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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?
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.
Yes you can, but you have to see the radius of convergence of the series. In other worde for what values of $z$ the series will converge. For example the series
$$ \sum_{k=0}^{\infty} A^k t^k $$
will converge under the condition
$$ ||tA||<1 \implies |t|<\frac{1}{||A||},$$
where $||A||$ is the norm of the matrix $A$.
Long story short: this is fine, no monkey business necessary, as long as $|z|$ is sufficiently small.
Long story long: This will certainly work for $z$ with a sufficiently small modulus. It comes down to the following:
Let $\|\cdot\|$ be any sub-multiplicative matrix-norm. If $\sum a_n z^n$ is a power series with radius of convergence $R>0$, then $\sum a_n X^n$ will converge for the matrix $X$ whenever $\|X\| < R$ (or equivalently, whenever $\rho(X) < R$).
Now, your square-root series has a radius of convergence $1$. So, as long as $\|zX\| < 1$ for some sub-multiplicative matrix norm, the sum $\sum a_n (zX)^n$ will converge.
You can then confirm (using the Cauchy product) that $(\sum a_n z^n)^2 = 1 + z$. Since such manipulation is allowed when the sum is absolutely convergent, you'll find that $(\sum a_n (zX)^n)^2 = I + zX$, which is to say that this series will give you some square root of the matrix $I + zX$.
Note that if $z = 1$, the condition $|X_{ij}| < 1$ is insufficient because the max-norm is not sub-multiplicative.
If you know the Jordan canonical form of a matrix $A$, and if the set of roots $R=\{\lambda_{1},\cdots,\lambda_{k}\}$ of the characteristic polynomial for $A$ does not include $0$, then you can choose a branch of the square root $\sqrt{z}$ which is holomorphic on a neighborhood of $R$, and you can compute a square root using the Jordan form using a finite number of number of powers. For example, consider $\sqrt{\lambda I+N}$ where $N^{k}=0$. If $F(z)=\sqrt{z}$, then $$ F(\lambda I+N)=F(\lambda)I+\frac{F'(\lambda)}{1!}N+\cdots \frac{F^{(k-1)}(\lambda)}{(k-1)!}N^{k-1}. $$ The series expansion for $F(A)$ terminates after taking enough powers that the off diagonal terms $N$ vanish. In terms of the Jordan blocks, consider an example of a $4\times 4$ Jordan block: $$ \sqrt{\left[\begin{array}{cccc} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda = \end{array}\right]} = \left[\begin{array}{cccc} F(\lambda) & F'(\lambda) & F''(\lambda)/2! & F'''(\lambda)/3! \\ 0 & F(\lambda) & F'(\lambda) & F''(\lambda)/2! \\ 0 & 0 & F(\lambda) & F'(\lambda) \\ 0 & 0 & 0 & F(\lambda) \end{array}\right] $$ This works for any $\lambda \ne 0$. The function $F$ is then applied to each Jordan block separately.
The only thing you have to watch out for with $F(z)=\sqrt{z}$ in general (not in your case because of considering only $I+zX$ for $z$ small) is where the roots of the Jordan form cross a fixed branch cut as $z$ moves around (then the result is discontinuous in $z$, as with the usual square root,) or when one of the eigenvalues of $I+zX$ is $0$, which is not okay unless there are no non-zero off-diagonal elements for the Jordan block with eigenvalue $0$--this is because $F'(0)$ is not defined. Regardless of the possibility discontinuity in $z$, so long as the above is defined, you will get $F(A)^{2}=A$. In your case, take $F(w)=\sqrt{1+zw}$, using your favorite branch of the square root.
