Let $G$ be a irreducible, rowstochastic $n \times n$ matrix and let $A = (G-I)$ and $B = (G-\lambda I)$, where $|\lambda|\leq 1$ is an arbitrary eigenvalue of $G$. I am interested in learning about the kernel of the matrix $W = AB$.
I know that $Bx = 0 \implies ABx = 0$, such that $\text{ker}(B)\subseteq \text{ker}(AB)$. I am wondering, however, whether it is generally true that both kernels are equal?
If this is the case, it should hold that $\text{im}(B) \cap \text{ker}(A) = \{0\}$. From the Perron-Frobenius theorem, I know that $\text{ker}(A)$ only contains multiples of the unit vector $\iota$. So it should hold that $\text{im}(B) \cap \iota = \{0\}$, or equivalently, $\iota$ belongs to the left kernel of $B$.
But how do I proceed from here?
Edit: I should have mentioned that the diagonal elements of $G$ are all zero, as it is a rownormalized adjacency matrix.
Since $\ker(A)\subseteq\ker(BA)=\ker(AB)$, if you want $\ker(AB)=\ker(B)$, it is necessarily true that $\ker(A)\subseteq\ker(B)$. As $G$ is irreducible and row stochastic, $\ker(A)=\operatorname{span}(\{e\})$. But then $\ker(A)\subseteq\ker(B)$ only if $\lambda=1$, i.e. only if $A=B$.
Conversely, if $A=B$, since $G$ is irreducible, $0$ is a simple eigenvalue of $A$. It follows that $\ker(AB)=\ker(A^2)=\ker(A)$.