ℝ} =span{1}, im d = ℝ[t],$ since $d(p0t+p1t22+p2t33+⋯)= p0+p1t+p2t2 +p3t3+⋯.$ 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.