proof of rank-nullity theorem

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

$$dimV=dimKerT+dimImgT$$

The images of a basis of $V$ will span $ImgT$, and hence $ImgT$ is finite-dimensional. Choose then a basis $w_1,\mathrm{\dots },w_n$ of $ImgT$ and choose preimages $v_1,\mathrm{\dots },v_n\in U$ such that

$$w_i=T(v_i),i=1\mathrm{\dots }n$$

Choose a basis $u_1,\mathrm{\dots },u_k$ of $KerT$. The result will follow once we show that $u_1,\mathrm{\dots },u_k,v_1,\mathrm{\dots },v_n$ is a basis of $V$.

Let $v\in V$ be given. Since $T(v)\in ImgT$, by definition, we can choose scalars $b_1,\mathrm{\dots },b_n$ such that

$$T(v)=b_1{w}_1+\mathrm{\dots }{b}_n{w}_n.$$

Linearity of $T$ now implies that $T(b_1{v}_1+\mathrm{\dots }+b_n{v}_n-v)=0,$ and hence we can choose scalars $a_1,\mathrm{\dots },a_k$ such that

$$b_1{v}_1+\mathrm{\dots }+b_n{v}_n-v=a_1{u}_1+\mathrm{\dots }{a}_k{u}_k.$$

Therefore $u_1,\mathrm{\dots },u_k,v_1,\mathrm{\dots },v_n$ span $V$.

Next, let $a_1,\mathrm{\dots },a_k,b_1,\mathrm{\dots },b_n$ be scalars such that

$$a_1{u}_1+\mathrm{\dots }+a_k{u}_k+b_1{v}_1+\mathrm{\dots }+b_n{v}_n=0.$$

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

$$b_1{w}_1+\mathrm{\dots }+b_n{w}_n=0,$$

and since $w_1,\mathrm{\dots },w_n$ are linearly independent that

$$b_1=b_2=\mathrm{\dots }=b_n=0.$$

Consequently

$$a_1{u}_1+\mathrm{\dots }+a_k{u}_k=0$$

as well, and since $u_1,\mathrm{\dots },u_k$ are also assumed to be linearly independent we conclude that

$$a_1=a_2=\mathrm{\dots }=a_k=0$$

also. Therefore $u_1,\mathrm{\dots },u_k,v_1,\mathrm{\dots },v_n$ are linearly independent, and are therefore a basis. Q.E.D.

Title	proof of rank-nullity theorem
Canonical name	ProofOfRanknullityTheorem
Date of creation	2013-03-22 12:25:13
Last modified on	2013-03-22 12:25:13
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	4
Author	rmilson (146)
Entry type	Proof
Classification	msc 15A03