Linear Tranformation

$$\newcommand{\0}{\textbf 0} \newcommand{\1}{\textbf 1} \newcommand{\x}{\boldsymbol x} \newcommand{\y}{\boldsymbol y} \newcommand{\z}{\boldsymbol z} \newcommand{\u}{\boldsymbol u} \newcommand{\v}{\boldsymbol v} \newcommand{\Ker}{\text{ker}} \renewcommand{\Im}{\text{Im}} \newcommand{\Span}{\text{span}} \newcommand{\Norm}{\unlhd} \newcommand{\Null}{\text{nullity}} \newcommand{\Rank}{\text{rank}} $$

1. Linear Transformation

Definition. Linear transformation

Let $V$ and $W$ be vector spaces (over $F$). We call a function $T:V\to W$ a linear transformation from $V$ to $W$ if $\forall \x,\y \in V,\forall c\in F$.

$$\begin{aligned} T(\x+\y)&=T(\x)+T(\y)\\ T(c\x)&=cT(\x) \end{aligned} $$

If $T$ is a linear transformation, we often just call $T$ linear. It could be checked that $T$ is linear iff for any $\{\x_k\}_{k=1}^n$ and $\{a_k\}_{k=1}^n$, we have

$$T(\sum_{k=1}^n a_k\x_k)=\sum_{k=1}^na_kT(\x_k) $$

Definition. Identity transformation
For vector space $V$ over $F$, the identity transformation $I_V$ is defined as follows

$$\begin{aligned} I_V:V&\to V\\ \v&\mapsto \v \end{aligned} $$

We can write $I$ instead of $I_V$

Definition. Zero transformation
For vector space $V$ and $W$ over $F$, the zero transformation $T_0$ is defined as follows

$$\begin{aligned} T_0:V&\to W\\ \x&\mapsto \0 \end{aligned} $$

Definition. Null space / kernel, range
Let $V$ and $W$ be two vector spaces. $T:V\to W$ is linear. We define the null space $N(T)$ / kernel $\Ker(T)$ of $T$ as

$$\Ker(T):=\{\x\in V\mid T(\x)=\0\}=:T^{-1}(\{\0\}) $$

The range / image of $T$ is defined as $\Im(T):=\{T(\v)\mid \v\in V\}=:T(V)$

Theorem. Let $V$ and $W$ be vector spaces and $T:V\to W$ is linear. Then $\Ker(T)$ is a subspace of $V$, and $\Im(T)$ is a subspace of $W$.
Pf. Easy to show using definition of subspaces.

From the perspective of groups, $N$ is the kernel of some group homomorphism from $G$ is equivalent to $N\Norm G$. Here $V$ is abelian, so every kernel should be corresponding a subspace of $V$.

****Theorem**. Let $V$ and $W$ be vector spaces and $T:V\to W$ is linear. If $\beta=\{\v_k\}_{k=1}^n$ is a basis for $V$, then

$$\Im(T)=\Span(T(\beta))=\Span(\{T(\v_1), T(\v_2), \ldots, T(\v_n)\}) $$

Definition. Nullity / rank
Let $V$ and $W$ be two vector spaces. $T:V\to W$ is linear. We define the nullity (denoted by $\Null(T)$ ) and rank of $T$ (denoted by $\Rank(T)$) as follows

$$\begin{aligned} \Null(T)&:=\dim(\Ker(T))\\ \Rank(T)&:=\dim(\Rank(T)) \end{aligned} $$

Theorem. Dimension theorem