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It’s from the book “Linear Algebra and its Applications” by Gilbert Strang, page 260.

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

Nonnegative matrix $A$ has the largest eigenvalue $\lambda_1<1$.

Then, the book says $(I-A)^{-1}$ has the same eigenvector, with eigenvalue $1/(1-\lambda_1)$.

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

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2 Answers 2

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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, 2012 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, 2012 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, 2012 at 11:06
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    $\begingroup$ @EuYu Hope I can revive this post for a quick question. What happened when going from the second implication to the third in your answer. To be more clear, how did u go from $A^{-1}A\mathbf{v} = \lambda A^{-1}\mathbf{v}$ to $A^{-1}\mathbf{v} = \frac{1}{\lambda}\mathbf{v}$? Is it because we know that $A^{-1}\mathbf{v}=\lambda^{-1}\mathbf{v}$? Thanks in advance. $\endgroup$ Nov 27, 2018 at 7:32
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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, 2018 at 7:51
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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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