이 포스트에서 $V, W$는 모두 $F$-벡터공간으로 취급한다.
The nullity and rank
Definition 2. Let $T \in \mathcal{L}(V, W)$. If $N(T)$ and $R(T)$ are finite-dimensional, then we define
(1) the nullity of $T$, denoted nullity($T$) := dim($N(T)$),
(2) the rank of $T$, denoted rank($T$) := dim($R(T)$).
$N(I_V) = \{\mathbf{0}\}, R(I_V) = V$, 그리고 $N(T_0) = V, R(T_0) = \{\mathbf{0}\}$ 임을 생각해 볼 때, 직관적으로 nullity가 클수록 rank는 작아지고, 역으로 nullity가 작을수록 rank는 커진다는 것을 눈치챌 수 있다. 이러한 nullity와 rank의 관계는 다음의 정리에서 깔끔하게 정리된다.
Theorem 1 (Dimension Theorem)
Theorem 1. (Dimension Theorem) Let $T \in \mathcal{L}(V, W)$.
If $V$ is fintie-dimensional, then nullity($T$) + rank($T$) = dim($V$).
Proof. Let $\beta = \{v_1, ..., v_n\}$ and $\alpha = \{v_1, ..., v_k\}$ be bases for $V$ and $N(T)$, respectively. By Theorem 2, $R(T) = $ <$\{T(v_1), ..., T(v_n)\}$>. We claim that $S = \{T(v_{k+1}), ..., T(v_n)\}$ is a basis for $R(T)$.
Note that $T(v_i) = \mathbf{0} (i = 1, ..., k)$. Thus $R(T) = $ <$\{T(v_1), ..., T(v_n)\}$> = <$\{T(v_{k+1}), ..., T(v_n)\}$> = $S$. So $S$ generates $R(T)$.
Suppose that $\sum_{i=k+1}^n a_iT(v_i) = \mathbf{0}$. Then we have $\sum_{i=k+1}^n T(a_iv_i) = \mathbf{0} \Longrightarrow \sum_{i=k+1}^n a_iv_i \in N(T) \Longrightarrow \sum_{i=k+1}^n a_iv_i = \sum_{j=1}^k b_jv_j \\ \Longrightarrow \sum_{j=1}^k b_jv_j + \sum_{i=k+1}^n (-a_i)v_i = \mathbf{0}.$ Since $\beta$ is linearly independent, $a_i = 0 (i = k+1, ..., n)$. Thus $S$ is a basis for $R(T)$.
Hence we have nullity($T$) + rank($T$) = $k + (n - k) = n$ = dim($V$). $\blacksquare$
Theorem 2
Theorem 2. Let $T \in \mathcal{L}(V, W)$. Then $T$ is injective $\Longleftrightarrow$ $N(T) = \{\mathbf{0}\}.$
Proof. Let $x, y \in N(T)$. Then $T(x) = T(y) = \mathbf{0} = T(\mathbf{0}) \Longrightarrow x = y = \mathbf{0}.$
Suppose that $T(x) = T(y)$ for $x, y \in V$. Then $T(x) - T(y) = T(x - y) = \mathbf{0} \Longrightarrow x - y = \mathbf{0} \Longrightarrow x = y$. Thus $T$ is injective. $\blacksquare$
Theorem 3
Theorem 3. Let $V$ and $W$ have equal (finite) dimension, and let $T \in \mathcal{L}(V, W)$. Then the following are equivalent:
(a) $T$ is injective.
(b) $T$ is surjective.
(c) rank($T$) = dim($V$).
Proof. Assume that $T$ is injective $\Longleftrightarrow N(T) = \{\mathbf{0}\} \Longleftrightarrow $ nullity($T$) = 0 $\Longleftrightarrow$ rank($T$) = dim($V$) = dim($W$) $\Longleftrightarrow$ $R(T) = W \Longleftrightarrow$ $T$ is surjective. $\blacksquare$
주의할 것은 두 vector space가 동일한 finite dimension을 가지지 않는다면 위 명제들은 일반적으로 동치가 아니다.
Theorem 4
Theorem 4. Suppose that $\{v_1, ..., v_n\}$ is a basis for $V$. For $w_i (i = 1, ..., n) \in W$, $! \exists \, T \in \mathcal{L}(V, W)$ such that $T(v_i) = w_i (i = 1, ..., n)$.
Proof. Define $T : V \rightarrow W$ by $T(x) = \sum_{i=1}^{n} a_iw_i \forall x \in V$ where $x = \sum_{i=1}^{n} a_iv_i$.
Let $u, v \in V$ and $c \in F$. Then $u = \sum_{i=1}^{n} b_iv_i$ and $v = \sum_{i=1}^{n} c_iv_i.$ Hence $$T(cu + v) = T(\sum_{i=1}^{n} (cb_i + c_i)v_i) = \sum_{i=1}^{n} (cb_i + c_i)w_i \\ = c \sum_{i=1}^{n} b_iw_i + \sum_{i=1}^{n} c_iw_i = cT(u) + T(v).$$ So $T$ is linear, and clearly $T(v_i) = w_i (i = 1, ..., n).$
Let $U \in \mathcal{L}(V, W)$ such that $U(v_i) = w_i (i = 1, ..., n)$. $\forall x \in V, U(x) = U(\sum_{i=1}^{n} a_iv_i) = \sum_{i=1}^{n} a_iU(v_i) = \sum_{i=1}^{n} a_iw_i = T(x).$ Hence $U = T$. $\blacksquare$
이 정리로부터 finite-dimensional에 한해서 두 개의 linear transformation을 비교할 때 basis만 체크하면 된다는 결론을 얻는다.
Corollary
Corollary. Suppose that $\{v_1, ..., v_n\}$ is a basis for $V$. If $U, T \in \mathcal{L}(V, W)$ such that $U(v_i) = T(v_i) (i = 1, ..., n)$, then $U = T$.
