# Injective and Surjective Linear Maps Examples 1

Recall from the Injective and Surjective Linear Maps page that a linear map $T : V \to W$ is said to be **injective** if:

- $T(u) = T(v)$ implies that $u = v$.

- $\mathrm{null} (T) = \{ 0 \}$.

Furthermore, the linear map $T : V \to W$ is said to be surjective if:**

- If for every $w \in W$ there exists a $v \in V$ such that $T(v) = w$.

- $\mathrm{range} (T) = W$.

We will now look at some examples regarding injective/surjective linear maps.

## Example 1

**Let $T \in \mathcal L ( \wp (\mathbb{R}), \mathbb{R})$ be defined by $T(p(x)) = \int_0^1 2p'(x) \: dx$. Prove whether or not $T$ is injective, surjective, or both.**

We will first determine whether $T$ is injective. Suppose that $p(x) \in \wp (\mathbb{R})$ and $T(p(x)) = 0$. Then we have that:

(1)Note that if $p(x) = C$ where $C \in \mathbb{R}$, then $p'(x) = 0$ and hence $2 \int_0^1 p'(x) \: dx = 0$. Hence $\mathrm{null} (T) \neq \{ 0 \}$ and so $T$ is not injective.

We will now determine whether $T$ is surjective. Suppose that $C \in \mathbb{R}$. We want to determine whether or not there exists a $p(x) \in \wp (\mathbb{R})$ such that:

(2)Take the polynomial $p(x) = \frac{C}{2}x$. Then $p'(x) = \frac{C}{2}$ and hence:

(3)Therefore $T$ is surjective.

## Example 2

**Suppose that $S_1, S_2, ..., S_n$ are injective linear maps for which the composition $S_1 \circ S_2 \circ ... \circ S_n$ makes sense. Prove that $S_1 \circ S_2 \circ ... \circ S_n$ is injective.**

Let $u$ and $v$ be vectors in the domain of $S_n$, and suppose that:

(4)From the equation above we see that $S_n (u) = S_n(v)$ and since $S_n$ injective this implies that $u = v$. Since the remaining maps $S_1, S_2, ..., S_{n-1}$ are also injective, we have that $u = v$, so $S_1 \circ S_2 \circ ... \circ S_n$ is injective.

## Example 3

**Let $T$ be a linear map from $V$ to $W$, and suppose that $T$ is injective and that $\{ v_1, v_2, ..., v_n \}$ is a linearly independent set of vectors in $V$. Show that $\{ T(v_1), ..., T(v_n) \}$ is a linearly independent set of vectors in $W$.**

Consider the following equation (noting that $T(0) = 0$):

(5)Now since $T$ is injective, this implies that $a_1v_1 + a_2v_2 + ... + a_nv_n = 0$. However, $\{ v_1, v_2, ..., v_n \}$ is a linearly independent set in $V$ which implies that $a_1 = a_2 = ... = a_n = 0$. Therefore $\{ T(v_1), T(v_2), ..., T(v_n) \}$ is a linearly independent set in $W$.

## Example 4

**Let $T$ be a linear map from $V$ to $W$ and suppose that $T$ is surjective and that the set of vectors $\{ v_1, v_2, ..., v_n \}$ spans $V$. Show that $\{ T(v_1), T(v_2), ..., T(v_n) \}$ spans $W$.**

Let $w \in W$. Since $T$ is surjective, then there exists a vector $v \in V$ such that $T(v) = w$, and since $\{ v_1, v_2, ..., v_n \}$ spans $V$, then we have that $v$ can be written as a linear combination of this set of vectors, and so for some $a_1, a_2, ..., a_n \in \mathbb{F}$ we have that $v = a_1v_1 + a_2v_2 + ... + a_nv_n$ and so:

(6)Therefore any $w \in W$ can be written as a linear combination of $\{ T(v_1), T(v_2), ..., T(v_n) \}$ and so $\{ T(v_1), T(v_2), ..., T(v_n) \}$ spans $W$.