# proof of rank-nullity theorem

Let $T:V\rightarrow W$ be a linear mapping, with $V$ finite-dimensional. We wish to show that

 $\dim V=\dim\mathop{\mathrm{Ker}}T+\dim\mathop{\mathrm{Img}}T$

The images of a basis of $V$ will span $\mathop{\mathrm{Img}}T$, and hence $\mathop{\mathrm{Img}}T$ is finite-dimensional. Choose then a basis $w_{1},\ldots,w_{n}$ of $\mathop{\mathrm{Img}}T$ and choose preimages $v_{1},\ldots,v_{n}\in U$ such that

 $w_{i}=T(v_{i}),\quad i=1\ldots n$

Choose a basis $u_{1},\ldots,u_{k}$ of $\mathop{\mathrm{Ker}}T$. The result will follow once we show that $u_{1},\ldots,u_{k},v_{1},\ldots,v_{n}$ is a basis of $V$.

Let $v\in V$ be given. Since $T(v)\in\mathop{\mathrm{Img}}T$, by definition, we can choose scalars $b_{1},\ldots,b_{n}$ such that

 $T(v)=b_{1}w_{1}+\ldots b_{n}w_{n}.$

Linearity of $T$ now implies that $T(b_{1}v_{1}+\ldots+b_{n}v_{n}-v)=0,$ and hence we can choose scalars $a_{1},\ldots,a_{k}$ such that

 $b_{1}v_{1}+\ldots+b_{n}v_{n}-v=a_{1}u_{1}+\ldots a_{k}u_{k}.$

Therefore $u_{1},\ldots,u_{k},v_{1},\ldots,v_{n}$ span $V$.

Next, let $a_{1},\ldots,a_{k},b_{1},\ldots,b_{n}$ be scalars such that

 $a_{1}u_{1}+\ldots+a_{k}u_{k}+b_{1}v_{1}+\ldots+b_{n}v_{n}=0.$

By applying $T$ to both sides of this equation it follows that

 $b_{1}w_{1}+\ldots+b_{n}w_{n}=0,$

and since $w_{1},\ldots,w_{n}$ are linearly independent that

 $b_{1}=b_{2}=\ldots=b_{n}=0.$

Consequently

 $a_{1}u_{1}+\ldots+a_{k}u_{k}=0$

as well, and since $u_{1},\ldots,u_{k}$ are also assumed to be linearly independent we conclude that

 $a_{1}=a_{2}=\ldots=a_{k}=0$

also. Therefore $u_{1},\ldots,u_{k},v_{1},\ldots,v_{n}$ are linearly independent, and are therefore a basis. Q.E.D.

Title proof of rank-nullity theorem ProofOfRanknullityTheorem 2013-03-22 12:25:13 2013-03-22 12:25:13 rmilson (146) rmilson (146) 4 rmilson (146) Proof msc 15A03