1
$\begingroup$

suppose I have an asymptotic expansion for a matrix-valued function $\psi : \mathbb{C} \to \mathbb{C}^{2 \times 2}$ : $$\psi(\lambda) \sim I + \frac{m_1}{\lambda} + \frac{m_2}{\lambda^2} + \cdots \ \ \ \ \lambda \to \infty$$ where $m_i$ are constant $2 \times 2$ matrices. we also know that $\psi(\lambda)$ is invertible for all $\lambda$. My question is : Given $m_is$, How can I write an asymptotic expansion for $\psi^{-1}(\lambda)$ as $\lambda \to \infty$, I need the first $4$ terms of this expansion.

Any help is really appreciated, Thanks

$\endgroup$
1

1 Answer 1

4
$\begingroup$

First note that for small $A$ (small here means that $\rho(A) < 1$, where $\rho$ denotes the spectral radius), we have (von Neumann's series) $$ (I - A)^{-1} = \sum_{k=0}^\infty A^k\tag{$*$} $$ In our case $$ A = I - \psi(\lambda) \sim -\frac{m_1}\lambda - \frac{m_2}{\lambda^2} - \cdots, \quad \lambda \to \infty $$ Using this in $(*)$, we get $$ \psi(\lambda)^{-1} \sim I - \frac{m_1}\lambda - \frac{m_2 - m_1^2}{\lambda^2} - \frac{m_3 - m_1m_2 - m_2m_1 + m_1^3}{\lambda^3} - \frac{m_4 - m_2^2 - m_1m_3 - m_3m_1 + m_2m_1^2 + m_1m_2m_1 + m_2m_1^2 - m_1^4}{\lambda^4} - \ldots, \quad \lambda \to \infty $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .