There is an eigenvalue $\lambda_1$ in matrix A, which is $3x3$, with corresponding eigenvector $x_a = 1/\sqrt{2}(1,0,1)$. The same eigenvalue $\lambda_1$ can be found in the subspace of matrix A, which is $2x2$, with corresponding eigenvector $x_b = 1/\sqrt{2}(1,1)$.

Question: Could a zero be added to eigenvector $x_b$ such that it equals $x_a$?

Matrix A = \begin{pmatrix} 1 & 0 & \lambda \\ 0 & 0 & 0 \\ \lambda & 0 & 1 \\ \end{pmatrix} and the subspace of Matrix A = \begin{pmatrix} 1 & \lambda \\ \lambda & 1 \ \\ \end{pmatrix}


Depends on how you define equal, obvious $x_a \ne x_b$ in the exact sense since they DEFINITELY aren't the same thing (one has 3 dimensions the other has 2)!


That doesn't mean they are not exhibiting some similar behavior. If you restate "equal" as "equal under a particular model" then the question becomes interesting again. It might be the case that there is a function $g$ out there such that $g(x_b) = g(x_a)$ where $g$ encodes some information about the matrix you are working with and the matching eigenvalues.

  • $\begingroup$ ok, thank you... do you by any chance know how this might apply in perturbation theory and the degenerate subspace? $\endgroup$ – CuriousGeorge119 Mar 19 '16 at 23:26
  • 1
    $\begingroup$ not off the top of my head head you should add those in as tags, and write the matrices too for more information $\endgroup$ – frogeyedpeas Mar 19 '16 at 23:27
  • $\begingroup$ Maybe you can take a look at the edit if you get a chance $\endgroup$ – CuriousGeorge119 Mar 20 '16 at 17:44

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.