Question

It can be shown that the algebraic multiplicity of an eigenvalue λ is always greater than or equal to the dimension of the eigenspace corresponding to λ. Find a value of h in the matrix A below such that the eigenspace for λ = -2 is two-dimensional:

Answer 1

you should understand about algebraic and geometric of an eigenvalue in eigenspace:
here is example. if characteristic equation of a matrix has the form (L -1)^5 ---> L =1
so L is only one eigenvalue of the matrix, and 5 is algebraic number of that eigenvalue,
After putting L =1 back to L I -A to count eigenvectors, you may have 1,2,3,4 or at most 5 eigenvectors from that L. No more. Those new numbers (1,2,3,4,5) is geometric number of eigenvalue in eigenspace, those eigenvectors will span eigenspace and their dimension cannot be greater than 5, right? how can you get 6 from at most 5 variables in at most 5 equations when counting dimension of eigenspace?
That's what I understand about this topic, give you mine and how to arrange it is your duty.
This is a "prove or disprove" problem, there is no form for that kind of question
Hope this helps