# 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.