# If A has a finite number of distinct eigenvectors then each eigenvector must have a distinct eigenvalue. How to prove that?

Preface: This is an exercise question, I have no idea how to prove this.

Question: In this question we will be exploring eigenvectors and eigenvalues. Let A ∈ R n×n. Recall that a vector v ∈ R n is an eigenvector of A if there exists λ ∈ R such that λv = Av, and we call λ an eigenvalue of A, associated with the vector v. If we have some eigenvector v of A note that av is also trivially an eigenvector for any a ∈ R. Because of this v and av are not considered different eigenvectors: i.e. colinear eigenvectors are not considered distinct. Prove the following: Proposition 1. If A has a finite number of distinct eigenvectors then each eigenvector must have a distinct eigenvalue

• Is this a translation from another language? May 8, 2019 at 8:18
• I corrected it, is it ok? May 8, 2019 at 8:23

Assume two vectors $$v,w$$ have the same eigenvalue $$\lambda$$ and are not colinear (as otherwise we wouldn't count them as two).
Then also $$\alpha v + \beta w$$ has eigenvalue $$\lambda$$. Can you construct an infinite amount of pairwise non-colinear vectors in this way?