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If the equation has a repeated root, can you tell without evaluating in the matrix if that repeated root corresponds to more than one eigenvector?

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  • $\begingroup$ In general, you only know that the geometric multeplicity does not exceed the algebraic multeplicity of an eigenvalue. $\endgroup$ – AnyAD Feb 22 at 20:05
  • $\begingroup$ The general topic here is the Jordan normal form, which exists for all square matrices (over $\mathbb C$), unlike diagonalization. $\endgroup$ – Lee Mosher Feb 22 at 20:30
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No. Consider $$ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, $$ and $$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. $$ They have the same characteristic polynomial, but not the same eigenvectors.

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