The following is a theorem from Reed & Simon's Methods of Modern Mathematical Physics, Volume I. Here, we are working in a complex Hilbert space $(\mathcal{H}, (\cdot, \cdot))$, and $\mathscr{L}(\mathcal{H})$ is the space of bounded linear operators $\mathcal{H} \to \mathcal{H}$.

Statement of theorem:

Let $A \in \mathscr{L}(\mathcal{H})$ be positive (i.e., $(x, Ax) \ge 0$, all $x \in \mathcal{H}$). Then there is a unique positive $B \in \mathscr{L}(\mathcal{H})$ such that $B^2 = A.$

The proof in the text begins as follows:

"It is sufficient to consider the case where $||A|| \le 1$. First observe that

$$||I- A|| = \operatorname{sup}_{||\phi|| = 1}|((I-A)\phi, \phi)| \le 1.$$

Next we use that fact that $\sqrt{1-z} = 1 + \sum_{k=1}^\infty c_kz^k$ converges absolutely for complex $z$ satisfying $|z| \le 1$, where the constants $c_k$ are known explicitly. This fact implies that the series $1 + \sum_{k=1}^\infty c_k(I- A)^k$ converges in norm to an operator $B$. Since the convergence is absolute, we can square the series and rearrange terms, which proves that $B^2 = A \dots$

I am able to verify most of the statements in this passage. However:

My question is, how does one successfully work out the calculation for squaring the series?

I have been trying for awhile to do the formal multiplication $$(1 + c_1(I - A) + c_2(I-A)^2 + \cdots )(1 + c_1(I - A) + c_2(I-A)^2 + \cdots ),$$

using the fact that $0 = 1 + \sum_{k=1}^\infty c_k$. But I am just getting a bunch of messy terms. Is there a good trick to use, or an easier way to realize that $B^2 = A$?

Hints or solutions are greatly appreciated.


2 Answers 2


From $\sqrt{1-z} = 1 + \sum_{k=1}^\infty c_kz^k$ for $|z|\leq1$, you get $$ 1-z=\left(1 + \sum_{k=1}^\infty c_kz^k\right)\,\left(1 + \sum_{k=1}^\infty c_kz^k\right) =\sum_{k,j=0}^\infty c_kc_jz^{k+j}\\ =\sum_{\ell=0}^\infty\left(\sum_{k=0}^\ell c_kc_{\ell-k}\right)\,z^\ell. $$ This shows that $c_0=1$, $c_0c_1+c_1c_0=-1$, and $\sum_{k=0}^\ell c_kc_{\ell-k}=0$ for all $\ell\geq2$.

Now, when you square $B$, the same sums between the coefficients $c_k$ arise, and so $$ B^2=I-(I-A)=A. $$


There is more known about this special case of Newtons binomial series,

$$\sqrt{1-z}=\sum_{k=0}^\infty \binom{1/2}{k} (-z)^k$$

for $|z|<1$, and since for $k\ge 1$

$$\binom{1/2}{k}=\frac{\tfrac12(\tfrac12-1)(\tfrac12-2)...(\frac12-k+1)}{k!} =(-1)^{k-1} \frac{(2k-3)(2k-5)...3\cdot 1}{2^k k!}=\frac{(-1)^{k-1}}{4^k(2k-1)}\binom{2k}{k}$$

this is

$$\sqrt{1-z}=1-\sum_{k=1}^\infty \frac1{2k-1}\binom{2k}{k} \left(\frac z4\right)^k$$

The point is that in the first form, squaring in the sense of the Cauchy product reproduces $1-z$, and this stay true if you replace $z$ by a matrix of norm smaller $1$


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