Proofs from Rank-nullity

Let $k=\dim\ker(T)$ and choose a basis

for $\ker(T)$. Extend it to a basis of $V$:

The vectors $T(v_1),\ldots,T(v_r)$ span $\operatorname{im}(T)$. Indeed, if $x\in V$, write

Applying $T$ gives

because $T(u_i)=0$ for each $i$.

The vectors $T(v_1),\ldots,T(v_r)$ are linearly independent. If

then

so $c_1v_1+\cdots+c_rv_r\in\ker(T)$. It is therefore a linear combination of $u_1,\ldots,u_k$. Since the full list

is linearly independent, $c_1=\cdots=c_r=0$. Thus $T(v_1),\ldots,T(v_r)$ is a basis of the image, so $\operatorname{rank}(T)=r$. The basis of $V$ has $k+r$ vectors, giving