Suppose an n$\times$n matrix has n-1 distinct eigenvalues, $\{\lambda_1,\lambda_2,..,\lambda_{n-1}\}$. The eigenvalue $\lambda_{n-1}$ has algebraic multiplicity 2 and geometric multiplicity 1.The corresponding eigenvector set is $\{v_1,v_2,..,v_n\}$, where $v_{n-1}$ and $v_n$ are associated with eigenvalue $\lambda_{n-1}$ and $(A-\lambda_{n-1}I)v_{n-1}=0$ and $(A-\lambda_{n-1}I)^2v_{n}=0$. Show that the eigenvector set is linearly independent.

  • $\begingroup$ There's something missing, setting $v_{n-1}=v_n$ satisfies all the hypothesis. $\endgroup$ – Git Gud Nov 25 '14 at 22:27
  • $\begingroup$ Usually one might require $v_{n-1} \ne 0$ and $(A - \lambda_{n-1} I) v_n \ne 0$. $\endgroup$ – Robert Israel Nov 25 '14 at 23:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.