Injective and Surjective Linear Maps

# 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 One-to-One 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 anti-differentiable 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 $T$.