126 lines
No EOL
4.8 KiB
Markdown
126 lines
No EOL
4.8 KiB
Markdown
# Linear transformations
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## Definition
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> *Definition*: let $V$ and $W$ be vector spaces, a mapping $L: V \to W$ is a **linear transformation** or **linear map** if
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>
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> $$
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> L(\lambda \mathbf{v}_1 + \mu \mathbf{v}_2) = \lambda L(\mathbf{v}_1) + \mu L(\mathbf{v}_2),
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> $$
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>
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> for all $\mathbf{v}_{1,2} \in V$ and $\lambda, \mu \in \mathbb{K}$.
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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**.
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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** .
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## The image and kernel
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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 $W$.
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> *Definition*: let $L: V \to W$ be a linear transformation. The **kernel** of $L$, denoted by $\ker(L)$, is defined by
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>
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> $$
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> \ker(L) = \{\mathbf{v} \in V \;|\; L(\mathbf{v}) = \mathbf{0}\}.
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> $$
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The kernel is therefore a set consisting of vectors in $V$ that get mapped into the zero vector of $W$.
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> *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
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>
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> $$
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> L(S) = \{\mathbf{w} \in W \;|\; \mathbf{w} = L(\mathbf{v}) \text{ for } \mathbf{v} \in S \}.
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> $$
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>
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> The image of the entire vector space $L(V)$, is called the **range** of $L$.
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With these definitions the following theorem may be posed.
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> *Theorem*: if $L: V \to W$ is a linear transformation and $S$ is a subspace of $V$, then
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>
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> 1. $\ker(L)$ is a subspace of $V$.
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> 2. $L(S)$ is a subspace of $W$.
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??? note "*Proof*:"
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Let $L: V \to W$ be a linear transformation and $S$ is a subspace of $V$.
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To prove 1, let $\mathbf{v}_{1,2} \in \ker(L)$ and let $\lambda, \mu \in \mathbb{K}$. Then
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$$
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L(\lambda \mathbf{v}_1 + \mu \mathbf{v}_2) = \lambda L(\mathbf{v}_1) + \mu L(\mathbf{v}_2) = \lambda \mathbf{0} + \mu \mathbf{0} = \mathbf{0},
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$$
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therefore $\lambda \mathbf{v}_1 + \mu \mathbf{v}_2 \in \ker(L)$ and hence $\ker(L)$ is a subspace of $V$.
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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
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$$
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\lambda \mathbf{w}_1 + \mu \mathbf{w}_2 = \lambda L(\mathbf{v}_1) + \mu L(\mathbf{v}_2) = L(\lambda \mathbf{v}_1 + \mu \mathbf{v}_2),
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$$
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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 $W$.
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## Matrix representations
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> *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
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>
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> $$
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> L(\mathbf{x}) = A \mathbf{x},
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> $$
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>
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> for all $x \in \mathbb{R}^n$. With the $i$th column vector of $A$ given by
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>
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> $$
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> \mathbf{a}_i = L(\mathbf{e}_i),
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> $$
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>
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> for a basis $\{\mathbf{e}_1, \dots, \mathbf{e}_n\} \subset \mathbb{R}^n$ and $i \in \{1, \dots, n\}$.
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??? note "*Proof*:"
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For $i \in \{1, \dots, n\}$, define
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$$
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\mathbf{a}_i = L(\mathbf{e}_i),
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$$
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and let
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$$
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A = (\mathbf{a}_1, \dots, \mathbf{a}_n).
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$$
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If $\mathbf{x} = x_1 \mathbf{e}_1 + \dots + x_n \mathbf{e}_n$ is an arbitrary element of $\mathbb{R}^n$, then
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$$
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\begin{align*}
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L(\mathbf{x}) &= x_1 L(\mathbf{e}_1) + \dots + x_n L(\mathbf{e}_n), \\
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&= x_1 \mathbf{a}_1 + \dots + x_n \mathbf{a}_n, \\
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&= A \mathbf{x}.
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\end{align*}
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$$
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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.
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> *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 $E$,
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>
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> $$
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> \mathbf{e}_i = S \mathbf{f}_i,
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> $$
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>
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> for $i \in \mathbb{N}; i\leq n$.
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>
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> If $A$ is the matrix representing $L$ with respect to $E$, and $B$ is the matrix representing $L$ with respect to $F$, then
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>
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> $$
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> B = S^{-1} A S.
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> $$
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??? note "*Proof*:"
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Will be added later.
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> *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 $B = S^{-1} A S$.
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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. |