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

  • $\begingroup$ Is this a translation from another language? $\endgroup$ May 8, 2019 at 8:18
  • $\begingroup$ I corrected it, is it ok? $\endgroup$
    – harsh
    May 8, 2019 at 8:23

1 Answer 1


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?

  • $\begingroup$ if we prove the following does it mean the same thing: eigenvectors from different eigenvalues are linearly independent ? $\endgroup$
    – harsh
    May 8, 2019 at 8:25

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