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Prove that any eigenvectors of $A^{-1}$ must be in some subspace of $adj(A)$

Since $A^{-1}$=$\frac{1}{detA}ajdA$, so if $\lambda$ is eigenvalue of $A^{-1}$ so it exist $x\not=0$ such that $A^{-1}x=\lambda x$ then $\frac{1}{detA}ajdAx=\lambda x$ since $\frac{1}{det}$ is some scalar then $adjAx=detA\lambda x$ and if I put $\lambda´=detA\lambda$ then $adjAx=\lambda^´ x$, so they have the same eigenvectors, I think that is easy way to show but what you think?

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  • $\begingroup$ What's "some subspace of $adj(A)$ means? $\endgroup$ – Vim Sep 8 '18 at 9:12
  • $\begingroup$ I think that I need to write eigensubspace $\endgroup$ – Marko Škorić Sep 8 '18 at 9:14
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    $\begingroup$ As you have noted $A^{-1}$ is just a non-zero multiple of $\text{adj}(A)$ so they trivially have the same eigenvectors don't they? $\endgroup$ – Vim Sep 8 '18 at 9:15
  • $\begingroup$ Yes you are right $\endgroup$ – Marko Škorić Sep 8 '18 at 9:17

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