The theorem goes:
Let $V$ and $W$ be vector spaces and $T:V \rightarrow W$ is a linear transformation.
If $V$ is finite dimensional, then $\operatorname{nullity}(T)+\operatorname{rank}(T)=\operatorname{dim}(V)$
How would you prove this?
[Math] Simplest Proof of the Dimensions theorem
abstract-algebralinear algebralinear-transformations
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Best Answer
With the incomplete base theorem : take $(v_1,...,v_p)$ a base of $\ker T\subset V.$ Now complete it in a base $(v_1,...,v_p,v_{p+1},...,v_n)$ of $V.$ You will have that $$T(V)=\underset{=\{0\}}{\underbrace{T(v_1)\oplus T(v_2)\oplus...\oplus T(v_p)}}\oplus T(v_{p+1})\oplus...\oplus T(v_n)=T(v_{p+1})\oplus...\oplus T(v_n)$$ and as $T$ is injective on $\mathrm{span}\,(v_{p+1},...,v_n)$ you get that $(T(v_{p+1}),...,T(v_n))$ is a base of $T(V).$ Finally, $$\dim\ker T+\mathrm{rank}\,T=p+(n-(p+1)+1)=n=\dim V.$$