Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is possible to have two hermitian operators $A$ et $B$, with :

$B^2 = \mathbb{I}d$

$[A,B] = i * \mathbb{I}d$

where $i$ is the usual (complex) square root of $(-1)$, and $\mathbb{I}d$ is the identity operator ?

(I think that A is necessarily not bounded, due to the last condition)

If it is possible, may we exhibit a explicit representation of these operators $A$ and $B$ ?

share|cite|improve this question
You are correct that $A$ would have to be unbounded because $B$ is bounded (e.g., see here). But I don't know the answer. – Jonas Meyer Jan 5 '12 at 11:04
For what it's worth, replacing $B$ with $\frac{B+1}{2}$ and $A$ with $2A$, the problem is the same if we ask whether we can have $B^2=B$, i.e., $B$ is an orthogonal projection. – Jonas Meyer Jan 5 '12 at 11:24
@JonasMeyer I did not follow you, if I take $B = (B'+1)/2$, it gives $B'² = 3 \mathbb{I}d - 2 B'$ and not $B'² = B'$ ??? – Trimok Jan 5 '12 at 11:48
@Trimok: he meant the opposite way around: $[(B+1)/2]^2 = (B^2 + 1)/4 + 2B / 4 = (B+1)/2$. And $[2A,(B+1)/2] = [A,B] = i$. So you can ask the equivalent question of $C^2 = C$ and $[D,C] = i$. – Willie Wong Jan 5 '12 at 11:56
BTW, in view of that first comment by @Jonas perhaps you can drop the (linear-algebra) and (matrices) tags and replace by (functional-analysis)? – Willie Wong Jan 5 '12 at 12:01
up vote 4 down vote accepted

The commutator $[A,B]$ is proportional to $\mathbb{I}$ and therefore commutes with everything. So $$A-BAB=AB^2-BAB=[A,B]B=B[A,B]=BAB-B^2A=BAB-A$$ or $$A=BAB.$$ But then $$[A,B]=AB-BA=(BAB)B-BA=BA-BA=0.$$ Btw. from this it follows that the result generalizes to all commutators with $$[A,B]=f(B).$$

share|cite|improve this answer
OK, nice, +1 for you – Trimok Jan 5 '12 at 13:02
@Trimok: Also, I suggest that you change the name of the question to "An idempotent operator problem". There is not so much hermitian about it, except that the commutator smells like quantum mechanics. – NikolajK Jan 5 '12 at 13:20
Sorry, but what is the procedure to accept an answer? – Trimok Jan 5 '12 at 18:05
OK, found...... – Trimok Jan 5 '12 at 18:18
Was your question motivated by a Putnam or IMO problem? – NikolajK Jan 5 '12 at 18:21

This is impossible. In fact, I will prove that if $A,B$ are operators in a real or complex algebra such that $B$ has finite multiplicative order ( that is, $B^{k} = I$ for some positive integer $k),$ then we can't have $AB -BA = cI$ for any non-zero scalar $c.$ For otherwise, an induction argument (due to Wielandt, and to be found in the link given in Jonas's comment) and which does not require any boundedness assumption, shows that we have $A(c^{-1}B)^{n} - (c^{-1}B)^{n}A = n(c^{-1}B)^{n-1}$ for every positive integer $n$. Taking $n =k$ yields a contradiction, since then the left side is $0,$ but the right side is non-zero.

An alternative approach is to note that if $AB - BA = I,$ then $B$ is not an algebraic element (when $B$ is Hermitian, this implies that the spectrum of $B$ must be infinite). By an algebraic operator, we mean an operator $T$ such that $f(T) =0$ for some monic polynomial $f(z) \in \mathbb{C}[z].$ Note that an algebraic operator has a unique minimum polyomial, that is a unique monic polynomial $p(z) \in \mathbb{C}[z]$ of least degree subject to $p(T) =0.$ For suppose that $B$ is algebraic, and let $q(z)$ be the minimum polynomial for $B,$ say of degree $r.$ Then $r >1$ since $AB \neq BA.$ Then we have $0 = Aq(B) - q(B)A = (r-1)B^{r-1} +$ (some linear combination of lower powers of $B$), so that $h(B) =0$ for some monic monic polynomial $h(z) \in \mathbb{C}[z]$ of degree $r-1,$ contrary to the fact that the minimum polynomial of $B$ has degree $r.$

share|cite|improve this answer
But A is not bounded, so the demonstration you made here does not work ??? – Trimok Jan 5 '12 at 12:51
It does work, but I have rewritten it in more generality in any case. – Geoff Robinson Jan 5 '12 at 17:02
Understood..., there is a clear contradiction for n=1 or n= 2, for instance. – Trimok Jan 5 '12 at 18:38
Sorry, but it seems impossible to put the flag "answer to the question" to several persons, which seems not very clever... – Trimok Jan 5 '12 at 18:46
Maybe it's an other question, but do you think the conclusion will be the same if B was bounded (but not a root of unity). – Trimok Jan 5 '12 at 19:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.