[선형대수학] 차원 정리 (Dimension Theorem)

$\underline{Thm}$ (Dimension Theorem)

Let $T \colon V \rightarrow W$ be linear. If $dim(V) < \infty$, then $dim(V)=nullity(T)+rank(T)$.

 

$\underline{Proof}$

Since $N(T)$ is a subspace of $V$, $N(T)$ is finite-dimensional & $dim(N(T))=nullity(T)=k \leq n$.

If $k=n$, then $N(T)=V$, and so $R(T)=\{0\}$; thus $dim(V)=n=n+0=nullity(T)+rank(T)$.

Suppose $k<n$. Let $\beta{'}=\{ v_{1},\cdots,c_{n} \}$ be a basis for $N(T)$.

Since $\beta^{'}$ is linear independent, we can chose $v_{k+1},\cdots,c_{n} \in V \backslash N(T)$ so that $\beta \{ v_{1}, \cdots,v_{n} \}$ is a basis for $V$.

Note $R(T)=span(\{T(v_{1}),\cdots,T(v_{n}) \})=span(\{T(v_{k+1}),\cdots,T(v_{n})\})=span(S)$.

It remains to show that $S$ is linear independent.

Suppose $\displaystyle b_{k+1},\cdots,b_{n} \in \mathbb{F}\,\&\sum_{i=k+1}^{n} b_{i}v_{i}=0$.

By the linearity of $T$, we have $\displaystyle T(\sum_{i=k+1}^{n} b_{i}v_{i})=0$, and so $\displaystyle \sum_{i=k+1}^{n} b_{i}v_{i} \in N(T)$.

So, $\exists \! c_{1},\cdots,c_{k} \in \mathbb{F}$ such that $\displaystyle \sum_{i=k+1}^{n} b_{i}v_{i}=\sum_{i=1}^{k} c_{i}v_{i}$.

That is, $(-c_{1})v_{1}+\cdots+(-c_{k})v_{k}+b_{k+1}v_{k+1}+\cdots+b_{n}v_{n}=0$.

Thus, $S$ is linear independent. We now concolude that $rank(T)=n-k$; that is $dim(V)=n=nullity(T)+rank(T)$.