How to prove that if $AB = A + B$ then $\frac{\lambda}{\lambda - 1}$ is an eigenvalue for $A$?

Question

I'm struggling with a problem in linear algebra where I have to prove that:

Given $AB = A + B$ and $ 1 \notin \sigma (A) \cup \sigma(B)$ where $\sigma$ represent the spectrum of a Matrix and $\lambda \in \sigma(B)$ then the quotient : $\frac{\lambda}{\lambda - 1}$ is an eigenvalue for A ( $\frac{\lambda}{\lambda - 1} \in \sigma(A)$ )

I have proved before this that A is invertible iff B is invertible, it has also asked me to prove that :

$$\prod\limits_{\lambda_i \in \sigma(A)}(\lambda_i - 1) \prod\limits_{\lambda_i \in \sigma(B)}(\lambda_i - 1) = 1$$

But I couldn't find a way to do it, how can I prove both questions ?

Answer 1

$A(B-I) = B$. Suppose $Bv = \lambda v$, then $A (\lambda-1) v = \lambda v$, or $Av = {\lambda \over \lambda -1} v$.

Answer 2

To see $$\prod\limits_{\lambda_i \in \sigma(A)}(\lambda_i - 1) \prod\limits_{\lambda_i \in \sigma(B)}(\lambda_i - 1) = 1$$

fix an $i$ and look at $$\begin{split} (\lambda_i-1)\left(\frac{\lambda_i}{\lambda_i-1}-1\right) &=\frac{\lambda_i^2}{\lambda_i-1}-\left(\lambda_i+\frac{\lambda_i}{\lambda_i-1}\right)+1\\ &=\frac{\lambda_i^2}{\lambda_i-1}-\left(\frac{\lambda_i(\lambda_i-1)}{\lambda_i-1}+\frac{\lambda_i}{\lambda_i-1}\right)+1\\ &=\frac{\lambda_i^2}{\lambda_i-1}-\frac{\lambda_i^2-\lambda_i}{\lambda_i-1}-\frac{\lambda_i}{\lambda_i-1}+1\\ &=\frac{\lambda_i^2}{\lambda_i-1}-\frac{\lambda_i^2}{\lambda_i-1}+\frac{\lambda_i}{\lambda_i-1}-\frac{\lambda_i}{\lambda_i-1}+1\\ &=1 \end{split}$$

There's a little more to show. You'll need to also show that every eigenvalue of $A$ has the form $\frac{\lambda}{\lambda-1}$ where $\lambda$ is an eigenvalue of $B$.

Answer 3

To add to above answers, using that $1$ is not an eigenvalue of $B$, one can show that: \begin{align*} AB=A+B\Rightarrow A(B-I)=B\Rightarrow A=\frac{B}{B-I}\Rightarrow (B-I)A=B\Rightarrow BA=A+B\end{align*} This proves that the implication $\lambda\in\sigma(A)\Rightarrow\frac{\lambda}{1-\lambda}\in\sigma(B)$ is actually an equivalence. Using the identity in N8tron's comment, the product formula is now very easy to prove.

Answer 4

Here's an even easier answer.

Let $x$ be the eigenvector associated with $\lambda$. Then $$ \lambda Ax = ABx = (A+B)x = Ax + Bx = Ax + \lambda x. $$ This implies that $$ (\lambda - 1) Ax = \lambda x $$ from which the results follows with $x$ being an eigenvector for $A$ with eigenvalue $\lambda / (\lambda - 1)$.