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 7: Vector spaces and linear transformations

Let $𝔽$ be a field. A vector space over $𝔽$ is an abelian group $V$ with a function $$\begin{array}{ccc}𝔽\times V& ⟶& V\\ (c,v)& ⟼& cv\end{array}$$ such that

(a)	If $c_1,c_2\in 𝔽$ and $v\in V$ then $c_1\left(c_{2}v\right)=\left(c_1{c}_2\right)v,$
(b)	If $v\in V$ then $1v=v,$
(c)	If $c_1,c_2\in 𝔽$ and $v\in V$ then $(c_1+c_2)v=c_{1}v+c_{2}v,$
(d)	If $c\in 𝔽$ and $v_1,v_2\in V$ then $c(v_1+v_2)=c{v}_1+c{v}_2\text{.}$

Column vectors of length 3, is a vector space over $𝔽,$ $${𝔽}^3=M_{3\times 1}\left(𝔽\right)=\left\{\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}{a}_i\in 𝔽\}$$ with $$\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)+\left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right)=\left(\begin{array}{c}{a}_1+b_1\\ a_2+b_2\\ a_3+b_3\end{array}\right)\text{and}c\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)=\left(\begin{array}{c}c{a}_1\\ c{a}_2\\ c{a}_3\end{array}\right)\text{.}$$

Vectors in ${ℝ}^3$ is a vector space over $ℝ,$ $$\begin{array}{c} (0,3,2) (2,3,2) (2,0,0) v1 v2 v1+v2 3(v1+v2) \end{array}$$

Let $V$ be a vector space over $𝔽\text{.}$ A subspace of $V$ is a subset $U\subseteq V$ such that

(a)	If $u_1,u_2\in U$ then $u_1+u_2\in U\text{.}$
(b)	If $u\in U$ then $-u\in U\text{.}$
(c)	$0\in U\text{.}$
(d)	If $u\in U$ and $c\in 𝔽$ then $cu\in U\text{.}$

Let $V$ be a vector space over $𝔽\text{.}$ Let $U$ and $W$ be subspaces of $V\text{.}$ Then $$\begin{array}{ccc}{\displaystyle U\cap W}& =& {\displaystyle \{v\in V\hspace{0.17em}|\hspace{0.17em}v\in U\hspace{0.17em}\text{and}\hspace{0.17em}v\in W\}\text{and}}\\ {\displaystyle U+W}& =& {\displaystyle \{u+w\hspace{0.17em}|\hspace{0.17em}u\in U\hspace{0.17em}\text{and}\hspace{0.17em}w\in W\}}\end{array}$$ are subspaces of $V\text{.}$

Let $V$ be a vector space over $𝔽\text{.}$ Let $U,W$ be subspaces of $V\text{.}$ The subspaces $U,W$ are complementary if they satisfy

(a)	$U\cap W=\left\{0\right\},$
(b)	$U+W=V\text{.}$

Write $V=U\oplus W$ if $U$ and $W$ are complementary subspaces of $V\text{.}$

Linear transformations are for comparing vector spaces.

Let $𝔽$ be a field and let $A$ and $V$ be vector spaces over $𝔽\text{.}$ A linear transformation from $A$ to $V$ is a function $f:A\to V$ such that

(a)	If $a_1,a_2\in A$ then $f(a_1+a_2)=f\left(a_1\right)+f\left(a_2\right),$
(b)	If $c\in 𝔽$ and $a\in A$ then $f\left(ca\right)=cf\left(a\right)\text{.}$

Let $f:A\to V$ be a linear transformation. Then

(a)	$f\left(0\right)=0,$ and
(b)	if $a\in A$ then $f(-a)=-f\left(a\right)\text{.}$

		Proof.
	Assume $f:A\to V$ is a linear transformation. To show: (a) $f\left(0\right)=0\text{.}$ (b) If $a\in A$ then $f(-a)=-f\left(a\right)\text{.}$ (a) To show: $f\left(0\right)=0\text{.}$ $$f\left(0\right)=f(0+0)=f\left(0\right)+f\left(0\right)\text{.}$$ Add $-f\left(0\right)$ to each side. So $0=f\left(0\right)\text{.}$ (b) Assume $a\in A\text{.}$ To show: $f(-a)=-f\left(a\right)\text{.}$ To show: (ba) $f\left(a\right)+f(-a)=0,$ (bb) $f(-a)+f\left(a\right)=0\text{.}$ (ba) $f\left(a\right)+f(-a)=f(a+(-a))=f\left(0\right)=0\text{.}$ (bb) This follows from (a) since $V$ is a commutative group. $\square$

Let $f:A\to V$ be a linear transformation. The kernel of $f,$ or null space, of $f$ is $$\text{ker}\hspace{0.17em}f=\{a\in A\hspace{0.17em}|\hspace{0.17em}f\left(a\right)=0\}\text{.}$$ The image of $f$ is $$\text{im}\hspace{0.17em}f=\left\{f\left(a\right)\hspace{0.17em}\right|\hspace{0.17em}a\in A\}\text{.}$$

Let $f:A\to V$ be a linear transformation.

(a)	$\text{ker}\hspace{0.17em}f$ is a subspace of $A\text{.}$
(b)	$\text{im}\hspace{0.17em}f$ is a subspace of $V\text{.}$
(c)	$\text{ker}\hspace{0.17em}f=\left\{0\right\}$ if and only if $f$ is injective.
(d)	$\text{im}\hspace{0.17em}f=V$ if and only if $f$ is surjective.

		Proof.
	(a) To show: $\text{ker}\hspace{0.17em}f$ is a subspace of $A\text{.}$ To show: (aa) If $a_1,a_2\in \text{ker}\hspace{0.17em}f$ then $a_1+a_2\in \text{ker}\hspace{0.17em}f\text{.}$ (ab) $0\in \text{ker}\hspace{0.17em}f\text{.}$ (ac) If $a\in \text{ker}\hspace{0.17em}f$ then $-a\in \text{ker}\hspace{0.17em}f\text{.}$ (ad) If $c\in 𝔽$ and $a\in \text{ker}\hspace{0.17em}f$ then $ca\in \text{ker}\hspace{0.17em}f\text{.}$ (aa) Assume $a_1,a_2\in \text{ker}\hspace{0.17em}f\text{.}$ To show: $a_1+a_2\in \text{ker}\hspace{0.17em}f\text{.}$ To show: $f(a_1+a_2)=0\text{.}$ We know: $f\left(a_1\right)=0$ and $f\left(a_2\right)=0\text{.}$ So $f(a_1+a_2)=f\left(a_1\right)+f\left(a_2\right)=0+0=0\text{.}$ (ab) To show: $0\in \text{ker}\hspace{0.17em}f\text{.}$ To show: $f\left(0\right)=0\text{.}$ $f\left(0\right)=f(0+0)=f\left(0\right)+f\left(0\right)\text{.}$ Add $-f\left(0\right)$ to each side. So $0=f\left(0\right)\text{.}$ (ac) To show: If $a\in \text{ker}\hspace{0.17em}f$ then $-a\in \text{ker}\hspace{0.17em}f\text{.}$ Assume $a\in \text{ker}\hspace{0.17em}f\text{.}$ To show: $-a\in \text{ker}\hspace{0.17em}f\text{.}$ To show: $f(-a)=0\text{.}$ We know: $f\left(a\right)=0\text{.}$ $f(-a)=f((-1)·a)=(-1)·f\left(a\right)=(-1)·0=0\text{.}$ (ad) To show: if $c\in 𝔽$ and $a\in \text{ker}\hspace{0.17em}f$ then $ca\in \text{ker}\hspace{0.17em}f\text{.}$ Assume $c\in 𝔽$ and $a\in \text{ker}\hspace{0.17em}f\text{.}$ To show: $ca\in \text{ker}\hspace{0.17em}f\text{.}$ To show: $f\left(ca\right)=0\text{.}$ We know: $f\left(a\right)=0\text{.}$ $f\left(ca\right)=cf\left(a\right)=c·0=0\text{.}$ So $\text{ker}\hspace{0.17em}f$ is a subspace of $A\text{.}$ (c) To show: $\text{ker}\hspace{0.17em}f=\left\{0\right\}$ if and only if $f$ is injective. To show: (ca) If $\text{ker}\hspace{0.17em}f=\left\{0\right\}$ then $f$ is injective. (cb) If $f$ is injective then $\text{ker}\hspace{0.17em}f=\left\{0\right\}\text{.}$ (ca) Assume $\text{ker}\hspace{0.17em}f=\left\{0\right\}\text{.}$ To show: $f$ is injective. To show: If $a_1,a_2\in A$ and $f\left(a_1\right)=f\left(a_2\right)$ then $a_1=a_2\text{.}$ Assume $a_1,a_2\in A$ and $f\left(a_1\right)=f\left(a_2\right)\text{.}$ To show: $a_1=a_2\text{.}$ To show: $a_1-a_2=0\text{.}$ Since $$\begin{array}{ccc}{\displaystyle f(a_1-a_2)}& =& {\displaystyle f(a_1+(-1)a_2)}\\ & =& {\displaystyle f\left(a_1\right)+f\left((-1)a_2\right)}\\ & =& {\displaystyle f\left(a_1\right)+(-1)f\left(a_2\right)}\\ & =& {\displaystyle f\left(a_1\right)-f\left(a_2\right)}\\ & =& {\displaystyle f\left(a_1\right)-f\left(a_1\right)\text{(since}\hspace{0.17em}f\left(a_1\right)=f\left(a_2\right)\text{)}}\\ & =& {\displaystyle 0,}\end{array}$$ then $a_1-a_2\in \text{ker}\hspace{0.17em}f\text{.}$ Since $\text{ker}\hspace{0.17em}f=\left\{0\right\},$ then $a_1-a_2=0\text{.}$ So $f$ is injective. (cb) To show: If $f$ is injective then $\text{ker}\hspace{0.17em}f=\left\{0\right\}\text{.}$ Assume $f$ is injective. To show: $\text{ker}\hspace{0.17em}f=\left\{0\right\}\text{.}$ To show: If $a\in A$ and $f\left(a\right)=0$ then $a=0\text{.}$ Assume $a\in A$ and $f\left(a\right)=0\text{.}$ To show: $a=0\text{.}$ We know: $f\left(a\right)=f\left(0\right)\text{.}$ Since $f$ is injective, $a=0\text{.}$ $\square$

Notes and References

These are a typed copy of Lecture 7 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 9, 2011.

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