Inverse matrix's eigenvalue?

It's from the book "linear algebra and its application" by gilbert strang, page 260.

$(I-A)^{-1}$=$I+A+A^{2}+A^{3}$+...

Nonnegative matrix A has the largest eigenvalue $\lambda_{1}$<1.

Then, the book says, $(I-A)^{-1}$ has the same eigenvector, with eigenvalue $\frac{1}{1-\lambda_{1}}$.

Why? Is there any other formulas between inverse matrix and eigenvalue that I don't know?

If you are looking at a single eigenvector $v$ only, with eigenvalue $\lambda$, then $A$ just acts as the scalar $\lambda$, and any reasonable expression in $A$ acts on $v$ as the same expression in $\lambda$. This works for expressions $I-A$ (really $1-A$, so it acts as $1-\lambda$), its inverse $(I-A)^{-1}$, in fact for any rational function of $A$ (if well defined; this is where you need $\lambda_1<1$) and even for $\exp A$.
A matrix $A$ has an eigenvalue $\lambda$ if and only if $A^{-1}$ has eigenvalue $\lambda^{-1}$. To see this, note that $$A\mathbf{v} = \lambda\mathbf{v} \implies A^{-1}A\mathbf{v} = \lambda A^{-1}\mathbf{v}\implies A^{-1}\mathbf{v} = \frac{1}{\lambda}\mathbf{v}$$
If your matrix $A$ has eigenvalue $\lambda$, then $I-A$ has eigenvalue $1 - \lambda$ and therefore $(I-A)^{-1}$ has eigenvalue $\frac{1}{1-\lambda}$.
• One more question! How can I show that I-A has an eigenvalue 1-$\lambda_{1}$? – email Nov 15 '12 at 10:55
• Let $\mathbf{v}$ be an eigenvector of $A$ under $\lambda$. Then $$(I-A)\mathbf{v} = I\mathbf{v} - A\mathbf{v} = \mathbf{v} - \lambda\mathbf{v} = (1-\lambda)\mathbf{v}$$ – EuYu Nov 15 '12 at 10:59
• @S.Crim We have $\lambda A^{-1}\mathbf{v} = A^{-1}A\mathbf{v} = \mathbf{v}$. We know that $\lambda \neq 0$ since $A$ is invertible, so we can divide through by $\lambda$ to get the desired result. – EuYu Nov 27 '18 at 7:51