If $P$ is self-adjoint and $P^2$ is a projection, when is $P$ a projection

functional-analysishilbert-spacesoperator-theory

Let $H$ be a Hilbert space and $P:H \to H$ a bounded linear operator.

I am aware that if $P$ is self-adjoint and idempotent then it's a (orthogonal) projection.

My question is: if $P$ is self-adjoint and $P^2$ is a projection, is $P$ necessarily a projection? If not always, when?

Definitions:

  1. $P$ is self-adjoint $:\iff$ $P = P^*$, that is, $\langle Px, y \rangle = \langle x, P^*y \rangle \forall x,y \in H$ (if $H$ is a complex Hilbert space than it suffices to check this for $y=x$)
  2. $P$ is idempotent $:\iff$ $P = P^2$

My initial thoughts:

  • maybe not in general. consider $H = \mathbb{R}^2$ with $P$ the left-shift operator ($P$ defined by the matrix $((0, 0)^T, (1, 0)^T)$). Then $P^2$ projects everything to zero, but $P$ is not a projection.
  • so: maybe this needs stricter assumptions.
    • must $P^2 \ne 0$?
    • maybe the space must be infinite dimensional?

Best Answer

While the comments gave the answer already I, nonetheless, want to give a proof which does not rely on eigenvalues or diagonalizability, thus settling your question for arbitrary Hilbert spaces. A corollary of the following theorem is that any $P\in\mathcal B(\mathcal H)$ self-adjoint with $P^2=P^4$ is a projection if and only if $P\geq 0$; however, I prefer the version below because it shifts the assumptions on $P$ to the statement.

Theorem. Let a Hilbert space $\mathcal H$ as well as $P\in\mathcal B(\mathcal H)$ be given. Then $P$ is an orthogonal projection if and only if $P$ is positive semi-definite and satisfies $P^2=P^4$.

NB: For arbitrary Hilbert spaces "positive semi-definite" is defined as $A=A^*$ and $\langle x,Ax\rangle\geq 0$ for all $x\in\mathcal H$. Now if $\mathcal H$ is a complex Hilbert space, then it is enough to define $A\geq 0$ as $\langle x,Ax\rangle\geq 0$ because this automatically implies that $A$ is self-adjoint.

Proof of theorem. "$\Rightarrow$": Let $P$ be an orthogonal projection, i.e. $P=P^2=P^*P$. Obviously $P^4=(P^2)(P^2)=PP=P^2$. Moreover $\langle x,Px\rangle=\langle x,P^*Px\rangle=\langle Px,Px\rangle=\|Px\|^2\geq 0$ for all $x\in\mathcal H$, that is, $P\geq 0$ (because $P=P^*$ by assumption).

"$\Leftarrow$": Because $P\geq 0$, in particular $P$ is self-adjoint so all we have to show is $P=P^2$. Note that $P^2=P^*P\geq 0$ which implies that $P^2$ has a unique square root; (by uniqueness) this is obviously given by $P$. Analogously one sees that the square root of $P^4\geq 0$ is given by $P^2$. Again because the square root is unique we conclude$ P=\sqrt{P^2}=\sqrt{P^4}=P^2$. $\quad\square$

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