1
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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? $\endgroup$ – Nate Eldredge Nov 10 '16 at 21:13
  • 1
    $\begingroup$ @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. $\endgroup$ – mrka Nov 11 '16 at 3:05
  • $\begingroup$ Do you know whether this is true in finite dimensions? $\endgroup$ – Nate Eldredge Nov 11 '16 at 3:14
  • 1
    $\begingroup$ @NateEldredge I do not know for the answer for the finite dimensional case. $\endgroup$ – mrka Nov 11 '16 at 4:23
1
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Nov 10 '16 at 21:10
  • $\begingroup$ Oh, I didn't notice that his definition of "positive" was nonstandard! $\endgroup$ – Nik Weaver Nov 11 '16 at 0:11
  • $\begingroup$ 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.) $\endgroup$ – Nate Eldredge Nov 11 '16 at 1:24
  • $\begingroup$ Yes, I noticed that too. Indeed, it is the unique "positive" square root. $\endgroup$ – Nik Weaver Nov 11 '16 at 5:43
1
$\begingroup$

The answer to both questions is yes.

First a word about terminology: I agree with Nate Eldredge and Nik Weaver that what you defined should not be called positive, for this clashes with an existing (more common) notion of positivity. To avoid any confusion, let us say that an operator $T$ on a real Hilbert space $\mathcal H_{\mathbb{R}}$ is:

  • quasi-positive if $\langle Tx,x\rangle \geq 0$ holds 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 $B_{\mathbb{R}}(\mathcal H_{\mathbb{R}})$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.