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Suppose matrix $A$ has eigenvalue 1 with corresponding eigenvector $\mathbb{x}$. If $BA$ is to have eigenvalue of 1 with the same eigenvector, what would be the requirement or condition for it?

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Well, then also $\,B\,$ must have the eigenvalue 1 with exactly the same eigenvector $\,x\,$...About your second question: it is a little weird, but in the bottom line is: of course no, as this would contradict directly the definition of eigevector. – DonAntonio Dec 27 '12 at 13:33
(For the sake of continuity, the second question was whether a single eigenvector could have multiple eigenvalues) – Eric Stucky Dec 27 '12 at 14:54

We would need to have $(BA)\mathbb{x}=\lambda\mathbb{x}$. We know that matrix multiplication is associative and that $A\mathbb{x}=\mathbb{x}$, so it should be that $\lambda\mathbb{x}=(BA)\mathbb{x}=B(A\mathbb{x})=B\mathbb{x}$. We can see that by definition $\mathbb{x}$ would also have to be an eigenvector of $B$.

Running the equality in the other direction we can see that the converse is also true; that is, if $\mathbb{x}$ is an eigenvector of both $A$ and $B$ then it is an eigenvector of $AB$.

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