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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}

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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)!

But...

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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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