# Distinct Eigenvalues and Linearly Independent Eigenvectors

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 (*)?

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 $\lambda_1 = \lambda_2 = 0$.
• Why is $\lambda_1 v_1 + \lambda_2 v_2$ linearly independent of $v_3$? – usainlightning Aug 23 '15 at 22:07
• $v_3$ is meant to be an eigenvector to an eigenvalue $\alpha_2 \ne \alpha_1$, so this follows from the lemma (*) in your post. – Dominik Aug 23 '15 at 22:09
• Where you wrote $\lambda_2 v_2 + \lambda_2 v_2 + \lambda_3 v_3 = 0$ I assume you meant something else? – usainlightning Aug 23 '15 at 22:27
• I don't understand why setting $\lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 = 0$ implies that $\lambda_3=0$ – usainlightning Aug 23 '15 at 23:00