Let $H$ be a $\mathbb C$-Hilbert space and $A\in\mathfrak L(H)$ with $A^2=A=A^\ast$.
Let $\lambda\in\mathbb C$ and $x,y\in H$ with $(\lambda-A)x=y$. How can we solve this for $x$?
I've stumbled about this post, which is showing how e can do this. However, I don't get one step in the description.
First, we can multiply both sides by $A$. Using that $A^2=A$, we obtain $$(\lambda-1)Ax=Ay\tag1.$$ But now the author of the post is multiplying both sides of $(1)$ by $A-1$. I don't get that since with this multiplication, bot sides of $(1)$ should simply vanish. For example, $$(A-1)Ay=Ay-A^2y=Ay-Ay=0\tag2.$$ What am I missing here?
The author didn‘t multiply both sides of (1) with $A - 1$ - as you pointed out correctly this gives $0= 0$, but he multiplied the equation he started with $(\lambda - A)x = y$ with $A-1$.