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Can I have a concise proof of the following:

$$\operatorname{dim} V =\operatorname{dim\,ker}T+\operatorname{dim \, range}T$$

I have read a few proofs of this, and they are all so long, I always forget how it is done within a few days. I could 'see' why it was true prior to seeing the proofs, so it isn't a lack of intuition.

Understand
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    I don't care much, but can the duplicate ribbon be removed, obviously asking for(and no offence intended) an 'alternative' proof implies it cannot be a duplicate, but further than that, my intuition is perfectly fine. – Understand Feb 25 '15 at 15:01

2 Answers2

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Divide the Basis $\lbrace b_1, b_2, \ldots \rbrace$ of $V$ in 2 sets.

1) All basis Elements for which we have $T(b)=0$ 2) All basis Elements for which $T(b) \neq 0$.

These sets have empty intersection and their union is the whole basis of $V$.

But the Elements of 1) span the Kernel of T and the Elements of 2) span the range of $T$.

Near
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Take a basis $B'$ of $\ker T$ and complete it to a basis $B=B'\cup B_0$ of $V$. Define $\tilde{B}=\left\{Tv\mid v\in B_0\right\}$, and check that $\tilde{B}$ is an independent spanning set, hence a basis.

Guy
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