### Proof of the linear system equivalence

Hello,
In this proof, if I understand correctly, we are trying to show that x(I - theta H) = a (even though the proof writes 'want to show that pi (I - G) = 0 iff x(I - G) = 0, the ultimate goal is to show the formula, it seems). However, the last step of the proof uses x(I - theta H) = a, which makes it a circular argument. As far as I know, one is not allowed to use the formula we want to prove in the actual proof.
Any thoughts?

Not quite: we are trying to show that if $$x$$ is the solution of the linear system $$x(I-\theta H)=a$$, then it satisfies $$x(I-G)=0$$. Therefore, to show the latter, we are allowed to use the definition of $$x$$.

Okay, so my question becomes, where did we prove $$x(I-\theta H) = a$$? It seems that this formula just appears without justification.

@noureddine_gueddach1 said:
Okay, so my question becomes, where did we prove $$x(I-\theta H) = a$$? It seems that this formula just appears without justification.

This comes from the first step of the linear-system method, which is to compute the solution of $$x(I-\theta H) = a$$ (i.e. we're trying to find a vector $$x$$ such that $$x(I-\theta H) = a$$ ).

I understand that this comes from the the said method, but what's the justification of this method? I mean this step has to derive from something more 'fundamental', doesn't it?

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