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I was thinking about the proof of the rank-nullity theorem and I thought about proving it as follows. I just wondered whether this proof worked?

Lemma. If $V$ is a finite-dimensional $F$-vector space and $U\leq V$, then $V/U$ is finite dimensional.

$\hspace{16.5mm}$ Moreover, we have that $\dim{V/U}=\dim{V}-\dim{U}$.

Theorem (Rank-Nullity). If $\alpha:V\to W$ is linear with $V$ finite-dimensional, then $$\dim{V}=\dim(\text{im }\alpha)+\dim(\ker \alpha)$$

Proof. By the first isomorphism theorem we have $$V/\ker{\alpha} \cong \text{im }{\alpha}.$$ Taking dimensions and applying the lemma we get $$\dim V - \dim(\ker\alpha)=\dim(\text{im }\alpha)$$ which on rearrangement yields the result. //

user168365
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    Not only this works, it is more natural from an algebraic point of view. Moreover, it gives a version of the theorem that applies to infinite-dimensional vector spaces as well (i.e. the first isomorphism theorem). I am a fan of this approach. – Giuseppe Negro Aug 06 '14 at 20:00
  • Looks perfectly fine to me. – Hayden Aug 06 '14 at 20:00
  • Related to my previous comment: http://math.stackexchange.com/q/752056/8157 – Giuseppe Negro Aug 06 '14 at 20:04
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    I notice also that the identity $$\dim(U+V)=\dim(U)+\dim(V)-\dim(U\cap V)$$ follows from the Second Isomorphism Theorem :)

    But that the third isomorphism theorem tells us nothing in terms of dimension!

     – user168365 Aug 07 '14 at 08:48

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