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?
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1Wikipedia has even two simple proofs. Here on this site there are more proofs, e.g., see here. – Dietrich Burde Dec 03 '15 at 20:43
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I looked for it here but I must have missed it. I didn't check wiki though, thanks for the tip! – Cayde Dec 03 '15 at 21:01
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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, $$\dim\ker T+\mathrm{rank}\,T=p+(n-(p+1)+1)=n=\dim V.$$
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1I'd prefer to lift a basis of $T(V)$, though, and augment it with a basis of $\ker T$ and verify that together these are a linearly independent generating family. Also it should be noted that finite-dimensionality is not really needed. – Hagen von Eitzen Dec 03 '15 at 20:57
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But $T(V)\subset W$ whereas $\ker T\subset V$ ? And finite-dimensionality is needed, except if you assume the Zorn lemma (which is another story...). – Balloon Dec 03 '15 at 20:58
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May I see your method, Hagen? I'm looking for different ways to do it (didn't like the books method). – Cayde Dec 03 '15 at 21:01
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@Baloown By "lift a basis of $T(V)$," I think he means choose a basis $w_1, \ldots, w_k$ for $T(V)$, and then choose vectors $v_1, \ldots, v_k \in V$ such that $T(v_i) = w_i$. "Pulling back a basis for $T(V)$ by $T$" is another way to say it. – Viktor Vaughn Dec 04 '15 at 00:33
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