Injective and Surjective Linear Maps Examples 4

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 $T \in \mathcal L (V, W)$ and let $W$ be a finite-dimensional vector space. Prove that $T$ is injective if and only if there exists a linear map $S \in \mathcal L (W, V)$ such that $ST = I$ where $I$ is the identity map on $V$.

$\Rightarrow$ Suppose that $T$ is an injective linear map. We note that if $\mathrm{dim} (V) > \mathrm{dim} (W)$ then no linear map from $V$ to $W$ is injective. Since $T$ is injective, we must have that $\mathrm{dim} (V) ≤ \mathrm{dim} (W)$. We're also given that $W$ is a finite-dimensional vector space which implies that them $V$ is also a finite-dimensional vector space.

Let $\{ v_1, v_2, ..., v_n \}$ be a basis of $V$. We saw in a previous example that if $\{ v_1, v_2, ..., v_n \}$ is linearly independent and if $T$ is injective then $\{ T(v_1), T(v_2), ..., T(v_n) \}$ is linearly independent in $W$. This set of has $n$ vectors in it, and hence can be extended to a basis of $W$, say $\{ T(v_1), T(v_2), ..., T(v_n), w_{n+1}, ..., w_m \}$.

We now define a linear map $S \in \mathcal L(V, W)$ as:

(1)

\begin{align} \quad S(T(v_1)) = v_1 \\ \quad S(T(v_2)) = v_2 \\ \quad \quad \vdots \quad \quad \\ \quad S(T(v_n)) = v_n \\ \quad S(w_{n+1}) = 0 \\ \quad S(w_{n+2}) = 0 \\ \quad \quad \vdots \quad \quad \\ \quad S(w_m) = 0 \end{align}

So for any vector $v = a_1v_1 + a_2v_2 + ... + a_nv_n$ we have that:

(2)

\begin{align} \quad S(T(a_1v_1 + a_2v_2 +... + a_nv_n)) = S(a_1T(v_1) + a_2T(v_2) + ... + a_nT(v_n)) \\ \quad = a_1S(T(v_1)) + a_2S(T(v_2)) + ... + a_nS(T(v_n)) = a_1v_1 + a_2v_2 + ... + a_nv_n \end{align}

Therefore $(ST)(v) = I(v) = v$ for every vector $v \in V$.

$\Leftarrow$ Suppose that there exists a linear map $S \in \mathcal L(W, V)$ such that $ST = I$ where $I$ is the identity map on $V$. Let $u, v \in V$ such that $u \neq v$. Then we have that:

(3)

\begin{align} \quad u = Iu = (ST)(u) \end{align}

We also have that:

(4)

\begin{align} \quad v = Iv = (ST)(v) \end{align}

Therefore $(ST)(u) \neq (ST)(v)$, which implies that $T(u) \neq T(v)$. Therefore $T$ is injective.

Example 2

Let $T \in \mathcal L (V, W)$ and let $V$ be a finite-dimensional vector space. Prove that $T$ is surjective if and only if there exists a linear map $S \in \mathcal L (W, V)$ such that $TS = I$ where $I$ is the identity map on $W$.

$\Rightarrow$ Suppose that $T$ is surjective. Then we must have that $\mathrm{dim} (W) ≤ \mathrm{dim} (V)$. We're already given that $V$ is finite-dimensional and so $W$ must also be finite-dimensional. Let $\{ w_1, w_2, ..., w_m \}$ be a basis of $W$. Since $T$ is surjective, then for each $j = 1, 2, ..., m$ there exists a vector $v_j \in V$ such that $w_j = T(v_j)$.

We can define a linear map $S \in \mathcal L (W, V)$ by:

(5)

\begin{align} \quad S(w_1) = v_1 \\ \quad S(w_2) = v_2 \\ \quad \quad \vdots \quad \quad \\ \quad S(w_m) = v_m \end{align}

Then we have that:

(6)

\begin{align} \quad T(S(w_1)) = T(v_1) = w_1 \\ \quad T(S(w_2)) = T(v_2) = w_2 \\ \quad \quad \vdots \quad \quad \\ \quad T(S(w_m)) = T(v_m) = w_m \end{align}

So for any vector $w = a_1w_1 + a_2w_2 + ... + a_mw_m$ we have that:

(7)

\begin{align} \quad T(S(a_1w_1 + a_2w_2 + ... + a_mw_m)) = T(a_1S(w_1) + a_2S(w_2) + ... + a_mS(w_m)) \\ = T(a_1v_1 + a_2v_2 + ... + a_mv_m) = a_1T(v_1) + a_2T(v_2) + ... + a_mT(v_m) = a_1w_1 + a_2w_2 + ... + a_mw_m \end{align}

Therefore $TS = I$ where $I$ is the identity map on $W$.

$\Leftarrow$ Suppose that there exists a linear map $S \in \mathcal L (W, V)$ such that $TS = I$ where $I$ is the identity map on $W$. Then for any vector $w \in W$ we have that:

(8)

\begin{align} \quad w = Iw = (TS)w = T(S(w)) \end{align}

Therefore, for every vector $w \in W$ there exists a vector $S(w) \in V$ such that $T(S(w)) = w$, so $T$ is surjective.