Any lemma for $(A+A^{-1})^{-1}$?

Question

I'm actually a little surprised since I wasn't able to find any nice property to compute $(A+A^{-1})^{-1}$ ... Anyone knows about a theoretical way to achieve this ? Like an specific inversion lemma?

Thanks!

Answer 1

Assuming that $A$ and $A+A^{-1}$ are invertible, this is $A (I+A^2)^{-1} = (I+A^2)^{-1} A$.

Answer 2

A related result is given in this answer, where we can show that: $$(A^{-1}+\epsilon A)^{-1}=A- \epsilon A^3+\epsilon^2 A^5+\ldots = \sum_{k=1}^\infty (-\epsilon)^{k-1}A^{2k-1}$$ If the series converges. Specifically, when the norm of the matrix $A$ is smaller than $1$, the series will converge for $\epsilon=1$, so that:

$$(A^{-1}+A)^{-1}=A- A^3+A^5+\ldots $$ Which means you can approximate your expression without any actual inversions.

Answer 3

You may find use in expanding the inversions in terms of their Taylor series, and this could possibly help you provide bounds on some aspects. Note that this approach will only work if you know that the series converges, e.g. (it is sufficient but not necessary) that you know that $A$ is diagonalizable and has all eigenvalues in the radius of convergence of the same function over the reals.

The Taylor series coefficients have the cute form of

$$ \frac{1}{2},0,-\frac{1}{4},\frac{1}{4},-\frac{1}{8},0,\frac{1}{16},-\frac{1}{16},\frac{1}{32},0,-\frac{1}{64},\frac{1}{64},-\frac{1}{128},0,\frac{1}{256},-\frac{1}{256},\frac{1}{512} \dots $$