We can see that if and only if . We deduce that we have an isomorphism between these two space.
Question:
If A and B are similar matrices
the geometric multiplicity of an eigenvalue of A is the same as its geometric multiplicity as an eigenvalue of B.
Proof:
since Q is invertible, it follows that
I dont get the bit about "since Q is invertible" I do not see how this is relevant
An invertible matrix is one-to-one. Consequently, the null spaces of A and QA, where Q is invertible, are the same. QAv is 0 if and only if Av is 0- v is in the null space of A. Similarly, AQv= 0 only if Qv is in the null space of A. Since Q is invertible, the dimension of "all v such that Qv is in the null space of A" is the same as the dimension of the null space of A.
Personally, I wouldn't prove this statement in that way. Instead (longer but more "fundamental"):
Suppose is an eigenvalue of A with geometric multiplicity n- that is, there exist n independent vectors, vectors such that for i= 1, 2, ..., n. Suppose B is similar to A: for some invertible matrix Q.
Then, for every i from 1 to n,
so that every is an eigenvector of B corresponding to eigenvalue . To show that they are independent, write . Since Q is invertible, it is one-to-one: we have . Since the vectors are independent, we have so the are n independent eigenvectors of B corresponding to eigenvalue .