Linear Maps

For vectors spaces $\mathbf{V}$ and $\mathbf{W}$ over the same field, the function $T : V \rightarrow W$ is a linear map if it satisfies the following:

$$ \text{L1. $\textbf{Preservation of Additive Structure}$. } T(v+u) = T(v) + T(u) \\

\text{L2. $\textbf{Homogeneity}$. } T(\lambda v) = \lambda T(v) \hspace{39pt} \\ $$

The set of all linear maps from $\mathbf{V}$ to $\mathbf{W}$ is denoted $\mathscr{L}(\mathbf{V}, \mathbf{W})$ and the set of all operators, linear maps from a vector space $\mathbf{V}$ to itself, is denoted $\mathscr{L}(\mathbf{V})$.

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Proposition. $\mathscr{L}(\mathbf{V}, \mathbf{W})$ is a subspace of $\mathbf{W} ^\mathbf{V}$.

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Proposition. If $T \in \mathscr{L}(\mathbf{V}, \mathbf{W})$ then $T(0_\mathbf{V}) = 0_\mathbf{W}$.

• Proof </aside>

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Lemma. For finite dimensional $\mathbf{V}$ and $\mathbf{W}$ with $\dim \mathbf{V} = \dim \mathbf{W}$, then for any $T, S \in \mathscr{L}(\mathbf{V}, \mathbf{W})$, $T$ has a left inverse iff it has a right inverse.

$$ ST=I_\mathbf{V} \iff TS = I_\mathbf{W} $$

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Lemma. For finite dimensional $\mathbf{V}$ and $\mathbf{W}$ over the same field,

$$ \dim \mathbf{V} > \dim \mathbf{W} \rightarrow \nexists T \in \mathscr{L}(\mathbf{V}, \mathbf{W}) : \text{$T$ is injective} $$

$$ \dim \mathbf{V} < \dim \mathbf{W} \rightarrow \nexists T \in \mathscr{L}(\mathbf{V}, \mathbf{W}) : \text{$T$ is surjective} $$

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Linear Map Lemma. For any basis $v_1, \dots, v_n$ of $\mathbf{V}$ and any list $w_1, \dots, w_n$ in $\mathbf{W}$, there exists a unique $T \in \mathscr{L}(\mathbf{V}, \mathbf{W})$ such that $T(v_i) = w_i$ for $i = 1, \dots, n$.

• Proof

In general, we are always able to conjure up a well-defined function which assigns a basis to a particular list. The power of this lemma comes from the fact that any such function is a linear map.

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Fundamental Theorem of Linear Maps. For any finite dimensional vector field $\mathbf{V}$ and linear map $T \in \mathscr(\mathbf{V}, \mathbf{W})$, $\text{range }T$ is also finite dimensional and the following formula is given:

$$ \dim V = \dim (\text{null } T) + \dim (\text{range }T) $$

• Proof </aside>

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Fredholm Alternative. For finite-dimensional vector spaces $\mathbf{V}$ and $\mathbf{W}$ with $\dim V = \dim W$, then for any $T \in \mathscr{L}(\mathbf{V}, \mathbf{W})$, the following is true.

$$ \text{$T$ is bijective $\iff$ $T$ is injective $\iff$ $T$ is surjective} $$

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Null Space

The null space of a linear map $T \in \mathscr{L}(\mathbf{V}, \mathbf{W})$ is the set of all vectors it maps to $0_\mathbf{W}$.

$$ \ker T = \text{null }T := \{v \in V:T(v) = 0_\mathbf{W} \} $$

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Proposition. For $T \in\mathscr{L}(\mathbf{V}, \mathbf{W})$, $\text{null }T \leq \mathbf{V}$.

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Lemma. For $T \in\mathscr{L}(\mathbf{V}, \mathbf{W})$, $T$ is injective iff its null space is trivial.

$$ T \text{ is injective} \iff \text{null }T = \{0_\mathbf{V}\} $$

• Proof </aside>

Range

The range of a linear map $T \in \mathscr{L}(\mathbf{V}, \mathbf{W})$ is the set of all vectors it maps elements of $V$ to.