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Say the null space corresponding to an eigenvalue is the xy plane. For some other eigenvalue, let's assume the null space is the yz plane.

Now, aj for an arbitrary scalar "a" is an eigen vector corresponding to both the eigenvalues. For proving linear independence of eigenvectors for different eigenvalues, we use the Vandermonde determinant after applying the map on a linear combination of the eigenvectors.

But say we chose j as an eigenvector for two eigenvalues, after applying the map, which eigenvalue will we choose? If we take the same eigenvalue twice, the determinant vanishes and hence we cannot prove linear independence.

Some of my peers suggested that I take different eigenvectors, eigenvectors which do not lie in the common set of two null spaces and then proceed, but isn't that what I have to prove in the first place?

PS: sorry I don't know Mathjax

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    $\begingroup$ What you describe in your first paragraph is impossible, because the null space (I assume you mean of $A - \lambda I$) for two different eigenvalues $\lambda$ cannot have a nonzero vector in common. Are you trying proof by contradiction? If so you should state that up front. $\endgroup$ – Rahul Feb 4 at 9:25
  • $\begingroup$ exactly what are you trying to prove? $\endgroup$ – David C. Ullrich Feb 4 at 14:45
  • $\begingroup$ In short, they can’t by definition. In the situation you describe, there’s some nonzero vector $v$ for which $Av=\lambda v$ and $Av=\mu v$, with $\lambda\ne\mu$. This is impossible. $\endgroup$ – amd Feb 5 at 1:42
  • $\begingroup$ Ok yeah, two eigenvalues cannot have an eigenvector in common except <b>0</b>. So I guess that if a null space is the xy, then yz cannot be another null space? $\endgroup$ – Ajun908 Feb 5 at 11:16

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