The is a simple iteration to compute $A^{-1}$ by applying Newton's method to
$$X^{-1} = A,$$
namely the iteration
$$X_{n+1} = X_n(2I – AX_n).$$
Is there is similarly simple iteration to solve the equation
$$X – \alpha X^{-1} = A$$
for $\alpha>0$?
I am only interested in the case where $A$ is symmetric positive definite and I also look for positive semidefinite answers $X$…
Best Answer
First, the coefficient $\alpha$ is redundant: w.l.o.g $\alpha=1$ since we can normalize it with the change of variables $\tilde{X}=\frac{1}{\sqrt{\alpha}}X$ and $\tilde{A}=\frac{1}{\sqrt{\alpha}}A$. From the discussion, the solution should not contain an inverse.
The equation $$ X+X^{-1}=A$$ is just a quadratic matrix equation $$ X^2-AX+I=0$$
An explicit solution is given by the formula for the scalar quadratic equation (which we can use in a matrix sense directly since all matrices commute) as essentially also pointed out in Jonas solution:
$$ X=\frac{1}{2}(A\pm \sqrt{A^2-4I})$$
To avoid inverses, we can use "two-variable method" for the square root, which needs to be scaled to be inside the applicable interval https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#A_two-variable_iterative_method
Initiate: $$ S=\frac{1}{\|A^2-4I\|}(A^2-4I)$$ $$ A_0=S$$ $$ C_0=S-I$$ Iterate: $$ A_{k+1}\leftarrow A_k-A_kC_k/2$$ $$ C_{k+1}\leftarrow C_k^2(C_k-3I)/4$$ until convergence.
Solution given by
$$ X=\frac{1}{2}(A+\sqrt{\|A^2-4I\|}A_k)$$
The norm can be chosen essentially arbitrary, but should be cheap to compute, so e.g. the Frobenius norm.
Applicability: This will only work if $A^2-4I$ positive definite, but if that's not satisfied the solution is complex so probably not what is of interest in this application. Roundoff errors can probably be substantial, since e.g. symmetry is not preserved. A standard quick fix trick which might help is to make $A_k$ symmetric in every step $A_k \leftarrow \frac{1}{2}(A_k+A_k^T)$.