57

From page 260 of Gilbert Strang's Linear Algebra and its Applications,

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

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?

Rodrigo de Azevedo
  • 23,223
email
  • 813
  • 2
  • 8
  • 11

2 Answers2

218

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

altwoa
  • 1,453
  • 4
  • 15
EuYu
  • 42,696
  • 3
    OMG! That's brilliant! Thanks a lot. – email Nov 15 '12 at 10:51
  • One more question! How can I show that I-A has an eigenvalue 1-$\lambda_{1}$? – email Nov 15 '12 at 10:55
  • 15
    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
  • Let $\vec{v}$ be an eigenvector of $A$ Then $(I-A)\vec{v}=I\vec{v}-A\vec{v}=\vec{v}-\lambda\vec{v}=(1-\lambda)\vec{v}$ – J L Nov 15 '12 at 10:59
  • 2
    Why I can't accept your answer? The box keeps saying me to wait in 6 minutes. – email Nov 15 '12 at 11:06
  • 1
    @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. – Charlie Shuffler Nov 27 '18 at 07:32
  • 5
    @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 07:51
  • @EuYu Perfect, thanks alot! – Charlie Shuffler Nov 27 '18 at 07:55
  • @EuYu, I would also like to revive this post and ask for a precision for the point of Charlie Shuffler: If we have $\textbf{A}^{-1}\textbf{A}\textbf{v} = \lambda \textbf{A}^{-1}\textbf{v}$, this could also be written $\textbf{I}\textbf{v} = \lambda \textbf{A}^{-1}\textbf{v}$, correct? Shouldn't we write then $\frac{1}{\lambda}\textbf{I}\textbf{v} = \lambda \textbf{A}^{-1}\textbf{v}$ ("matrix = matrix"). In other words how can we get rid of $\textbf{I}$ to have at end "vector = matrix"? Thanks a lot in advance – ecjb Aug 02 '19 at 14:43
  • 1
    @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. – EuYu Aug 03 '19 at 18:05
  • Thanks a lot for your explanations @EuYu! – ecjb Aug 03 '19 at 18:18
  • should be accepted answer! – Rémy Hosseinkhan Boucher Aug 07 '21 at 16:31
17

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 $\exp A$.

Marc van Leeuwen
  • 119,547