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Seems so obvious but I can't get it:

If unity is not an eigenvalue of $A$, then $(I - A)$ is nonsingular.

How can I prove this?

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It is the definition, when $(I-A)$ is singular it means $(1 \cdot I -A)$ has a non trivial kernel, which means there is a $v\neq 0 $ such that $$(I-A)v=0$$ which would mean, that $A$ has the eigenvalue 1.

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Let $1$ not be an eigenvalue of $A$ but $I-A$ is singular.

Thus, Using the Characteristic Polynomial Approach

\begin{align} \det(A-\lambda I)&=0\\ \text{We Know,}\\ \det(A-I)&=0\\ \end{align} Thus, $1$ is an eigenvalue of $A$.

Contradiction!

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