Find eigenvalue of $B= I + A^{-1}+A^{-2}+A^{-3}+...$

Solution 1:

Hint: show that $B=(I-A^{-1})^{-1}$, thus by spectral mapping theorem, the eigenvalues of $B$ are $(1-\lambda^{-1})^{-1}$ where $\lambda = 2, 3$ are eigenvalues of $A$.

Solution 2:

Note that if $\lambda$ is an eigenvalue of $A$ with corresponding eigenvector $v \neq 0$, so that $A v = \lambda v$, then $$ A^k v = \lambda^k v$$ for $k \in \mathbb Z$ (including negative $k$, in the event that $A$ is invertible). Now hit $B$ with the same vector $v$, so that $$ \begin{align*} B v &= (I + A^{-1} + A^{-2} + \dots ) v = Iv + A^{-1} v + A^{-2} v + \dots \\ &= \lambda^0 v + \lambda^{-1} v + \lambda^{-2} v + \dots = \Bigl( \sum_{k = 0}^\infty (\lambda^{-1})^k \Bigr) v = \frac{1}{1 - \lambda^{-1}} v. \end{align*} $$ Hence the eigenvalues of $B$ are $1/(1-\lambda^{-1})$ where $\lambda$ is an eigenvalue of $A$.

The same game with geometric series also informs the observations about $B = (I - A^{-1})^{-1}$; we have $$ B = \sum_{k = 0}^\infty (A^{-1})^k = (I - A^{-1})^{-1}. $$