# 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?

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

• OMG! That's brilliant! Thanks a lot. Nov 15, 2012 at 10:51
• 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, 2012 at 10:59
• Why I can't accept your answer? The box keeps saying me to wait in 6 minutes. Nov 15, 2012 at 11:06
• @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. Nov 27, 2018 at 7:32
• @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, 2018 at 7:51

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 $\exp A$.