Is the Rank-Nullity Theorem Always True for Linear Operators?

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SUMMARY

The Rank-Nullity Theorem states that for vector spaces V and W, and a linear operator T: V -> W, V is isomorphic to the direct sum of the image and kernel of T. However, this theorem does not hold in the case of internal direct sums when T is an operator from V to itself. A counterexample provided involves T: R^2 -> R^2, where T(e1)=0 and T(e2)=e1, demonstrating that ker(T) and im(T) can be equal, thus invalidating the internal direct sum assertion. The distinction between isomorphic and equal is crucial in understanding the theorem's limitations.

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geor
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Hello all,

In wikipedia, http://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem" a generalized rank-nulity theorem as below:

"If V, W are vector spaces and T : V -> W is a linear operator then
V is isomorphic with the direct sum of im(T) and ker(T)".

I had an exercise in Algebra which would be straightforward by using the theorem
above, but we had been given a somewhat complicated hint. When I mentioned this to the prof she said that this is not true and she also gave me the counter-example below:

T : R^2 -> R^2
T(e1)=0
T(e2)=e1

(R = the real numbers, e1, e2 the usual basis).

As she said, in this example, <e1>=ker(T)=Im(T) so the above theorem "is not true"..

I'm a bit confused, could you give some light please?!
I guess that this has to do with the fact that we say "isomorphic" and not "equal" (?!),
but still, that does not mean that the theorem is not correct.

In fact, the exercise we had to do was this:

If V is a v.s. and A: V -> V is a linear operator with Im(A^p) = Im( A^(p+1) ) for some p \in Z, prove <various stuff> and also prove that V = ker( A^p ) \directsum Im( A^p ).

Well, if you take in account that A^p is also a linear operator and by using the theorem above, this is straightforward.

Her proof is a almost a page...

Then I mentioned her this theorem and she said that it is not true and she gave me the above "counterexample"..

What do I miss here?!

Thanks in advance..
 
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Yup: "isomorphic" vs "equal to" is the problem. What wiki has is definitely true, as can be seen just by comparing dimensions. The direct sum in this case is the "exterior" direct sum of vector spaces, not the "interior" direct sum of subspaces. Your prof's counterexample shows that V is, in general, not an internal direct sum of kerT and imT when T is an operator V->V. On the other hand, notice that in that example kerT=~R and imT=~R, so that R^2=~kerT x imT.
 
Just repeating what dvs said, what wiki said is true, but it's not helpful to the problem at hand because it is a statement about the external direct sum while you are asked to prove a statement about the internal direct sum.

Your professor gave a counter-example showing that even in the special case V=W, the formula V\cong Im(T)\oplus Ker(T) (external direct sum) cannot be improved to V= Im(T)\oplus Ker(T) (internal direct sum).
 
Thanks a lot for the feedback! I see it now..
 

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