Suppose T is a linear map and dim(Im(T))=k. Prove that T has at most k+1 distinct eigenvalues.
for every distinct eigenvalue you have at least one eigenvector. let the set A contain one eigenvector for every non zero eigenvalue. eigenvectors for distinct eigenvalues are independent. because they are eigenvectors for nonzero eigenvalues you have
span(A) = span(T(A)) - so
so there are at most dim(Im(T)) non zero eigenvalues, or dim(Im(T))+1 eigenvalues