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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?

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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 $\exp A$.

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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 $\frac{1}{1-\lambda}$.

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    $\begingroup$ OMG! That's brilliant! Thanks a lot. $\endgroup$ – email Nov 15 '12 at 10:51
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    $\begingroup$ Let $\mathbf{v}$ be an eigenvector of $A$ under $\lambda$. Then $$(I-A)\mathbf{v} = I\mathbf{v} - A\mathbf{v} = \mathbf{v} - \lambda\mathbf{v} = (1-\lambda)\mathbf{v}$$ $\endgroup$ – EuYu Nov 15 '12 at 10:59
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    $\begingroup$ Why I can't accept your answer? The box keeps saying me to wait in 6 minutes. $\endgroup$ – email Nov 15 '12 at 11:06
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    $\begingroup$ @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. $\endgroup$ – EuYu Nov 27 '18 at 7:51
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    $\begingroup$ @ecjb You have $I\mathbf{v} = \mathbf{v}$ because the identity matrix, by definition, does nothing to the vector it acts on. It's just like how you can write $a = 1\times a$ for any real number $a$. A matrix acting on a vector returns a vector, not a matrix. More precisely, the product of a $m\times n$ matrix with a $n\times k$ matrix returns a $m \times k$ matrix. A (column) vector is just an $n\times 1$ matrix, so there's no contradiction here. $\endgroup$ – EuYu Aug 3 '19 at 18:05

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