Operator Positivity and Roots

Kreyszig Functional Analysis

Let $H$ denote an arbitrary complex Hilbert space. On the space of self-adjoint operators $H\to H$, we define a partial order where $T_1\le T_2$ iff $⟨T_{1}x,x⟩\le ⟨T_{2}x,x⟩$ for all $x\in H$. Noting that $0$ is a Hermitian operator $H\to H$, we say $T$ is positive iff $T\ge 0$; that is, a self-adjoint operator $T:H\to H$ is positive iff $⟨Tx,x⟩\ge 0$ for all $x\in H$.

Note that if $S$ and $T$ are self-adjoint, their product $ST$ is self-adjoint iff they commute ($ST=TS$) as $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }=TS$. Furthermore, note that under our system of definitions, linearity and boundedness are implied by “self-adjoint operator.”

Theorem 1. If two positive self-adjoint operators $S$ and $T$ on $H$ commute ($ST=TS$), their product $ST$ is positive.

Proof) We use an idiomatic decomposition that can express any positive operator as a sum of squares of other positive operators. We begin by taking $S_1=S/\Vert S\Vert$, which clearly satisfies $0\le S_1\le I$. Consider the sequence defined by

$$S_{k+1}=S_k(I-S_k)=S_k-S_k^2$$

Clearly, every $S_k$ is self-adjoint as, by construction, $S$ and all the $S_k$s commute with one another. Furthermore, we can show every $S_k$ satisfies $0\le S_k\le I$ by induction. This is because if $0\le S_k\le I$, for any $x\in H$,

$$⟨S_k(I-S_k)x,x⟩=⟨(I-S_k)x,S_{k}x⟩=⟨S_{k}x,x⟩-\Vert S_{k}x{\Vert }^2\le \Vert S_k\Vert \Vert x{\Vert }^2\le \Vert x{\Vert }^2=⟨Ix,x⟩$$

and thus $S_{k+1}\le I$. Furthermore, $S_k(I-S_k{)}^2\ge 0$ and $S_k^2(I-S_k)\ge 0$ as for any $x\in H$,

$$⟨S_k(I-S_k{)}^{2}x,x⟩=⟨S_k(I-S_k)x,(I-S_k)x⟩\ge 0,⟨S_k^2(I-S_k)x,x⟩=⟨(I-S_k)S_{k}x,S_{k}x⟩\ge 0$$

note we used the fact that $0\le S_k\le I$ implies $0\le (I-S_k)\le I$. This implies that $S_k(I-S_k)=S_k^2(I-S_k)+S_k(I-S_k{)}^2$ is also positive, and $S_{k+1}\ge 0$.

For any $n\in \mathbb{N}$, we can say $S/\Vert S\Vert =S_1=S_1^2+S_1^2+\dots +S_n^2+S_{n+1}$ by construction. Here, for any $x\in H$,

$$⟨S_{1}x,x⟩=\sum _{i=1}^n⟨S_i^{2}x,x⟩+⟨S_{n+1}x,x⟩=\sum _{i=1}^n\Vert S_{i}x{\Vert }^2+⟨S_{n+1}x,x⟩$$

For any fixed $x\in H$ the series $\sum \Vert S_{i}x{\Vert }^2$ is therefore monotonic and bounded by $⟨S_{1}x,x⟩$, implying convergence. This forces $\Vert S_{i}x{\Vert }^2\to 0$ as $i\to \infty$, implying $S_{i}x\to 0$. Thus, the operator series $\sum S_i^2$ strongly converges to $S_1$ as for any $x\in H$,

$$S_{1}x-\sum _{i=1}^n{S}_i^{2}x=S_{n+1}x,S_{1}x-\underset{n\to \infty }{\lim }\sum _{i=1}^n{S}_i^{2}x=\underset{n\to \infty }{\lim }{S}_{n+1}x=0$$

We therefore obtain a representation of $S$ as a sum of squares of positive operators under the strong operator topology. Each of the $S_i$s also commute with any operators $S$ commutes with, as they are simply polynomials of $S$. This decomposition makes Theorem 1 a straightforward corollary as follows:

Note that expressions of the form $A^{2}B$ where $B$ is positive and $A$ is self-adjoint are easy to deal with when $A$ and $B$ commute, as evidenced by our observations of $S_k^2(I-S_k)$ and $S_k(I-S_k{)}^2$. Our decomposition of $S$ allows us to put the operator product $ST$ in precisely such a form.

$$⟨STx,x⟩=\underset{n\to \infty }{\lim }⟨T\left(\sum _{i=1}^n{S}_i^2\right)x,x⟩=\underset{n\to \infty }{\lim }\sum _{i=1}^n⟨T{S}_{i}x,S_{i}x⟩\ge 0$$

This holds for all $x\in H$, and thus $ST\ge 0$.

Theorem 2. Let $T_i$ be a sequence of self-adjoint operators on $H$ such that

$$T_1\le T_2\le \dots \le T_n\le \dots \le K$$

where $K$ is some self-adjoint operator on $H$. Suppose any $T_i$ commutes with any other $T_j$ and with $K$. Then $(T_n)$ is strongly operator convergent with $T_n\to T$, where $T$ is some self-adjoint operator satisfying $T\le K$.

Proof) As always, it is easier to deal with differences than the sequence objects themselves. We take $S_i=K-T_i$. $S_i$ must clearly be self-adjoint and positive. Furthermore, all the $S_i$s clearly commute with each other. In order to prove Theorem 2, a necessary first step is to show $S_i$ is strongly operator convergent.

For any $x\in H$, we show that $S_{i}x$ is Cauchy. Take any $i<j$. By Theorem 1, $S_j(S_i-S_j)\ge 0$ and $S_i(S_i-S_j)\ge 0$, implying $S_i^2\ge S_i{S}_j\ge S_j^2$. Furthermore, for any such $(i,j)$,

$$\Vert S_{i}x-S_{j}x{\Vert }^2=\Vert S_{i}x{\Vert }^2+\Vert S_{j}x{\Vert }^2-2⟨S_i{S}_{j}x,x⟩\le \Vert S_{i}x{\Vert }^2-\Vert S_{j}x{\Vert }^2$$

It therefore suffices to show $\Vert S_{i}x{\Vert }^2$ is convergent. This is clearly true as $\Vert S_{i}x{\Vert }^2$ is a lower-bounded monotonically decreasing sequence, as shown by

$$\Vert S_{i}x{\Vert }^2=⟨S_i^{2}x,x⟩\ge ⟨S_{i+1}^{2}x,x⟩=\Vert S_{i+1}x{\Vert }^2$$

Therefore, $S_{i}x$ is Cauchy for all $x\in H$, implying $S_i$ is strongly operator convergent. This, in turn, implies $T_i\to T$ for some bounded linear operator $T$ (boundedness follows from the Uniform Boundedness Theorem). By the continuity of inner product, it is easy to see that $T$ must be self-adjoint and that $T\le K$.

Theorem 3. For any positive operator $T:H\to H$, there exists a unique positive operator $A:H\to H$ such that $T=A^2$. This operator $A$ is called the positive square root of $T$, and commutes with every bounded linear operator on $H$ which commutes with $T$.

Proof) Kreyszig presents an interesting constructive proof where a sequence of elementary corrections leads to the desired square root operator. Although the base iteration is a very rudimentary error-correction sequence, note there are some technical details (the speculation that $A_n\le A_{n+1}$, constraining $T\le I$) that aren’t very intuitive.

Clearly, by scaling it suffices to show Theorem 3 for positive operators $T$ such that $T\le I$. Consider the sequence $A_n$ where $A_0=0$ and $A_{n+1}=A_n+(T-A_n^2)/2$. $A_n$ is a polynomial of $T$ and is therefore self-adjoint. If $A_n$ is strongly operator convergent to some operator $A$, for any $x\in H$, $T=A^2$ and so we can speculate fairly confidently this $A$ Is our desired positive square root. We now flesh out this speculation.

Note that $A_n\le I$ as $I-A_n=I-A_{n-1}-T/2+A_{n-1}^2/2=(I-T)/2+(A_{n-1}-I{)}^2/2\ge 0$. Furthermore, we can see that $A_n\le A_{n+1}$ by induction and Theorem 1, as

$$A_{n+1}-A_n=(A_n-A_{n-1})-(A_n^2-A_{n-1}^2)/2=(A_n-A_{n-1})(I-(A_n+A_{n-1})/2)$$

and $I-(A_n+A_{n-1})/2\ge 0$ as $A_n\le I$. Therefore, $A_n$ is a monotonically increasing sequence bounded below by $A_0=0$ and above by $I$. This forces $A_n$ to strongly converge to some positive operator $0\le A\le I$ by Theorem 2. For any $x\in H$,

$$(Tx-A^{2}x)/2=\underset{n\to \infty }{\lim }(A_{n+1}-A_n)x=0\Rightarrow A^{2}x=Tx$$

Therefore, $A$ is our desired positive square root of $T$. Note that as every $A_n$ is a polynomial of $T$, it must commute with any bounded linear operator that commutes with $T$. By continuity, $A$ must as well.

We finally show uniqueness. Assume that $T$ has two distinct positive square roots $A$ and $B$. This implies $A^2=B^2=T$. Then, if we take any $x\in H$ and $y=(A-B)x$,

$$0=⟨(A^2-B^2)x,y⟩=⟨(A+B)y,y⟩=⟨Ay,y⟩+⟨By,y⟩$$

As $A$ and $B$ are positive, this means $⟨Ay,y⟩=⟨By,y⟩=0$ for all such $y$.

As $A$ is itself positive, it also has a positive square root $C$. This $C$ satisfies $0=⟨C^{2}y,y⟩=\Vert Cy{\Vert }^2\Rightarrow Cy=0$ for all $y\in (A-B)H$, which in turn implies $Ay=0$ for all $y\in (A-B)H$. Analogously, $By=0$ for all such $y$. For all $x\in H$, this means that

$$(A-B{)}^{2}x=0\Rightarrow 0=⟨(A-B{)}^{2}x,x⟩=\Vert (A-B)x{\Vert }^2\Rightarrow Ax=Bx$$

There is a contradiction, and the positive square root must be unique.