Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Have you read this? en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem –  wj32 Oct 22 '12 at 0:26
    
Show that using a basis of the nullspace of $T$ and a basis of the image of $T$ you can write down a basis of $V$. –  Qiaochu Yuan Oct 22 '12 at 0:26
    
@QiaochuYuan, I'm sorry, but I really don't know where to begin here. Could you be more explicit? –  Chris Oct 22 '12 at 0:42
    
@QiaochuYuan Do you mean that by forming each and adjoining them, we can reduce them to a linearly independent set, spanning each space? –  Chris Oct 22 '12 at 0:50

2 Answers 2

up vote 2 down vote accepted

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).

share|improve this answer
    
@GerryNyerson I don't really follow you. Why does this squeezing down leave equal dimensions (isn't the nullspace a subset of the range? Why would the overlap of their dimensions not exceed the dimension of the domain?) –  Chris Oct 22 '12 at 1:13
    
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.... –  Chris Oct 22 '12 at 1:28

Since $nullT \leq V$, you can pick a basis for $nullT$ and extend it to a basis for $V$. This is a hint.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.