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?

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    $\begingroup$ 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? $\endgroup$ Nov 10, 2016 at 21:13
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    $\begingroup$ @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, 2016 at 3:05
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    $\begingroup$ Do you know whether this is true in finite dimensions? $\endgroup$ Nov 11, 2016 at 3:14
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    $\begingroup$ @NateEldredge I do not know for the answer for the finite dimensional case. $\endgroup$
    – mrka
    Nov 11, 2016 at 4:23

2 Answers 2


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.

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    $\begingroup$ 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. $\endgroup$ Nov 10, 2016 at 21:10
  • $\begingroup$ Oh, I didn't notice that his definition of "positive" was nonstandard! $\endgroup$
    – Nik Weaver
    Nov 11, 2016 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.) $\endgroup$ Nov 11, 2016 at 1:24
  • $\begingroup$ Yes, I noticed that too. Indeed, it is the unique "positive" square root. $\endgroup$
    – Nik Weaver
    Nov 11, 2016 at 5:43

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