Let's consider a real square matrix $A$ with eigenvalues $\lambda_n$ and eigenvectors $\mathbb x_n$, i.e. $A \mathbb x_n = \lambda_n \mathbb x_n$.
Suppose there are some generalized eigenvectors $\mathbb y_n$ too, i.e. $A \mathbb y_n = \lambda_n \mathbb y_n + \mathbb x_n$.
Now let a vector product $(. ,. )$ be such that $(\mathbb y_1, \mathbb y_2) = (A \mathbb y_1, A \mathbb y_2) $. We can write down this product as: $$(\mathbb y_1, \mathbb y_2) = \lambda_1 \lambda_2 (\mathbb y_1, \mathbb y_2) + \lambda_1 (\mathbb y_1, \mathbb x_2) + \lambda_2 (\mathbb x_1, \mathbb y_2) + (\mathbb x_1, \mathbb x_2) $$
What I'm struggling to prove is that if $\lambda_1 \lambda_2 \neq 1$, then $(\mathbb y_1, \mathbb y_2) = 0$ and $\lambda_1 (\mathbb y_1, \mathbb x_2) + \lambda_2 (\mathbb x_1, \mathbb y_2) + (\mathbb x_1, \mathbb x_2) = 0$. Someone could help?