Injective and Surjective Linear Maps Examples 2

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

Recall that that if $V$ and $W$ are vector spaces then the set of all linear maps from $V$ to $W$, $\mathcal L(V, W)$ with addition defined by $(S + T)(v) = S(v) + T(v)$ and scalar multiplication defined by $(aT)(v) = aT(v)$ for all $S, T \in \mathcal L (V, W)$ and for all $a \in \mathbb{F}$, is a vector space. Suppose that $2 ≤ \mathrm{dim} (V) ≤ \mathrm{dim} (W)$. Prove that then the subset $\{ T \in \mathcal L (V, W) : T \: \mathrm{is \: not \: injective} \}$ is NOT a subspace of $\mathcal L (V, W)$.

To show that $U = \{ T \in \mathcal L (V, W) : T \: \mathrm{is \: not \: injective} \}$ is not a subspace of $\mathcal L (V, W)$ - we must either show that $0 \not \in \mathcal L (V, W)$ or show that $\{ T \in \mathcal L (V, W) : T \: \mathrm{is \: not \: injective} \}$ is not closed under addition or scalar multiplication.

Note that the zero linear map $0 : V \to W$ defined by $0(v) = 0$ which maps every vector $v \in V$ to $0 \in W$ is not injective because $0(v) = 0(w)$ does not imply that $v = w$. Therefore the zero map is not injective and:

(1)

\begin{align} \quad 0 \in \{ T \in \mathcal L (V, W) : T \: \mathrm{is \: not \: injective} \} \end{align}

So we have thus far failed to prove that $\{ T \in \mathcal L (V, W) : T \: \mathrm{is \: not \: injective} \}$ is not a subspace of $\mathcal L(V, W)$.

We will now check to see if $\{ T \in \mathcal L (V, W) : T \: \mathrm{is \: not \: injective} \}$ is closed under addition. Let $S, T \in \{ T \in \mathcal L (V, W) : T \: \mathrm{is \: not \: injective} \}$.

Let $\{ v_1, v_2, ..., v_n \}$ be a basis of $V$ and let $\{ w_1, w_2, ..., w_m \}$ be a basis of $W$. We note that $2 ≤ n ≤ m$. Define the linear map $S$ by:

(2)

\begin{align} \quad S(v_1) = 0 \\ \quad S(v_2) = w_2 \\ \quad \quad \vdots \quad \quad \\ \quad S(v_n) = w_n \end{align}

(3)

\begin{align} \quad T(v_1) = w_1 \\ \quad T(v_2) = 0 \\ \quad T(v_3) = w_3 \\ \quad \quad \vdots \quad \quad \\ \quad T(v_n) = wn \end{align}

Note that $S$ is not injective. To show this, suppose that $u, v \in V$. Then $u = a_1v_1 + a_2v_2 + ... + a_nv_n$ and $v = b_1v_2 + b_2v_2 + ... + b_nv_n$ for some set of scalars $a_1, a_2, ..., a_n, b_1, b_2, ..., b_n \in \mathbb{F}$, and so:

(4)

\begin{align} \quad S(u) = S(v) \\ \quad S(a_1v_1 + a_2v_2 + ... + a_nv_n) = S(b_1v_1 + b_2v_2 + ... + b_nv_n) \\ \quad a_1S(v_1) + a_2S(v_2) + ... + a_nS(v_n) = b_1S(v_1) + b_2S(v_2) + ... + b_nS(v_n) \\ \quad a_1(0) + a_2w_2 + ... + a_nw_n = b_1(0) + b_2w_2 + ... + b_nw_n \\ \quad a_2w_2 + ... + a_nw_n = b_2w_2 + ... + b_nw_n \end{align}

Note that the equation above does not imply that $u = v$. For example, if $u = a_1v_1 + a_2v_2 + ... + a_nv_n$ and $v = b_1v_1 + a_2v_2 + ... + a_nv_n$ where $a_1 \neq b_1$ then $S(u) = S(v)$ but clearly $u \neq v$.

We also note that $T$ is not injective. To show this, suppose that $u, v \in V$. Then once again, $u = a_1v_1 + a_2v_2 + ... + a_nv_n$ and $v = b_1v_1 + b_2v_2 + ... + b_nv_n$ for some set of scalars $a_1, a_2, ..., a_n, b_1, b_2, ..., b_n \in \mathbb{F}$, and so:

(5)

\begin{align} \quad T(u) = T(v) \\ \quad T(a_1v_1 + a_2v_2 + ... + a_nv_n) = T(b_1v_1 + b_2v_2 + ... + b_nv_n) \\ \quad a_1T(v_1) + a_2T(v_2) + ... + a_nT(v_n) = b_1T(v_1) + b_2T(v_2) + ... + b_nT(v_n) \\ \quad a_1w_1 + a_2(0) + a_3w_3 + ... + a_nw_n = b_1w_1 + b_2(0) + b_3w_3 + ... + b_nw_n \\ \quad a_1w_1 + a_3w_3 + ... + a_nw_n = b_1w_1 + b_3w_3 + ... + b_nw_n \end{align}

Note that the equation above does not imply that $u = v$. For example, if $u = a_1v_1 + a_2v_2 + ... + a_nv_n$ and $v = a_1v_1 + b_2v_2 + a_3v_3 + ... + a_nv_n$ where $a_2 \neq b_2$ then $S(u) = S(v)$ but clearly $u \neq v$.

Now consider the linear map $(S + T)(v) = S(v) + T(v)$ for all $v \in V$. For $u$ and $v$ defined as the linear combinations of the basis vectors previously, we have that:

(6)

\begin{align} \quad (S + T)(u) = (S + T)(v) \\ \quad S(u) + T(u) = S(v) + T(v) \\ \quad (a_2w_2 + ... + a_nw_n ) + (a_1w_1 + a_3w_3 + ... + a_nw_n) = (b_2w_2 + ... + b_nw_n) + ( b_1w_1 + b_3w_3 + ... + b_nw_n) \\ \quad a_1w_1 + a_2w_2 + 2a_3w_3 + ... + 2a_nw_n = b_1w_1 + b_2w_2 + 2b_3w_3 + ... + 2b_nw_n \end{align}

if we subtract both sides of the equations above, we have that:

(7)

\begin{align} \quad (a_1 - b_1)w_1 + (a_2 - b_2)w_2 + (2a_3 - 2b_3)w_3 + ... + (2a_n - 2b_n)w_n = 0 \end{align}

Since $\{ w_1, w_2, ..., w_m \}$ is a basis - it is also a linearly independent set. Note that $n ≤ m$ (from earlier), and so the set of vectors $\{ w_1, w_2, ..., w_n \}$ as a subset of the prescribed basis of $W$ is also linearly independent. This implies that:

(8)

\begin{align} \quad a_1 = b_1 \\ \quad a_2 = b_2 \\ \quad 2a_3 = 2b_2 \\ \quad \quad \vdots \quad \quad \\ \quad 2a_n = 2b_n \end{align}

Therefore $u = v$, so $S + T$ is injective. Therefore, $(S + T) \not \in \{ T \in \mathcal L (V, W) : T \: \mathrm{is \: not \: injective} \}$ so $\{ T \in \mathcal L (V, W) : T \: \mathrm{is \: not \: injective} \}$ is not closed under addition and hence is not a subspace of $\mathcal L (V, W)$.