4.15 The rank-nullity theorem

We’ll assume $V$ and $W$ are finite-dimensional, not that it matters. Here is an outline of how the proof is going to work.

1. 1.

   Choose a basis $𝒦={𝐤}_1,\mathrm{\dots },{𝐤}_m$ of $\ker T$

2. 2.

   Extend it to a basis $ℬ={𝐤}_1,\mathrm{\dots },{𝐤}_m,{𝐯}_1,\mathrm{\dots },{𝐯}_n$ of $V$ using Proposition 4.11.2. When we’ve done this, $dimV=m+n$ and we need only show $dim\mathrm{im}T=n$)

3. 3.

   Show that $T({𝐯}_1),\mathrm{\dots },T({𝐯}_n)$ is a basis of $\mathrm{im}T$.

The only part needing elaboration is the last part. First, let’s show that the sequence $T({𝐯}_1),\mathrm{\dots },T({𝐯}_n)$ spans $\mathrm{im}T$. Any element of the image is equal to $T(𝐯)$ for some $𝐯\in V$. We have to show that any such $T(𝐯)$ lies in the span of the $T({𝐯}_i)$s.

Since $ℬ$ is a basis of $V$ we may write $𝐯$ as ${\sum }_{i=1}^m{\lambda }_i{𝐤}_i+{\sum }_{i=1}^n{\mu }_i{𝐯}_i$ for some scalars ${\lambda }_i,{\mu }_i$. Then

	$T(𝐯)$	$=T\left({\displaystyle \sum _{i=1}^m}{\lambda }_i{𝐤}_i+{\displaystyle \sum _{i=1}^n}{\mu }_i{𝐯}_i\right)$
		$={\displaystyle \sum _{i=1}^m}{\lambda }_{i}T({𝐤}_i)+{\displaystyle \sum _{i=1}^n}{\mu }_{i}T({𝐯}_i)$	by linearity
		$={\displaystyle \sum _{i=1}^n}{\mu }_{i}T({𝐯}_i)$	$\text{as }{𝐤}_i\in \ker T$
		$\in \mathrm{span}(T({𝐯}_1),\mathrm{\dots },T({𝐯}_n))$

as required.

Now let’s show $T({𝐯}_1),\mathrm{\dots },T({𝐯}_n)$ is linearly independent. Suppose

$$\sum _{i=1}^n{\mu }_{i}T({𝐯}_i)=0,$$

so that we need to show the ${\mu }_i$ are all 0. Using linearity,

$$T\left(\sum _{i=1}^n{\mu }_i{𝐯}_i\right)=0$$

which means ${\sum }_{i=1}^n{\mu }_i{𝐯}_i\in \ker T$. As $𝒦$ is a basis for $\ker T$, we can write

$$\sum _{i=1}^n{\mu }_i{𝐯}_i=\sum _{i=1}^m{\lambda }_i{𝐤}_i$$

for some scalars ${\lambda }_i$. But $ℬ$, being a basis, is linearly independent and so all the scalars are 0. In particular all the ${\mu }_i$ are 0, which completes the proof. ∎