Hi zook,

I'll try to get your going in the right direction.

i) Remember that for to be an eigenvector of we must have two things: (1) and there is (in whatever field you're working with, presumably the real numbers) such that This means that if we want to show is an eigenvector of , we must establish (1) and (2).

To prove (1) we note that because it is an eigenvector of and so cannot be zero (by definition of eigenvector). Now also note that is assumed nonsingular/nondegenerate and so because

To prove (2) we us . This implies Since is an eigenvector of with corresponding eigenvalue we have Pull the constant through to get Now divide by

Does this help? Let me know if anything is unclear. Good luck!