[선형대수학] 영공간, 치역, 널리티, 랭크 (Null space, Range, Nullity, Rank)

$\underline{Def}$

Let $V,W$ be vector spaces, and let $T \colon V \rightarrow W$ be linear.

We define the null space $N(T)$, the range $R(T)$ of $T$.

$$ N(T)=\{ x \in V \colon Tx=0 \} $$

$$ R(T)=\{ Tx \in W \colon x \in V \} $$

Null space and range are called kernel and image respectively.

 

$\underline{Thm}$

Let $V, W$ & $T$ be an in the previous definition.

Then $N(T), R(T)$ are subspaces of $V, W$, respectively.

 

 

$\underline{Def}$

Let $V, W$ & $T$ be as in the above theorem. Suppose $N(T), R(T)$ are finite-dimensional.

Then we define the nullity, the rank of $T$.

$$ nullity(T)=dim(N(T)) \in \mathbb{N}_{0} $$

$$ rank(T)=dim(R(T)) \in \mathbb{N}_{0} $$

 

 

$\underline{Ex}$

Let $V,W$ be vector spaces. Let $I \colon V \rightarrow V$ be the identity map & $T_{0} \colon V \rightarrow W$ the zero map.

Then $N(I)= \{ 0 \},\, R(I)=V,\, N(T_{0})=V,\, R(T_{0})=\{ 0 \}$.

 

 

$\underline{Thm}$

Let $V,W$ be vector spaces, and let $T \colon V \rightarrow W$ be a linear map. If $\beta = \{ v_{1}, \cdots, v_{n} \}$ is a basis for $V$, then $R(T)=span(T(\beta))=span(\{T(v_{1}),\cdots,T(v_{n}) \})$.