Injective and Surjective Linear Maps Examples 3

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 more examples regarding injective/surjective linear maps.

Example 1

Let $\{ v_1, v_2, ..., v_n\}$ be a linearly independent set of vectors in $V$ and let $T \in \mathcal L (V, W)$ be an injective linear map. Prove that then $\{ T(v_1), T(v_2), ..., T(v_n) \}$ is a linearly independent set of vectors in $W$.

For $a_1, a_2, ..., a_n \in \mathbb{F}$, consider the following vector equation:

(1)

\begin{align} \quad a_1T(v_1) + a_2T(v_2) + ... + a_nT(v_n) = 0 \\ \quad T(a_1v_1) + T(a_2v_2) + ... + T(a_nv_n) = 0 \\ \quad T(a_1v_1 + a_2v_2 + ... + a_nv_n) = 0 \\ \quad T(a_1v_1 + a_2v_2 + ... + a_nv_n) = T(0) \end{align}

Now since $T$ is injective, we have that then $a_1v_1 + a_2v_2 + ... + a_nv_n = 0$. We are given that $\{ v_1, v_2, ..., v_n \}$ is a linearly independent set of vectors 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 of vectors in $W$.

Example 2

Reprove that if $T \in \mathcal L (V, W)$ then the linear map $T$ is injective if and only if $\mathrm{null} (T) = \{ 0 \}$.

$\Rightarrow$ Suppose that $T$ is injective. We want to show that $\mathrm{null} (T) = \{ 0 \}$.

Let $v \in \mathrm{null} (T)$. Then $T(v) = 0 = T(0)$. Now since $T$ is injective, we have that $T(v) = T(0)$ implies that $v = 0$, so $v \in \{ 0 \}$. Furthermore, if we let $v \in \{ 0 \}$ then $v = 0$ and clearly $0 \in \mathrm{null} (T)$ since $T(0) = 0$. Therefore $\mathrm{null} (T) = \{ 0 \}$.

$\Leftrightarrow$ Suppose now 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 we have that $T(u) - T(v) = 0$ so $T(u - v) = 0$ which implies that $(u - v) \in \mathrm{null} (T) = \{ 0 \}$ so $u - v = 0$ and so $u = v$. Therefore $T$ is injective.