• geor

#### geor

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.

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?!

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).