Let A be square matrix such that $$|A-xI|=x^4×(x-1)^2×(x-2)^3$$ If $rank(A^3)=rank(A^4)<rank(A^2)$ then geometric multiplicity of Eigen value 0 is?
I found that rank of A = 9 - dim eigenspace(0).
dim eigenspace(0) given by dim of kernel of $A−0I$, or the dimension of ker(A).
Since the rank of $A^3 = A^4$ I deduced that there could be a jordan block of size 3 in the matrix. But I could not find a reference to this point anywhere in the internet. Is this a valid inference?
This, if true would make geom multiplicity of zero = 2 and Rank of A = 7
Am I correct on this point.