Is $pq+qp$ ever a projection?

There is a question over on Mathoverflow to which I added this answer.

Is it an answer? i.e. if $$p$$ and $$q$$ are Hilbert space projections, $$p=p^2=p^*$$ and $$q=q^2=q^*$$, is $$pq+qp$$ ever a non-zero Hilbert space projection?

• What did you try ? Oct 27, 2021 at 11:31
• If $q = 1 -p$ then $pq + qp$ is a projection Oct 27, 2021 at 11:33
• @InfiniteLooper as per the MO post, I added the important assumption that $pq+qp$ be non-zero. Oct 27, 2021 at 11:34
• If $q=1$, then $pq+qp=2p$ is not a projection !
– Fred
Oct 27, 2021 at 11:36
• @Fred thanks, the questions is if $pq+qp$ is ever a projection. Oct 27, 2021 at 11:37

No.

Suppose that $$p,q,r:=pq+qp$$ are all projections in a vector space, i.e. $$pp=p$$, $$qq=q$$, $$(pq+qp)^2 = pq+qp$$ (We do not need to require that they are orthogonal or that we are on a hilbert space!).

part 1: Let $$s$$, $$t$$ be arbitrary operators with $$sq = tp$$. Then we have \begin{align} sq = tp &\implies sq = tpp = sqp \\ &\implies sq = sqq = sqpq \\ &\implies sq (qp + pq) = sqp + sqpq = 2sq \\ &\implies 2sq = sq (qp+pq) = sq(qp+pq)^2 = 2sq (qp+pq) = 4sq \\ &\implies sq = 0. \end{align}

part 2: By assumption, we have $$(pq+qp)(pq+qp)=pq+qp$$, which is equivalent to $$((pq+qp)p -p)q = (q - (pq+qp)q)p$$. Applying part 1 yields $$((pq+qp)p - p)q=0$$. Then we have \begin{align} pq = (pq+qp) pq = pqpq + qpq &\implies qpq = pq - pqpq \\ &\implies qpq = pq - p (pq - pqpq) = pqpq \\ &\implies pq = (pq+qp) pq = pqpq + qpq = 2qpq \\ &\implies 2qpq = pq = 2q(2qpq ) = 4qpq \\ &\implies qpq = 0. \end{align} By exchanging $$p$$ and $$q$$ we also get $$pqp =0$$. In summary, we have $$pq+qp = (pq + qp)(pq+qp) = pqpq + pqp + qpq + qpqp = 0,$$ which is what we wanted to show.

(Note: A previous version of this answer was incorrect.

The above uses a large number of elementary operations and only relies on $$rr=r$$-type substitution, which makes me think that there is probably a simpler proof).

• Nice manipulations. Oct 29, 2021 at 13:50
• One question, possibly ridiculous: How can you guarantee that $p$ and $q$ are subprojections of $pq+qp$? Oct 29, 2021 at 22:44
• @JPMcCarthy I hope I find some time in the next days to think about a fix. Nov 2, 2021 at 11:28
• I think maybe using Halmos two projections theory it is possible to show that if $pq+qp$ is a projection then $\operatorname{ran}(pq+qp)\subset \operatorname{ran}p$ and by symmetry show that $\operatorname{ran}(pq+qp)\subset \operatorname{ran}q$ and then it should follow quickly... but too busy to check details fully right now. Nov 2, 2021 at 11:50
• @JPMcCarthy I think I found a different fix myself, but this time I should check it more carefully. Nov 2, 2021 at 11:53

While I like the generality and simplicity of the accepted answer, here is an approach that does use e.g. self-adjointness (not just that $$p$$ and $$q$$ are elements of a complex algebra) to say slightly more.

If $$p$$ and $$q$$ are self-adjoint projections on Hilbert space $$H$$ and $$r := pq + qp$$ is a positive operator with finite spectrum, then either $$r = 0$$ or $$r$$ is twice a nonzero projection. (The second conclusion holds if and only if $$pH \cap qH$$ is nonempty, when these hypotheses imply that $$p$$ and $$q$$ commute and $$r$$ is twice the projection onto $$pH \cap qH$$.) In particular, $$pq + qp$$ is never a nonzero projection.

Note that if $$r$$ is not assumed positive (and it is not clear to me how this assumption on $$r$$ might be reflected in a hypothesis that is more clearly about the "relative position" of $$p$$ and $$q$$), it is quite possible for $$r$$ to be something else; e.g. if $$p$$ and $$q$$ are the rank one projections onto the spans of $$(1,0)$$ and $$(1,1)$$ in $$\mathbb{C}^2$$, then $$r$$ has $$(1 - \sqrt{2})/2$$ as an eigenvalue and in particular is not positive.

I write $$\sigma(x)$$ for the spectrum of an operator $$x$$ on Hilbert space.

Sketch of proof. Letting $$x := p + q$$ we have $$r = x^2 - x$$. Our hypotheses on $$p$$ and $$q$$ imply that $$x$$ is positive with norm at most $$2$$, and that $$r$$ is self-adjoint with norm at most $$2$$, so that $$\sigma(x) \subseteq [0,2]$$ and $$\sigma(r) \subseteq [-2,2]$$. Using the hypothesis that $$r$$ is positive we deduce $$\sigma(r) \subseteq [0,2]$$, and the spectral mapping theorem for polynomials in self-adjoint operators (which tells us that $$t \mapsto t^2 - t$$ has to map $$\sigma(x) \subseteq [0,2]$$ into $$[0,2]$$) then implies that $$\sigma(x)$$ is a finite subset of $$\{0\} \cup [1,2]$$, which we will use shortly.

There are mutually orthogonal projections $$\{x_{\lambda}: \lambda \in \sigma(x)\}$$ such that $$x = \sum_{\lambda \in \sigma(x)} \lambda x_{\lambda} = \sum_{\lambda \in \sigma(x) \setminus\{0\}} \lambda x_{\lambda}$$.

Consider the element $$c = (p-q)^2$$, which a short calculation shows commutes with both $$p$$ and $$q$$. Our hypotheses on $$p$$ and $$q$$ imply that $$c = 2x - x^2$$, so in terms of the spectral resolution of $$x$$ we have $$c = \sum_{\lambda \in \sigma(x)} (2 \lambda - \lambda^2) x_{\lambda} = \sum_{\lambda \in \sigma(x) \setminus \{0,2\}} (2 \lambda - \lambda^2) x_{\lambda}$$. Because the mapping $$t \mapsto 2t - t^2$$ is one-to-one and nonzero on $$[1,2)$$, the numbers $$2 \lambda - \lambda^2$$ are distinct and nonzero for distinct elements of $$\sigma(x) \setminus \{0,2\}$$, and there is therefore a polynomial $$f$$ satisfying $$f(0) = 0$$ and $$p(2 \lambda - \lambda^2) = \lambda$$ for all $$\lambda \in \sigma(x) \setminus \{0,2\}$$. For this $$f$$ we have $$f(c) = \sum_{\lambda \in \sigma(x) \setminus \{0,2\}} \lambda x_{\lambda}$$, and because this element is in the subalgebra of $$\mathcal{B}(H)$$ generated by $$c$$, this element commutes with both $$p$$ and $$q$$.

If $$2 \not \in \sigma(x)$$ then the element $$f(c)$$ just constructed (which commutes with both $$p$$ and $$q$$) is none other than $$x$$. In this case, from $$xp=px$$ we deduce that $$pq=qp$$, and hence that $$pq$$ is a projection, so that $$r = 2pq$$ is twice a projection; but our hypothesis $$2 \not \in \sigma(x)$$ and the spectral mapping theorem together imply that $$2 \not \in \sigma(r)$$; thus $$\sigma(r) = \{0\}$$, and hence $$r = 0$$.

If $$2 \in \sigma(x)$$, then the element $$f(c)$$ constructed above (which commutes with $$p$$ and $$q$$) is $$x - 2 x_2$$, and we will be able to deduce that $$x$$ commutes with $$p$$ and $$q$$ if we can show that $$x_2$$ commutes with both $$p$$ and $$q$$. To see this, first note that $$x_2$$ coincides with the projection onto $$pH \cap qH$$ (proof: if $$y \in pH \cap qH$$, then $$(p+q)y = py + qy = 2y$$ so that $$y \in x_2$$, and conversely, if $$y$$ is a nonzero element of $$x_2 H$$, then equality holds in $$\|2y\| = \|py + qy\| \leq \|py\| + \|qy\|$$ so that $$py$$ and $$qy$$ are multiples of $$y$$, with both multiples being $$1$$ as $$p$$ and $$q$$ are projections, so that $$py=qy=y$$ and $$y \in pH \cap qH$$). And because the projection onto $$pH \cap qH$$ is given by either of the strong limits $$(qpq)^n$$ (which clearly commutes with $$q$$) or $$(pqp)^n$$ (which clearly commutes with $$p$$), as in e.g. this MSE answer, we deduce that $$x_2$$ commutes with $$p$$ and $$q$$. Thus $$x = f(c) + 2 x_2$$ also commutes with both $$p$$ and $$q$$, and as in the other case we deduce from $$xp=px$$ that $$p$$ and $$q$$ commute, so that $$qp$$ is a projection (in fact, the projection $$x_2$$) and $$r = 2 x_2$$. End of proof.

Remark: the "golden ratio" $$\phi$$ (which appears in the MO answer that is cited in this question) appears explicitly if we run through the above argument in the case that $$r$$ is assumed to be a projection. Indeed, the general appeals to spectral theory can be replaced with "bare hands" calculation from $$r = x^2 - x$$ and $$r^2 = r$$ that the positive operator $$x$$ must be a root of the polynomial $$t(t-1)(t-\phi)$$, and one can produce from this polynomial relation a specific quadratic polynomial $$f$$ such that $$x = f(c)$$ (in fact, $$x = f(c)$$ for $$f(t) = (3 - 2 \phi)^{-1} (t^2 + 2(1 - \phi) t)$$). So it is possible to prove that $$x$$ commutes with $$c$$ very explicitly in this case (e.g. the possibility that $$2 \in \sigma(x)$$, which required special treatment above, does not occur in this case). But at first glance, even in this specific case, my approach still uses spectral consequences of the sum $$x=p+q$$ of positive elements being positive. So maybe this argument would not apply in a general algebra (?), although it certainly works in any $$C^*$$-algebra.