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Matrices A and B are invertible and have the same eigenvector v for different corresponding eigenvalues. Show that the inverse of (AB) also has eigenvector v and find the corresponding eigenvalue.

I'm stuck on this problem and would really appreciate some help, thanks!

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Hint. So, let's say $Av = \lambda v$, $Bv = \mu v$. Now compute $$ (AB)v = A(Bv) = A(\mu v) = \mu Av = \ldots $$ and multiply both sides by $(AB)^{-1}$.

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  • $\begingroup$ Thank you, I got it! $\endgroup$ – Myra Nov 13 '18 at 20:18
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let $B(\vec{v}) = b\vec{v} B^{-1}B(\vec{v}) = \vec{v} = B^{-1}(b\vec{v})\implies \frac{1}{b}$ is eigenvalue for eigenvector $\vec{v}$ for $B^{-1}$ similarly $\frac{1}{a}$ is eigenvalue for eigenvector $\vec{v}$ for $A^{-1}$

now $B^{-1}A^{-1}\vec{v} = \frac{1}{a} B^{-1}\vec{v} = \frac{1}{ab}\vec{v}$

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