Consider the power series expansion $$ \sqrt{1-z} = 1+\sum\limits_{k=1}^\infty c_k z^k, $$ converging absolutely in the ball $|z| \leq 1$. Let $H$ be a Hilbert space and $A \in \mathcal L(H)$ a bounded linear operator with $\|\mathrm{Id}-A\| \leq 1$. Then the series $$ 1 + \sum\limits_{k=1}^\infty c_k (I-A)^k $$ converges in norm to some $B \in \mathcal L(H)$. How to show that $B^2 = A$?

Clearly, we have $$ \biggl| \bigl(1-\|I-A\|\bigr) - \bigl(1+\sum_{k=1}^N c_k\|I-A\|^k \bigr)^2 \biggr| \to 0, \quad N \to +\infty, $$ but I need to show somehow that $$ \biggl\| (I-(I-A)) - \bigl(I+\sum\limits_{k=1}^Nc_k(I-A)^k\bigr)^2 \biggr\| \to 0, \quad N \to + \infty. $$ If it is necessary, $A = A^*$ and $A \geq 0$.

  • $\begingroup$ Shouldn't your definition of the square root $B$ be the limit of $I + \sum_{k=1}^N c_k ||I-A||^k$? $\endgroup$ Nov 2, 2014 at 13:30
  • $\begingroup$ That is his definition of the square root, and that's why he wants to prove that $B^2=A$, since it's not obviously seen from the limit definition. $\endgroup$
    – quangtu123
    Nov 2, 2014 at 14:04

2 Answers 2


I can help a liitle in the finite dimensional case.

$\sqrt{z}=1+\sum_{k=1}^\infty c_k(1-z)^k$ converges for $|1-z|\le 1$

If $A$ is diagonalizable with the base $\{v_i\}$ and corresponding eigenvalues $\lambda_i$ (since $||I-A||\le1$ we have $|1-\lambda_i|\le1$), consider:

$$\left(\sum_{k=1}^\infty c_k(I-A)^k\right)^2v_i=\left(\sum_{k=1}^\infty c_k(1-\lambda_i)^k\right)^2 v_i=\lambda_iv_i$$

therefore $\left(\sum_{k=1}^\infty c_k(I-A)^k\right)^2=B^2$ also take $\{v_i\}$ as a diagonalized base with the eigenvalues of that of $A$, therefore we conclude that $B=A$.

If $A$ is not diagonalizable, there exist a sequence $A_i$ of diagonalizable matrices which converges to $A$. We have $B_i^2=\left(\sum_{k=1}^\infty c_k(I-A_i)^k\right)^2=A_i$ converges to $A$ so we just need to prove that $B_i$ converges to $B$. We have $||B-B_i||=||\sum_{k=1}^\infty \left[c_k[(I-A)^k-(I-A_i)^k]\right]||\le\sum_{k=1}^\infty |c_k|\varepsilon^k$ for $i$ large enough. Since $|c_k|$ is bounded (why?-I'm not sure at this), $||B-B_i||$ go to $0$ as $i\rightarrow\infty$. So $B_i\rightarrow B$, so $B_i^2\rightarrow B^2=A$.

  • $\begingroup$ $|c_k|$ are bounded by the Cauchy inequalities. So, your proof generalizes to the case of compact operators as limits of matrices, if I'm not mistaken $\endgroup$
    – Appliqué
    Nov 2, 2014 at 15:42

We have the following identity $$ 1 - z = 1+\sum\limits_{N=1}^\infty \bigl( \sum\nolimits_{i+j=N} c_i c_j \bigr) z^N $$ near zero in $\mathbb C$, where $c_0 = 1$, which implies $$ \sum\nolimits_{i+j=N} c_ic_j = \begin{cases}-1, & N = 1, \\ 0, & N \geq 2. \end{cases} $$ Now we have $$ B^2 = I+\sum_{N=1}^\infty \bigl(\sum\nolimits_{i+j=N} c_i c_j\bigr) (I-A)^N = I-I+A=A. $$


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