48
$\begingroup$

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?

$\endgroup$

2 Answers 2

196
$\begingroup$

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

$\endgroup$
12
  • 3
    $\begingroup$ OMG! That's brilliant! Thanks a lot. $\endgroup$
    – email
    Nov 15, 2012 at 10:51
  • 13
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 5
    $\begingroup$ @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
14
$\begingroup$

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 $\exp A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.