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This is regarding the equality of nullity between A and PAP(inverse).

If my understanding is correct then the thing should be according to the diagram below

V---------------->R1(isomorphic to V)

| |

| |

\/ \/

W----------------->R2(isomorphic to W)

So A : V->W

P: V->R1

P: W->R2

This second thing I am not convinced since the size of W is less than V because of the nullity in this case is assumed to be greater than 1.

If my understanding is correct what one does is that sincce R1 is isomorphic to V so

R1---->R2 is R1->V->W->R2. Which means take bases in R1 and apply P(inverse) which is correct since inverse is defined for the isomorphic transformation and then applying A u land up all the vectors from V which u had got from R1 to 0 in W. Then u apply P to all the vectors and claim that none of them go to 0 in R2 other than the 0s of W. What is confusing is that how do u define the same linear transformation in both the two vector spaces of two different dimensions in general and if possible how do u prove that only the 0s in W go to 0s in R2 and no body else.

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