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One of the results in Axler (Linear Algebra Done Right) is that

$\mathrm{dim}V = \text{dim null }T + \text{dim range }T$

for a linear map $T: V \rightarrow W$. (He means by range what I usually mean by image, i.e. the set of $w \in W: w = T(v)$ for $v \in V$.)

I really don't know why this should be true, and the proof doesn't enlighten me. I looked at his examples of linear maps and generated some new ones, but this still isn't clear. What's going on here?

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    @QiaochuYuan Do you mean that by forming each and adjoining them, we can reduce them to a linearly independent set, spanning each space?2012-10-22

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I usually think of a linear transformation as squeezing its domain down to its range (or image, if you prefer) while squeezing its nullspace down to zero. The theorem tells you the two amounts of squeezing are the same; the squeezing of the domain is the dimension of the domain minus the dimension of the range, and the squeezing of the nullspace is the dimension of the nullspace minus the dimension of the zero-space (which is zero).

EDIT: Maybe a numerical example will make my visualization clearer. Suppose the domain has dimension 17, and the range has dimension 12. Where did those extra 5 dimensions in the domain go? They must have gone to zero. So the dimension of the nullspace --- that's the 5 --- plus the dimension of the range --- that's the 12 --- equals the dimension of the domain --- the 17. The 17 gets squeezed down to 12 by virtue of the 5 getting squeezed down to zero.

This doesn't remotely resemble a proof, it's for motivation purposes only (but I thought that's what was being requested).

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    Okay, I wrote the above too quickly; what I meant is, what is the special reason for which we can cordon off the nullspace, and then consider the dimension of the range independently? (I still don't understand the "squeezing" idea whatsoever.) Basically, when I turned the page and saw this, I felt like this result came out of nowhere. I still do....2012-10-22
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Since $nullT \leq V$, you can pick a basis for $nullT$ and extend it to a basis for $V$. This is a hint.