quick plz!!! subspace and span plz help!!!?

Say dim[null(A)0 = k. So there is a basis for this subspace {e1,....ek}. I claim the set B =
{V^-1e1, ...V^-1ek} is a basis for null(AV). If I can show this, then dim(null(AV)) = k also.

First, each vector in B belongs to null(AV): AV(V^-1ej) = Aej = 0.
Next the set B consists of linearly independent vectors. Suppose some linear combination of
these did equal zero. We would have:
a1(V^-1e1) _ ...+ak(V^-1ek) = 0. The scalars move past the matrix V^-1 and it is a linear transformation, hence V^-1(a1e1 + ... + akek) = 0. But V^-1 is invertible, so this implies
a1e1 + ... + akek = 0. Because {e1,...ek} is a basis the only way a linear combination of its vectors can equal zero is with zero coefficients, so a1 = 0, a2=0, ...ak=0. This proves the linear independence of the vectors in B. It remains to show that B is a spanning set for null(AV). let x be any vector in null(AV). So AVx = 0. Then A(Vx) = 0 and Vx belongs to null(A). Since {e1...ek} is a basis for null(A), there is a linear combination such that Vx = b1e1 + ... + bkek. Apply V^-1 on both sides to obtain:
x = a1(V^-1e1) +...+ ak(V^-1ek), and x is a linear combination of the V^-1ej vectors. This completes the proof.