Show that $λ^{−1}$ is an eigenvalue for $A^{-1}B$

Question

The given starting point is $Av = λBv$. $A$ and $B$ are two invertible matrices and $v$ is a non-zero vector and $λ$ is a non-zero scalar. I thought about the following:

1. $Av = λBv$
2. $A = λB$ (divide by v)
3. $AB^{-1} = λI$ (multiply by $B^{-1}$)

I am stuck and don't know how to arrive at $λ^{-1} = A^{-1}B$. What would be the needed steps?

Answer 1

$$A^{-1}Bv = \lambda^{-1}v $$so $\lambda^{-1}$ is an eigenvalue of $A^{-1}B$ with eigenvector $v$ (it was non-zero).

Edit: $$Av = \lambda Bv $$ multiply both sides by $A^{-1}$ from left $$v = \lambda A^{-1}Bv $$divide by $\lambda$ (it's non-zero because $v\neq 0$) $$A^{-1}Bv = \lambda^{-1}v $$

Answer 2

Here is a simple proof. Since A is invertible multiply both side from the left with $A^{-1}$. You'll get $$Iv=A^{-1}\lambda B v$$ Or $$v=\lambda A^{-1}Bv$$.

Here is where you can see that $\lambda^{-1}$ is eigenvalue of $A^{-1}B$.

Since $$v=\lambda A^{-1}Bv$$ and $\lambda \neq 0$, just rewrite $$A^{-1}Bv=\lambda v$$. This is precisely the definition of eigenvalue.

Answer 3

$$Av = λBv$$ $$A^{-1}Av = A^{-1}λBv$$ $$v = λA^{-1}Bv$$ $$λ^{-1}v = A^{-1}Bv$$

Thus $λ^{-1}$ is an eigenvalue for $ A^{-1}B$