# Linear Maps Review

We will now review some of the recent content regarding linear maps.

- Recall from the Linear Maps that a
**Linear Map**or**Linear Transformation**from a vector space $V$ to another vector space $W$ is a function $T : V \to W$ with the property that $T(au + bv) = aT(u) + bT(v)$ for all vectors $u, v \in V$ and for all scalars $a, b \in \mathbb{F}$.

- The simplest example of a linear map is the zero map which takes every vector $v \in V$ and maps it to the zero vector in $W$, that is $0(v) = 0$ for all $v \in V$. Another example of a linear map is the identity map $I : V \to V$ which takes every vector $v \in V$ and maps it to back to itself, that is, $T(v) = v$ for all $v \in V$.

- A more complex example of a linear map is that differentiation of polynomials map $T : \wp (\mathbb{R}) \to \wp (\mathbb{R})$ which takes every polynomial and maps it to its derivative, that is, $T(p(x)) = p'(x)$ for every $p(x) \in \wp (\mathbb{R})$.

- The set of all linear maps from a vector space $V$ to another vector space $W$ is denoted $\mathcal L (V, W)$.

- On the Addition, Multiples, and Compositions of Linear Maps page we saw that if $S$ and $T$ are both linear maps from $V$ to $W$ then their sum $S + T : V \to W$ defined by $(S + T)(v) = S(v) + T(v)$ is also a linear map from $V$ to $W$.

- Furthermore, we saw that if $T$ is a linear map from $V$ to $W$ and if $a \in \mathbb{F}$ then the scalar product $aT : V \to W$ defined by $(aT)(v) = aT(v)$ is also a linear map from $V$ to $W$.

- We also looked at compositions of linear maps. If $T \in \mathcal L (U, V)$ and $S \in \mathcal L (V, W)$ then the composition $S \circ T : U \to W$ defined by $S \circ T = ST(v) = S(T(v))$ is also a linear map, but from $U$ to $W$.

- On the Null Space of a Linear Map page we defined the
**Null Space**of a linear map as the set of vectors in $V$ which are mapped to $0$ in $W$, that is:

(1)

\begin{align} \quad \mathrm{null} (T) = \{ v \in V : T(v) = 0 \} \end{align}

- For example, the null space of the zero map is $V$ since every vector $v \in V$ is mapped to the zero vector. As another example, the null space of the identity map is $\{ 0 \}$ since $T(v) = v$ implies that only the zero vector is mapped to $0$.

- We also saw that $\mathrm{null} (T)$ is always subspace of the domain vector space $V$.

- On the Range of a Linear Map page we defined the
**Range**of a linear map as the set of vectors in $W$ which are mapped to from [[$ V $], that is:

(2)

\begin{align} \quad \mathrm{range} (T) = \{ T(v) : v \in V \} \end{align}

- For example, the range of the zero map is $\{ 0 \}$ since every vector $v \in V$ is mapped to $0 \in W$. The range of the identity map is $V$ since $T(v) = v$ for every vector $v \in V$.

- We also saw that $\mathrm{range} (T)$ is always a subspace of the codomain vector space $W$.

- On the Injective and Surjective Linear Maps page we defined a linear map $T \in \mathcal L (V, W)$ to be
**Injective**or**One-to-one**if whenever $T(u) = T(v)$ we have that $u = v$. One example of an injective linear map is the identity map.

- We saw that a linear map $T$ is injective if and only if $\mathrm{null} (T) = \{ 0 \}$.

- We also defined a linear map $T \in \mathcal L (V, W)$ to be
**Surjective**or**Onto**if $\mathrm{range} (T) = W$, that is, for every vector $w \in W$ there exists a vector $v \in V$ such that $T(v) = w$. One example of a surjective linear map is the differentiation of polynomials map.

- On The Dimension of The Null Space and Range we noted that if $T \in \mathcal L (V, W)$ and $V$ is a finite-dimensional vector space then the following formula for the dimension of $V$ holds:

(3)

\begin{align} \quad \mathrm{dim} (V) = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} ( \mathrm{range} (T)) \end{align}

- As one important corollary to this formula, we noted that if $\mathrm{dim} V > \mathrm{dim} W$ then $T$ cannot be injective. This is because if $\mathrm{dim} V > \mathrm{dim} W$ then:

(4)

\begin{align} \quad \mathrm{dim} (\mathrm{null} (T)) = \mathrm{dim} (V) – \mathrm{dim} (\mathrm{range} (T)) ≥ \mathrm{dim} V – \mathrm{dim} (W) > 0 \quad \Leftrightarrow \quad \mathrm{null} (T) \neq \{ 0 \} \end{align}

- Similarly, if $\mathrm{dim} V then $T$ cannot be surjective. This is because if $\mathrm{dim} V then:

(5)

\begin{align} \quad \mathrm{dim} (\mathrm{range} (T)) = \mathrm{dim} (V) – \mathrm{dim} (\mathrm{dim} (\mathrm{null} (T)) ≤ \mathrm{dim} V

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