I understand that if $v_1,..,v_r$ are the eigenvectors that correspond to distinct eigenvalues then they are linearly independent (*)

However what if I have say two linearly independent eigenvectors corresponding to one eigenvalue and an eigenvector corresponding to another, with $A$ a $3$x$3$ matrix and $Av=\lambda v$. Are these three eigenvectors linearly independent? Does this follow from (*)?


1 Answer 1


The answer is yes. Let's assumme that $v_1, v_2$ are the eigenvectors that correspond to the same eigenvalue $\alpha_1$. Observe that $\lambda_1 v_1 + \lambda_2 v_2$ is an eigenvector for the eigenvalues $\alpha_1$ and is therefore linearly independent of our third vector $v_3$. This means that if $\lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 = 0$ we necessarily have $\lambda_3 = 0$. Now this implies $\lambda_1 v_1 + \lambda_2 v_2 = 0$, which by assumption yields $\lambda_1 = \lambda_2 = 0$.

Note that this is nothing else than observing that the sum of eigenspaces to different eigenvalues is a direct sum.

  • $\begingroup$ Why is $\lambda_1 v_1 + \lambda_2 v_2$ linearly independent of $v_3$? $\endgroup$ Aug 23, 2015 at 22:07
  • 1
    $\begingroup$ $v_3$ is meant to be an eigenvector to an eigenvalue $\alpha_2 \ne \alpha_1$, so this follows from the lemma (*) in your post. $\endgroup$
    – Dominik
    Aug 23, 2015 at 22:09
  • $\begingroup$ Where you wrote $\lambda_2 v_2 + \lambda_2 v_2 + \lambda_3 v_3 = 0$ I assume you meant something else? $\endgroup$ Aug 23, 2015 at 22:27
  • $\begingroup$ Indeed, there were some mistakes in the indices. I've corrrected it now. $\endgroup$
    – Dominik
    Aug 23, 2015 at 22:29
  • $\begingroup$ I don't understand why setting $\lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 = 0$ implies that $\lambda_3=0$ $\endgroup$ Aug 23, 2015 at 23:00

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