Hello all,

In wikipedia, http://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem" [Broken] 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?!

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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).