tugmiddelc0

2021-11-20

What is an inverse matrix's eigenvalue?

barcelodurazo0q

A matrix A has an eigenvalue $\lambda$ only if ${A}^{-1}$ has eigenvalue ${\lambda }^{-1}$. For example:
$Av=\lambda v⇒{A}^{-1}Av=\lambda {A}^{-1}v⇒{A}^{-1}v=\frac{1}{\lambda }v$
If matrix A has eigenvalue $\lambda$, then I-A has eigenvalue $1-\lambda$. Therefore, ${\left(I-A\right)}^{-1}$ has eigenvalue $\frac{1}{1-\lambda }$

Wasither1957

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 in $\lambda$. This works for expressions I−A, which is 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 $\lambda 1<1$).