Injectivity and surjectivity of $\lambda I-A$.

Question

Let us $A$ a square matrix, $\lambda\in \mathbb R^+$, $I$ identity matrix, R a operator, X Banach space.

If $$(\lambda I-A) Ru=u \ \ (u\in X)$$ and $$R(\lambda I-A) u=u \ \ (u\in X)$$

then can we obtain that $\lambda I-A$ is one-to-one and onto?

Answer 1

Note that $R$ is also a square matrix and $X$ is just finite-dimensioned space. Let $B=(\lambda I-A)$. If I understand correctly, the first equation holds for $\forall u$, which implies that $\det BR\ne 0$, therefore $\det B\ne 0$, which gives that $B$ is injective and surjective.