Suppose I have a matrix of the form
$$U\ =\ (I+z\thinspace X)^{\frac{1}{2}}$$
where $I$ is the $n\times n$ identity matrix, $z\in\mathbb{C}$ and $X$ is a $n\times n$ arbitrary complex matrix with entries following $|X_{ij}|\le1$. If $z$ has a small modulus ($|z|\ll1$), am I allowed to expand in a power series the square root matrix expression as
$$U\ =\ I \ +\ \frac{z}{2}X\ +\ …\thinspace ,$$
or is there some monkey business?
Best Answer
Yes you are allowed to do that. The power series $$ (1 + A)^{\frac12} = \sum_{n=0}^\infty \frac{(-1)^n(2n)!}{(1-2n)(n!)^2(4^n)}A^n = 1 + \textstyle \frac{1}{2}A - \frac{1}{8}A^2 + \frac{1}{16} A^3 - \frac{5}{128} A^4 + \dots, $$ converges at least in the region $\|A\|<1$. The convergence is locally uniform and "absolute", but the term "absolute" should now be "normal" because we use a norm for matrices instead of absolute value.