# Simplest Proof of the Dimensions theorem

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?

• Wikipedia has even two simple proofs. Here on this site there are more proofs, e.g., see here. Dec 3, 2015 at 20:43
• I looked for it here but I must have missed it. I didn't check wiki though, thanks for the tip! Dec 3, 2015 at 21:01

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, $$\dim\ker T+\mathrm{rank}\,T=p+(n-(p+1)+1)=n=\dim V.$$
• I'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. Dec 3, 2015 at 20:57
• 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...). Dec 3, 2015 at 20:58
• @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. Dec 4, 2015 at 0:33