Consider the iteration matrix for the general splitting method $M=I-N^{-1}A$ where $N$ is any invertible matrix. Show that if $\lambda =1$ is an eigenvalue of $M$. then $A$ cannot be invertible.
I tried: $$1) Mx = \lambda x$$ $$2) (I-N^{-1}A)x = x$$ $$3) x-N^{-1}Ax = x$$ $$4) 0 = N^{-1}Ax$$ $$5) 0 = Ax$$ But i'm not sure how this help show invertibility of A.
If $Mv = v$ for some $v \neq 0$ then $v - N^{-1} A v = v$ and so $N^{-1} A v = 0$. Multiplying by $N$ gives $A v = 0$, hence $A$ is not invertible.