Eigenvectors with distinct eigenvalues are linearly independant, but what about the case where eigenvalues are repeated.

An example is where we have matrix A = $$ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} $$

with eigenvalues $λ$ = 1 which is repeated twice so spectrum A is 1(2).

And we find that the associated augments matrix is $$ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} $$ And because neither component of the vector is contrained, the eigenvectors is of the form [u v]^T. We observe that [u v]^T = u[1 0]^T + v[0 1]^T. Where {[1 0]^T + [0 1]^T} is linearly independant. So from here how does this actually show that eigenvectors with repeated eigenvalues are not linearly dependant or otherwise?

  • $\begingroup$ From [u v]^T = u[1 0]^T + v[0 1]^T i would've said that no the eigenvectors not linearly independant because u and v are not trivial solutions, but im not sure. $\endgroup$ – Randy Rogers Aug 25 '18 at 11:27

If $v$ and $w$ are eigenvectors with distinct eigenvalues, then they are linearly independent, as you wrote. If the eigenvalues are the same, then they sometimes are linearly independent and they are sometimes linearly dependent. Take your matrix $A$, for instance. The vectors $v_1=(1,0)$, $v_2=(2,0)$ and $v_3=(0,1)$ are all eigenvectors with eigenvalue $1$. In this case $v_1$ and $v_2$ are linearly dependent, whereas $v_1$ and $v_3$ are linearly independent.

  • 1
    $\begingroup$ @Moo I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Aug 26 '18 at 6:32

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