이 포스트에서 $V, W$는 모두 $F$-벡터공간으로 취급한다.
The null space and range
Definition 1. Let $T \in \mathcal{L}(V, W)$.
(a) The null space (or kernel) $N(T)$ of $T$ is the set $N(T) = \{ x \in V \,|\, T(x) = \mathbf{0} \}.$
(b) The range (or image) $R(T)$ of $T$ is the set $R(T) = \{ T(x) \in W \,|\, x \in V \}$.
Theorem 1
Theorem 1. Let $T \in \mathcal{L}(V, W)$. Then $N(T) \leq V$ and $R(T) \leq W$.
Proof. Clearly, $N(T), R(T) \neq \emptyset$. Let $x, y \in N(T)$ and $z, w \in R(T)$. Then
(1) $T(ax + y) = aT(x) + T(y) = \mathbf{0} \Longrightarrow ax + y \in N(T)$,
(2) $\exists z_0, w_0 \in V$ such that $T(z_0) = z$ and $T(w_0) = w$
$\Longrightarrow$ $cz + w = cT(z_0) + T(w_0) = T(cz_0 + w_0) \Longrightarrow cz + w \in R(T)$.
Thus $N(T) \leq V$ and $R(T) \leq W$. $\blacksquare$
Theorem 2
Theorem 2. Let $T \in \mathcal{L}(V, W)$. If $\beta$ is a basis for $V$, then $R(T) = \langle T(\beta) \rangle$.
Proof. Since $T(\beta) \subseteq R(T)$, <$T(\beta)$> $\subseteq R(T)$.
Let $v \in R(T)$. Then $\exists x \in V$ such that $T(x) = v$. Since $x \in V$, $x = \sum a_iu_i$ for $a_i \in F, u_i \in \beta$. Then $v = T(x) = T(\sum a_iu_i) =\sum a_iT(u_i).$ Hence $v \in \langle T(\beta) \rangle$. Thus $R(T) \subseteq$ $\langle T(\beta) \rangle$ so $R(T) = \langle T(\beta) \rangle . \blacksquare$