62 lines
2.6 KiB
Markdown
62 lines
2.6 KiB
Markdown
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# 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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In the case that the vector spaces $V$ and $W$ are the same; $V=W$, a linear transformation $L: V \to V$ will be reffered to as a **linear operator** on $V$.
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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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