I am studying some [edit: bounded] operators that have the property that the partial isometry in the polar decomposition $T = U|T|$ is not only unitary but also self-adjoint.

Is this an already established class of operators? Is there an exact condition when $U$ in the polar decomposition is self-adjoint? (I mean, obviously it gives that the Duggal transform is equal to $T^*$...) How often does this happen? Can the partial isometry in the PD ever be self-adjoint but not unitary?

The operator I am working on is not normal, or even hyponormal.



  • $\begingroup$ But $|T|$ is self-adjoint, as well? Are you considering bounded or unbounded operators? $\endgroup$ – Roland Mar 12 '17 at 19:21
  • $\begingroup$ Yes, necessarily. But they don't commute, since T is not self-adjoint. Although clearly you get |T|U = T* in this case. Edit: only bounded operators. $\endgroup$ – Derek Thompson Mar 12 '17 at 19:23
  • $\begingroup$ But your vector space isn't necessarily finite-dimensional? $\endgroup$ – Roland Mar 12 '17 at 19:47
  • $\begingroup$ Right. My specific example is on H^2. $\endgroup$ – Derek Thompson Mar 12 '17 at 19:53
  • 1
    $\begingroup$ @Derek: Because the initial space of $U$ is the initial space of $T$, which implies that $T$ is injective if $U$ is, and the final space of $U$ is the final space of $T$, which implies that $T$ has dense range if $U$ is onto. (please use @ if you want to notify people of responses that are not on their posts. I looked here only because of your comment on my answer.) $\endgroup$ – Jonas Meyer Mar 14 '17 at 3:11

The class of operators you're describing is self-adjoint operators in Krein spaces. Your $U$ is nice enough to allow us to introduce some additional structure on our vector space where $T$ behaves nicely.

To sum-up your setting (and put it into a slightly different notation to have a better reference to the existing literature).

We have a Hilbert space ($\mathcal K, (\cdot,\cdot)$) and a self-adjoint operator $A (=|T|)$ as well as an operator $J (=U)$ which fulfills $J^*=J^{-1}=J$. Although it is not a self-adjoint operator, we are interested in some properties of $T=JA$, for instance its spectrum.

We can define an indefinite inner product on $\mathcal K$ by defining $[x,y]=(Jx,y)$. This is not positive definite, but fulfills all other properties of a scalar product. We call $(\mathcal K,[\cdot,\cdot])$ a Krein space.

We have $(Ax,y)=(JJAx,y)=[JAx,y]=[Tx,y],$ hence as $A$ is symmetric w.r.t. $(\cdot,\cdot)$, so is $T$ w.r.t. $[\cdot,\cdot]$. With some work (less if $T$ is bounded), we can show that the $[\cdot,\cdot]$-adjoint of $T$ is equal to $T$, hence $T$ is self-adjoint in the Krein space ($\mathcal K, [\cdot,\cdot]$).

For some literature on Krein spaces, there's this introduction.

Notable books are

T.A. Azizov, I.S. Iokhvidov: Linear Operators in spaces with an indefinite metric (1989)

J. Bognár: Indefinite Inner Product Spaces (1974)

There are actually not too many results on the general class of self-adjoint operators in Krein spaces, but some sub-classes of those, such as definitizable operators are particularily nice and some advancement on those has been made in the recent years. It's worth noting that if $J$ (your $U$) has spectrum which lies only on the positive real line, except for a finite number of negative eigenvalues with finite-dimensional eigenspaces, you are in the nice situation of a Pontrjagin space.


"Can the partial isometry in the PD ever be self-adjoint but not unitary?"

Yes, every partial isometry $V$ appears as the partial isometry in a PD, e.g. its own, $V=V|V|$ (in which case $|V|=V^*V$ is the orthogonal projection onto the initial space of $V$). In general, if $V$ is a self-adjoint partial isometry, and $P$ is a positive operator such that $\ker(P)=\ker(V)$, then the operator $T=VP$ is already in its polar decomposition.

A self-adjoint partial isometry is a self-adjoint operator $V$ satisfying the equation $V^3=V$, which is equivalent to $V$ being a self-adjoint operator with spectrum contained in $\{-1,0,1\}$, with $0$ being in the spectrum if and only if the operator is not unitary. Such an operator is unitary on its initial space, which in turn is the orthogonal direct sum of the eigenspaces for eigenvalues $1$ and $-1$. One way to see this last part is to note that if $U$ is a self-adjoint unitary operator, then $\frac12(1\pm U)$ are orthogonal projections. In other words, self-adjoint partial isometries are the same as operators of the form $P-Q$ where $P$ and $Q$ are orthogonal projections such that $PQ=0$. Such an operator is unitary if and only if $P+Q=I$.

  • $\begingroup$ Very helpful, thank you! $\endgroup$ – Derek Thompson Mar 14 '17 at 2:55

Obviously, if $T$ is self-adjoint, then $U$ commutes with $|T|$ and $U$ is unitary and also self-adjoint.


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