Table of Contents

Injective and Surjective Linear Maps
We will now look at two important types of linear maps  maps that are injective, and maps that are surjective, both of which terms are analogous to that of regular functions.
Injective Linear Maps
Definition: A linear map $T \in \mathcal L (V, W)$ is said to be Injective or OnetoOne if whenever $T(u) = T(v)$ ($u, v \in V$), then $u = v$. 
Therefore, a linear map $T$ is injective if every vector from the domain $V$ maps to a unique vector in the codomain $W$.
For example, consider the identity map $I \in \mathcal L (V, V)$ defined by $T(v) = v$ for all $v \in V$. This linear map is injective. Suppose that $T(u) = T(v)$. Then since $T(u) = u$ and $T(v) = v$ we have that $u = v$.
In general, to show that a linear map $T$ is injective we must assume that $T(u) = T(v)$ and then show this assumption implies that $u = v$. Though, there is another way…
The following lemma will provide us with an easy way to determine whether a linear map $T$ is injective or not.
Lemma 1: If $T \in \mathcal L (V, W)$, then the linear map $T$ is injective if and only if $\mathrm{null} (T) = \{ 0 \}$. 
 Proof: $\Rightarrow$ Suppose that $T$ is injective. We want to show that $\mathrm{null} (T) = \{ 0 \}$. We will do this by showing that $\{ 0 \} \subseteq \mathrm{null} (T)$ and $\mathrm{null} (T) \subseteq \{ 0 \}$.
 We already know that $\{ 0 \} \subseteq \mathrm{null} (T)$ since we've already verified that the zero vector $0_V \in V$ is mapped to the zero vector $0_W \in W$. Now suppose that $v \in \mathrm{null} (T)$. Then by the definition of the null space of $T$ we have that $T(v) = 0$. But we also have that $T(0) = 0$. Therefore $T(v) = T(0)$. Since $T$ is injective, this implies that $v = 0$, and so $v \in \mathrm{null} (T)$, so $\mathrm{null} (T) \subseteq \{ 0 \}$. Therefore we conclude that $\mathrm{null} (T) = \{ 0 \}$
 $\Leftarrow$ Suppose that $\mathrm{null} (T) = \{ 0 \}$. We want to show that $T$ is injective. Let $u, v \in V$ and suppose that $T(u) = T(v)$. Then $T(u)  T(v) = 0$. Since $T$ is a linear map we have that $T(u)  T(v) = T(u  v) = 0$. Now this implies that $(u  v) \in \mathrm{null} (T) = \{ 0 \}$ and so $u  v = 0$. Therefore $u = v$, and so by the definition, $T$ is injective. $\blacksquare$
Surjective Linear Maps
Definition: A linear map $T \in \mathcal L (V, W)$ is said 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 $w = T(v)$. 
One such example of a surjective linear map is the linear map for polynomial differentiation $T \in \mathcal L (\wp (\mathbb{R}), \wp (\mathbb{R}))$ defined by $T(p(x)) = p'(x)$. This map is surjective since any polynomial $q(x) = a_0 + a_1x + a_2x^2 + ..$ is antidifferentiable to a polynomial $p(x) \in \wp (\mathbb{R})$, and so for any $q(x)$ there exists a $p(x)$ such that $T(p(x)) = p'(x) = q(x)$.
In general, to show that a linear map $T$ is surjective we must show that for any vector $w \in W$ there exists a vector $v$ that maps to $w$ under $T$.