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Let $H$ be a Hilbert space, and $A\in B(H\to H)$ be a bounded non-negative operator (i.e. $\langle Ax,x\rangle \geq 0$ for all $x\in H$). The square root of $A$ is a bounded non-negative operator $B\geq $ such that $B^2=A$.

First, We can assume without loss of generality that $0\leq A\leq I$. Note that $B^2=A$ if and only if $$I-B=\frac{1}{2}((I-A)+(I-B)^2).$$ Hence, we definite inductively a sequence $C_n$ of operators as follows: $C_0:=0$, and $C_{n+1}:=1/2((I-A)+C_n^2))$. Then it is easy to see that $C_n$ converges to a bounded non-negetive operator $B$ in the strong operator topology and we also have $B^2=A$, thus the square root exists, but I don't know how to show that it is unique.

Xiang Yu
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2 Answers2

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Suppose $A$ is a bounded nonnegative operator on a Hilbert space. Let $(p_n)$ be a sequence of polynomials such that $$ p_n(x) \to \sqrt{x} $$ uniformly for $x$ in the interval $[0, \|A\|]$ (the Weierstass Approximation Theorem implies the existence of such a sequence of polynomials).

Now suppose $B$ is a nonnegative square root of $A$. Let $\mathcal{B}$ denote the norm closed algebra generated by $B$. Then $\mathcal{B}$ is a commutative $C^*$-algebra that contains $B$ and $A$ (because $A = B^2$). Thus there is a compact Hausdorff space $K$ such that $\mathcal{B}$ is isomorphic as a $C^*$-algebra to $C(K)$. This isomorphism preserves all $C^*$ properties. Thus $A$ corresponds to some nonnegative function $f \in C(K)$ taking values in $[0, \|A\|]$ and $B$ must correspond to the function $\sqrt{f}$ (which is the only nonnegative square root of $f$ in $C(K)$).

Because $p_n \circ f$ converges uniformly to $\sqrt{f}$ uniformly on $K$, we conclude that $p_n(A)$ converges in operator norm to $B$. But the polynomials $(p_n)$ were chosen independently of $B$. Thus $B$ is uniquely determined as a nonnegative square root of $A$.

Sheldon Axler
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Here is another proof adapted from my prof's lecture notes (maybe it's just a more noob-friendly explanation to Sheldon Axler's answer):

Step 1: Find a square root operator

Suppose that $A$ is the non-negative bounded linear operator on our Hilbert space, let $$ p_n(x)\to\sqrt{x} $$ be approximating polynomials due to Weierstrass Approximation Theorem, and let $$ Y=\lim_{n\to\infty}p_n(A)= \lim_{n\to\infty}\sum_{\lambda\in\sigma_p(A)}p_n(\lambda)P_\lambda , $$ where the last equal sign is due to Spectral Theorem:

  • $\sigma_p(A)$ is the set of eigenvalues, the so-called point-spectrum of $A$;
  • $P_\lambda$ is the orthogonal projection to the eigenspace of eigenvalue $\lambda$;
  • We can write $p_n(A)$ like this, because $A^m=(\sum_{\lambda\in\sigma_p(A)}\lambda P_\lambda)^m=\sum_{\lambda\in\sigma_p(A)}\lambda^mP_\lambda$, since $P_\lambda P_\mu = \delta_{\lambda, \mu} P_\lambda$. Also see this Wikipedia page).

It can be proven that $Y$ fulfills $Y^2=A$.

Step 2: Show that it is unique

Suppose $X$ is another non-negative bounded linear operator such that $X^2=A$, then $$ XA-AX=X^3-X^3=0 , $$ so the elements $A$ and $X$ commute with respect to function composition. This implies that $$ Xp_n(A)-p_n(A)X=0 \qquad \forall n , $$ and $$ XY-YX=\lim_{n\to\infty}\big(Xp_n(A)-p_n(A)X\big)=0 . $$

Claim: $\mathbf{(X − Y)^3 = 0}$

Because $$ (X − Y)X(X − Y) + (X − Y)Y(X − Y) \\ = (X^2-Y^2)(X-Y) = (A-A)(X-Y) = 0 $$ and both $(X − Y)X(X − Y)$ and $(X − Y)Y(X − Y)$ are non-negative operators, we have $$(X − Y)X(X − Y) = (X − Y)Y(X − Y) = 0$$ and thus $$ (X − Y)^3 = (X − Y)X(X − Y) − (X − Y)Y(X − Y) = 0 . $$ Final Steps

We also have $(X − Y)^4=0$, and $$ 0 = \|(X − Y)^4\| = \|X − Y\|^4 . $$ This is because $(X − Y)^*=X-Y$ self-adjoint, and $\|T^\ast T\|=\|T^\ast\|\|T\|$ for a bounded linear operator $T$ on a Hilbert space (the so-called $C^\ast$-identity).

So $X=Y$.

Ooooscar
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