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?

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

2 Answers 2



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).

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    $\begingroup$ Nice manipulations. $\endgroup$
    – TheSimpliFire
    Oct 29, 2021 at 13:50
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    $\begingroup$ One question, possibly ridiculous: How can you guarantee that $p$ and $q$ are subprojections of $pq+qp$? $\endgroup$
    – Aweygan
    Oct 29, 2021 at 22:44
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    $\begingroup$ @JPMcCarthy I hope I find some time in the next days to think about a fix. $\endgroup$
    – supinf
    Nov 2, 2021 at 11:28
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    $\begingroup$ 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. $\endgroup$ Nov 2, 2021 at 11:50
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    $\begingroup$ @JPMcCarthy I think I found a different fix myself, but this time I should check it more carefully. $\endgroup$
    – supinf
    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.


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