# Square root of normal positive operators over real Hilbert spaces

A bounded linear operator $A$ on a Hilbert space $H$ is called a positive operator if $\langle Ax, x\rangle \geq 0$ for all $x$ in $H$. It is known that, if $A$ is a positive operator on a Hilbert space $H$ over the complex field $\mathbb{C}$, then $A$ has unique positive square root.

My question is the following: Does a normal positive operator on an infinite dimensional Hilbert space over the real field $\mathbb{R}$ have a normal positive square root? If it exists, is it unique?

• Your definition of "positive operator" is in some sense not the right one in a real Hilbert space, since it doesn't imply self-adjointness. Are you sure it is the definition you want to work with? Nov 10, 2016 at 21:13
• @NateEldredge, Thank you for the reply. Yes, I would like to work with this definition. I am aware of the result that, self adjoint positive operators on the real Hilbert spaces have unique square roots. Please let me know some references related to this problem.
– mrka
Nov 11, 2016 at 3:05
• Do you know whether this is true in finite dimensions? Nov 11, 2016 at 3:14
• @NateEldredge I do not know for the answer for the finite dimensional case.
– mrka
Nov 11, 2016 at 4:23

The answer to both questions is yes.

First a word about terminology: as pointed out by Nate Eldredge and Nik Weaver in the comments, there are two notions of positivity at play here. To avoid any confusion, let us say that an operator $$T$$ on a real Hilbert space $$\mathcal H_{\mathbb{R}}$$ is:

• quasi-positive if $$\langle Tx,x\rangle \geq 0$$ for all $$x\in \mathcal H_{\mathbb{R}}$$;
• positive semidefinite if it is quasi-positive and self-adjoint.

In order to answer the question, we must pass to the complexification of $$\mathcal H_{\mathbb{R}}$$.

## The complexification.

A complex conjugation on a complex vector space $$V$$ is a conjugate linear map $$f : V \to V$$ which is equal to its own inverse:

• For all $$\lambda,\mu\in\mathbb{C}$$ and all $$x,y\in V$$ we have $$f(\lambda x + \mu y) = \overline\lambda f(x) + \overline\mu f(y)$$.
• For all $$x\in V$$ we have $$f(f(x)) = x$$.

The conjugation is usually written $$\bar{\ }\, : V \to V$$ instead of $$f$$, and the conjugate of an element $$x\in V$$ is written $$\overline x$$. We say an element $$x \in V$$ is real if $$\overline x = x$$ holds. The set of all real elements forms a real subspace of $$V$$, denoted by $$\text{Re}(V)$$. This is clearly not a complex subspace. In fact, every $$x \in V$$ can be written uniquely as $$x = a + ib$$ with $$a,b\in\text{Re}(V)$$.

If $$\mathcal H_{\mathbb{R}}$$ is a real Hilbert space, then it has a complexification, a complex Hilbert space $$\mathcal H_{\mathbb{C}}$$ together with a complex conjugation $$\bar{\ }\, : \mathcal H_{\mathbb{C}} \to \mathcal H_{\mathbb{C}}$$ such that

• The conjugation satisfies any (and therefore all) of the following equivalent properties:

• For all $$x,y\in\mathcal H_{\mathbb{C}}$$ we have $$\langle \overline x,\overline y\rangle = \overline{\langle x,y\rangle}$$.
• For all $$x,y\in\text{Re}(\mathcal H_{\mathbb{C}})$$ we have $$\langle x,y\rangle \in \mathbb{R}$$.
• For all $$a,b\in\text{Re}(\mathcal H_{\mathbb{C}})$$ we have $$\lVert a + ib\rVert^2 = \lVert a\rVert^2 + \lVert b\rVert^2$$.
• For all $$x\in\mathcal H_{\mathbb{C}}$$ we have $$\lVert \overline x\rVert = \lVert x\rVert$$.
• The real subspace $$\text{Re}(\mathcal H_{\mathbb{C}})$$ is Hilbert space isomorphic to $$\mathcal H_{\mathbb{R}}$$.

With these properties, the complexification is uniquely defined (up to isomorphism). Every operator $$T \in B_{\mathbb{R}}(\mathcal H_{\mathbb{R}})$$ extends uniquely to an operator in $$B_{\mathbb{C}}(\mathcal H_{\mathbb{C}})$$, and this extension gives us a unital, isometric (real) $$*$$-homomorphism $$\phi : B_{\mathbb{R}}(\mathcal H_{\mathbb{R}}) \to B_{\mathbb{C}}(\mathcal H_{\mathbb{C}}).$$ We get an induced complex conjugation $$\bar{\ }\,$$ on $$B_{\mathbb{C}}(\mathcal H_{\mathbb{C}})$$, which maps an operator $$S \in B_{\mathbb{C}}(\mathcal H_{\mathbb{C}})$$ to the operator $$x \mapsto \overline{S\,\overline x}$$. The image of the above homomorphism $$\phi$$ is precisely the real subspace consisting of all operators which are real with respect to the induced conjugation on $$B_{\mathbb{C}}(\mathcal H_{\mathbb{C}})$$.

For $$S,T \in B_{\mathbb{C}}(\mathcal H_{\mathbb{C}})$$ we have $$\overline{ST} = \overline S\,\overline T$$. The identity $$1$$ is real, so we find that $$S$$ is invertible if and only if $$\overline S$$ is inverible, in which case we have $$\overline{S}^{\,-1} = \overline{S^{-1}}$$. From this it follows that $$\sigma(\overline S) = \overline{\sigma(S)}$$ holds for any $$S\in B_{\mathbb{C}}(\mathcal H_{\mathbb{C}})$$. In particular, the spectrum of a real element is a self-adjoint subset of $$\mathbb{C}$$. (However it need not be real – a real square matrix can have complex eigenvalues!)

## Quasi-positive normal operators.

Note that an operator $$T \in B_{\mathbb{R}}(\mathcal H_{\mathbb{R}})$$ is quasi-positive if and only if its self-adjoint part $$\tfrac{1}{2}(T + T^*)$$ is positive semidefinite. This is because we have $$\langle Tx,x\rangle = \langle x,T^*x\rangle = \langle T^*x,x\rangle$$, hence $$\langle Tx,x\rangle = \langle T^*x,x\rangle = \left\langle \tfrac{1}{2}(T + T^*)x,x\right\rangle.$$ It follows that a normal operator $$T \in B_{\mathbb{R}}(\mathcal H_{\mathbb{R}})$$ is quasi-positive if and only if its complex spectrum $$\sigma_{\mathbb{C}}(T) := \sigma(\phi(T))$$ is contained in the closed right half plane $$\{\text{Re}(z) \geq 0\}$$. (Use the Gelfand representation of the commutative $$C^*$$-subalgebra $$C^*(\phi(T)) \subseteq B_{\mathbb{C}}(\mathcal H_{\mathbb{C}})$$ generated by $$\phi(T)$$.) This allows us to define quasi-positivity for complex normal operators: we say that a normal operator $$S \in B_{\mathbb{C}}(\mathcal H_{\mathbb{C}})$$ is quasi-positive if and only if $$\sigma(S)$$ is contained in the closed right half plane $$\{\text{Re}(z) \geq 0\}$$. The principal branch of the complex square root defines a continuous function from the closed right half plane to itself, so we may use the Gelfand representation of $$C^*(S)$$ to obtain a quasi-positive normal square root $$R$$ of any quasi-positive normal operator $$S$$. Now, if $$R'$$ is any quasi-positive normal square root of $$S$$, then we may use the Gelfand representation of $$C^*(R')$$ to prove that $$R = R'$$ must hold. (We have $$S \in C^*(R')$$, hence $$R \in C^*(S) \subseteq C^*(R')$$.) This shows that quasi-positive normal square roots are necessarily unique.

Next, let $$S \in B_{\mathbb{C}}(\mathcal H_{\mathbb{C}})$$ be quasi-positive, normal, and such that $$\sigma(S)$$ is self-adjoint. Now:

• For a complex polynomial $$p$$ in $$\lambda$$ and $$\overline\lambda$$, we define $$\overline p$$ to be the coefficient-wise complex conjugate of $$p$$, without interchanging the variables. Then we have $$\overline p\big(\overline\lambda\big) = \overline{p(\lambda)}$$.
• The principal branch $${\scriptstyle\surd} : \{\text{Re}(z) \geq 0\} \to \{\text{Re}(z) \geq 0\}$$ of the complex square root satisfies a similar, yet slightly different property: $${\scriptstyle\surd}\big(\overline{\lambda}\big) = \overline{{\scriptstyle\surd}(\lambda)}$$.

Hence, if $$\{p_n\}_{n=1}^\infty$$ is a sequence of complex polynomials in $$\lambda$$ and $$\overline\lambda$$ converging to $${\scriptstyle\surd}$$ uniformly on $$\sigma(S)$$, then we have $$\left|{\scriptstyle\surd}(\lambda) - \overline{p_n}(\lambda)\right| \: = \: \left|\overline{{\scriptstyle\surd}\big(\overline\lambda\big)} - \overline{p_n\big(\overline\lambda\big)}\right| \: = \: \left|{\scriptstyle\surd}\big(\overline\lambda\big) - p_n\big(\overline\lambda\big)\right|,$$ so it follows that the sequence $$\{\overline{p_n}\}_{n=1}^\infty$$ also converges to $${\scriptstyle\surd}$$ uniformly on $$\sigma(S)$$. Consequently, the sequence $$\{\tfrac{1}{2}(p_n + \overline{p_n})\}_{n=1}^\infty$$ also converges to $${\scriptstyle\surd}$$ uniformly on $$\sigma(S)$$. This is a sequence of real polynomials in $$\lambda$$ and $$\overline\lambda$$, so now we see that the unique quasi-positive normal square root of $$S$$ lies in the closed real subalgebra generated by $$S$$ and $$S^*$$. From this it follows that a real quasi-positive normal operator has a (unique) real quasi-positive normal square root.

In fact, the argument can be extended to show that every normal operator $$T \in B_{\mathbb{R}}(\mathcal H_{\mathbb{R}})$$ satisfying $$\sigma(T) \cap \mathbb{R}_{<0} = \varnothing$$ has a unique quasi-positive normal square root in $$B_{\mathbb{R}}(\mathcal H_{\mathbb{R}})$$.

Yes, we have a spectral theorem for operators on real Hilbert spaces. The multiplication operator version says that there is a Hilbert space isomorphism between $H$ and some real $L^2$ space which turns $A$ into multiplication by some function. If $A$ is positive, multiplication by the square root of the function is a positive square root of $A$. Working in the multiplication operator picture, it's easy to see that for any positive operator $B$ the operator $B^2$ has the same spectral subspaces, from which it easily follows that positive square roots are unique.

• I don't think it's that easy. With $H = \mathbb{R}^2$, the operator $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is normal and positive by this definition, but it isn't self-adjoint. It's not diagonalizable over the reals, so no isomorphism of real Hilbert spaces can make it into a multiplication operator. Nov 10, 2016 at 21:10
• Oh, I didn't notice that his definition of "positive" was nonstandard! Nov 11, 2016 at 0:11
• Incidentally, though, my operator $A$ does have a "positive" square root, namely $\frac{1}{\sqrt{2}} (A+I)$. (This is "positive" since $A+I$ is the sum of two "positive" operators.) Nov 11, 2016 at 1:24
• Yes, I noticed that too. Indeed, it is the unique "positive" square root. Nov 11, 2016 at 5:43