# How to find eigenvector of inverse of matrix product

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!

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 $$(AB)^{-1}$$.

• Thank you, I got it!
– Myra
Nov 13 '18 at 20:18

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