Question:

If A and B are similar matrices

the geometric multiplicity of an eigenvalue $\displaystyle \lambda$ of A is the same as its geometric multiplicity as an eigenvalue of B.

Proof:

$\displaystyle {\lambda} I - B = Q({\lambda} I -A)Q^{-1}$

since Q is invertible, it follows that $\displaystyle nullity({\lambda}I - B) = nullity({\lambda}I - A)$

I dont get the bit about "since Q is invertible" I do not see how this is relevant