# Linear transformations

## Definition

> *Definition*: let $V$ and $W$ be vector spaces, a mapping $L: V \to W$ is a **linear transformation** or **linear map** if 
>
> $$
>   L(\lambda \mathbf{v}_1 + \mu \mathbf{v}_2) = \lambda L(\mathbf{v}_1) + \mu L(\mathbf{v}_2),
> $$
>
> for all $\mathbf{v}_{1,2} \in V$ and $\lambda, \mu \in \mathbb{K}$. 

A linear transformation may also be called a **vector space homomorphism**. If the linear transformation is a bijection then it may be called a **linear isomorphism**.

In the case that the vector spaces $V$ and $W$ are the same; $V=W$, a linear transformation $L: V \to V$ will be referred to as a **linear operator** on $V$ or **linear endomorphism** . 

## The image and kernel

Let $L: V \to W$ be a linear transformation from a vector space $V$ to a vector space $W$. In this section the effect is considered that $L$ has on subspaces of $V$. Of particular importance is the set of vectors in $V$ that get mapped into the zero vector of $W$. 

> *Definition*: let $L: V \to W$ be a linear transformation. The **kernel** of $L$, denoted by $\ker(L)$, is defined by
>
> $$
>   \ker(L) = \{\mathbf{v} \in V \;|\; L(\mathbf{v}) = \mathbf{0}\}.
> $$

The kernel is therefore a set consisting of vectors in $V$ that get mapped into the zero vector of $W$. 

> *Definition*: let $L: V \to W$ be a linear transformation and let $S$ be a subspace of $V$. The **image** of $S$, denoted by $L(S)$, is defined by
>
> $$
>   L(S) = \{\mathbf{w} \in W \;|\; \mathbf{w} = L(\mathbf{v}) \text{ for } \mathbf{v} \in S \}.
> $$
>
> The image of the entire vector space $L(V)$, is called the **range** of $L$. 

With these definitions the following theorem may be posed.

> *Theorem*: if $L: V \to W$ is a linear transformation and $S$ is a subspace of $V$, then
>
> 1. $\ker(L)$ is a subspace of $V$.
> 2. $L(S)$ is a subspace of $W$. 

??? note "*Proof*:"

    Let $L: V \to W$ be a linear transformation and $S$ is a subspace of $V$. 

    To prove 1, let $\mathbf{v}_{1,2} \in \ker(L)$ and let $\lambda, \mu \in \mathbb{K}$. Then

    $$
        L(\lambda \mathbf{v}_1 + \mu \mathbf{v}_2) = \lambda L(\mathbf{v}_1) + \mu L(\mathbf{v}_2) = \lambda \mathbf{0} + \mu \mathbf{0} = \mathbf{0},
    $$

    therefore $\lambda \mathbf{v}_1 + \mu \mathbf{v}_2 \in \ker(L)$ and hence $\ker(L)$ is a subspace of $V$. 

    To prove 2, let $\mathbf{w}_{1,2} \in L(S)$ then there exist $\mathbf{v}_{1,2} \in S$ such that $\mathbf{w}_{1,2} = L(\mathbf{v}_{1,2})$ For any $\lambda, \mu \in \mathbb{K}$ we have

    $$
        \lambda \mathbf{w}_1 + \mu \mathbf{w}_2 = \lambda L(\mathbf{v}_1) + \mu L(\mathbf{v}_2) = L(\lambda \mathbf{v}_1 + \mu \mathbf{v}_2),
    $$

    since $\lambda \mathbf{v}_1 + \mu \mathbf{v}_2 \in S$ it follows that $\lambda \mathbf{w}_1 + \mu \mathbf{w}_2 \in L(S)$ and hence $L(S)$ is a subspace of $W$. 

## Matrix representations 

> *Theorem*: let $L: \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation, then there is an $m \times n$ matrix $A$ such that 
>
> $$
>   L(\mathbf{x}) = A \mathbf{x},
> $$
>
> for all $x \in \mathbb{R}^n$. With the $i$th column vector of $A$ given by
>
> $$
>   \mathbf{a}_i = L(\mathbf{e}_i),
> $$
>
> for a basis $\{\mathbf{e}_1, \dots, \mathbf{e}_n\} \subset \mathbb{R}^n$ and $i \in \{1, \dots, n\}$. 

??? note "*Proof*:"

    For $i \in \{1, \dots, n\}$, define

    $$
        \mathbf{a}_i = L(\mathbf{e}_i),
    $$

    and let

    $$
        A = (\mathbf{a}_1, \dots, \mathbf{a}_n).
    $$

    If $\mathbf{x} = x_1 \mathbf{e}_1 + \dots + x_n \mathbf{e}_n$ is an arbitrary element of $\mathbb{R}^n$, then

    $$
    \begin{align*}
        L(\mathbf{x}) &= x_1 L(\mathbf{e}_1) + \dots + x_n L(\mathbf{e}_n), \\
                    &= x_1 \mathbf{a}_1 + \dots + x_n \mathbf{a}_n, \\
                    &= A \mathbf{x}.
    \end{align*}
    $$

It has therefore been established that each linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ can be represented in terms of an $m \times n$ matrix.

> *Theorem*: let $E = \{\mathbf{e}_1, \dots, \mathbf{e}_n\}$ and $F = \{\mathbf{f}_1, \dots, \mathbf{f}_n\}$ be two ordered bases for a vector space $V$, and let $L: V \to V$ be a linear operator on $V$, $\dim V = n \in \mathbb{N}$. Let $S$ be the $n \times n$ transition matrix representing the change from $F$ to $E$, 
> 
> $$
> \mathbf{e}_i = S \mathbf{f}_i,
> $$ 
> 
> for $i \in \mathbb{N}; i\leq n$. 
> 
> If $A$ is the matrix representing $L$ with respect to $E$, and $B$ is the matrix representing $L$ with respect to $F$, then
>
> $$
>   B = S^{-1} A S.
> $$

??? note "*Proof*:"

    Will be added later.

> *Definition*: let $A$ and $B$ be $n \times n$ matrices. $B$ is said to be **similar** to $A$ if there exists a nonsingular matrix $S$ such that $B = S^{-1} A S$. 

It follows from the above theorem that if $A$ and $B$ are $n \times n$ matrices representing the same operator $L$, then $A$ and $B$ are similar.