[Math] Proof of Rank–nullity theorem

linear algebralinear-transformationsvector-spaces

I'm trying to understand the proof of Rank–nullity theorem,but there are parts that I don't understand:

Steinitz exchange lemma

If ${\displaystyle U=\{u_{1},\dots ,u_{m}\}}$ is a set of ${\displaystyle m}$ linearly independent vectors in a vector space ${\displaystyle V}$, and ${\displaystyle W=\{w_{1},\dots ,w_{n}\}}$ span ${\displaystyle V}$, then:

  1. ${\displaystyle m\leq n}$;
  1. There is a set ${\displaystyle W'\subseteq W}$ with ${\displaystyle |W'|=n-m}$ such that ${\displaystyle U\cup W'}$ spans ${\displaystyle V}$.

Rank–nullity theorem

Let ${\displaystyle V}, {\displaystyle W}$ be vector spaces, where ${\displaystyle V}$ is finite dimensional. Let ${\displaystyle T\colon V\to W}$ be a linear transformation. Then

$${\displaystyle \operatorname {Rank} (T)+\operatorname {Nullity} (T)=\dim V}$$

Proof

Let ${\displaystyle V,W}$ be vector spaces over some field ${\displaystyle \mathbb {F} }$ and ${\displaystyle T}$ defined as in the statement of the theorem with ${\displaystyle \dim V=n}$.

As ${\displaystyle \operatorname {Ker} T\subset V}$ is a subspace, there exists a basis for it. Suppose ${\displaystyle \dim \operatorname {Ker} T=k}$

I know that $\text{ker T}$ is a subset of a finite set $V$,and hence is finite,but how does this imply that $\text{ker T}$ does have a basis?

and let ${\displaystyle {\mathcal {K}}:=\{v_{1},\ldots ,v_{k}\}\subset \operatorname {Ker} (T)}$
be such a basis.

We may now, by the Steinitz exchange lemma, extend ${\displaystyle {\mathcal {K}}}$ with ${\displaystyle n-k}$ linearly independent vectors ${\displaystyle w_{1},\ldots ,w_{n-k}}$ to form a full basis of ${\displaystyle V}$.

Let

${\displaystyle {\mathcal {S}}:=\{w_{1},\ldots ,w_{n-k}\}\subset V\setminus \operatorname {Ker} (T)}$
such that

${\displaystyle {\mathcal {B}}:={\mathcal {K}}\cup {\mathcal {S}}=\{v_{1},\ldots ,v_{k},w_{1},\ldots ,w_{n-k}\}\subset V}$
is a basis for ${\displaystyle V}$.

What is $V,W,W',U$ here? Since the proof uses Steinitz exchange lemma,however I can't recognize $V,W,W',U$.

From this, we know that ${\displaystyle \operatorname {Im} T=\operatorname {Span} T({\mathcal {B}})=\operatorname {Span} \{T(v_{1}),\ldots ,T(v_{k}),T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} \{T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} T({\mathcal {S}})}$

Why $\text{Im}\; T=\text{span} \;T (\mathcal B)$?

And why $\operatorname {Span} \{T(v_{1}),\ldots ,T(v_{k}),T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} \{T(w_{1}),\ldots ,T(w_{n-k})\}$?

Thanks for your help.

Best Answer

let $v_1,v_2,...,v_n$ be a basis for $V$ then for all $v \in V$

$v=c_1v_1+...+c_nv_n$ $\Rightarrow$ $T(v)=T(C_1v_1+...+c_nv_n)=c_1T(v_1)+...+c_nT(v_n) \in \text{span} \;T (\mathcal B)$

similarly, $w \in \text{span} \;T (\mathcal B) \Rightarrow w=T(c_1v_1+...+c_nv_n) \in Im(T)$

that is why $Im(T)=\text{span} \;T (\mathcal B)$

Related Question