So, manipulating some (square) matrices, $A$ and $B$, I encountered an equation of the form:

$$A=\tfrac{1}{2}\{A,B\}$$ where "$\{\cdot,\cdot\}$" denotes the anticommutator between matrices $A$ and $B$: $\{A,B\}\equiv AB+BA$. In order to solve the equation for $B$, I want to show that $B$ must be equal to the identity (matrix): $$B=1\!\mathrm{l}\equiv \mathrm{diag}(1,1,\ldots,1).$$

Is it always true? I tried to look for a proof for the previous statement but I didn't succeed... In fact, I only tried to multiply the equation, on the left and/or on the right, by something like $A^{-1}, B^{-1},(AB)^{-1},\ldots$.

Is my statement right anyway? Does anyone have any suggestion on what to do with this equation?

Thank you all very much in advance!

EDIT: as I know that $B=1\!\mathrm{l}$ is not the unique solution to the equation $A=\frac{1}{2}\{B,A\}$, how can I find all the solutions for arbitrary dimensions matrices?

In my answer I provided a condition for the solution with $2$-by-$2$ matrices. There exists a general solution for higher conventionalities matrices?

  • 1
    $\begingroup$ $2A=AB+BA$ may be rearranged as $A(2I-B)=BA$. Hence if $A$ is invertible, then $det(2I-B)=det(B)$. Note: $A=0$ works for any $B$, so certainly $B=I$ is not necessarily true. $\endgroup$
    – vadim123
    Mar 18 '14 at 14:13
  • $\begingroup$ Thanks for your help, @vadim123! Anyway, now that I have the equation $\det (2I-B)=\det (B)$, how can I solve for $B$? Does actually a solution exist? $\endgroup$ Mar 18 '14 at 14:20
  • $\begingroup$ I know that $\det(I+B)$ equals $1+\det(B)$ plus many other traces of $B$ and its powers... Actually, the results is increasingly complicated as the dimension of the matrix $B$ increases! $\endgroup$ Mar 18 '14 at 14:24
  • $\begingroup$ Let us suppose that $A$ is invertible and exclude the trivial case where $A=0$. $\endgroup$ Mar 18 '14 at 14:35

The statement is ture if you mean $A=\frac12(AB+BA)$ for all $A$ (proof: consider, for each $i$, the matrix $A=E_{ii}$ whose only nonzero entry is a $1$ at the $(i,i)$-th position). If you mean $A=\frac12(AB+BA)$ for some $A$, the statement is not true except in the $1\times1$ case. For a counterexample, consider $A=B=E_{11}$.

  • $\begingroup$ Hi @user1551! Thank you for your contribution! I got to the same conclusions! Anyway, I can't understand why $A=B=E_{11}$ isn't a solution. Indeed, $E_{11}E_{11}=E_{11}$, hence $E_{11}=\frac{1}{2}(E_{11}+E_{11})$...! $\endgroup$ Mar 18 '14 at 15:50
  • 1
    $\begingroup$ @AstoundingJB Of course $B=E_{11}$ is a solution, but $E_{11}\ne I$ (unless in the 1x1 case). So, it is a counterexample to the statement that $B$ must be equal to the identity matrix. $\endgroup$
    – user1551
    Mar 18 '14 at 15:51
  • $\begingroup$ Oops! My apologies, I misunderstood your statement! You're absolutely right, $B=E_{11}$ is a solution when $A=E_{11}$ and it is a counterexample to my previous statement: $B=I$! Thank you again! $\endgroup$ Mar 18 '14 at 15:55

I propose the edit I added to my question as a possible answer. If you have any suggestion to improve this answer, please leave me a comment. Thank you all!

Following the hint given by vadim123, we can try to solve the equation by means of the determinants. First of all, we can rewrite the equation as:

$$2A = AB+BA\\ A(2\,1\!\mathrm{l}-B)=BA.$$ Than, taking the determinant of both sides: $$\det(A)\det(2\,1\!\mathrm{l}-B)=\det(A)\det(B)$$ which simplifies into $$\det(2\,1\!\mathrm{l}-B)=\det(B),$$ provided $A$ is invertible. Now, taking a look at the matrix cookbook, equations ($25-27$), we can find that, for $n=2$, that is for $A,B$ $2$-by-$2$ matrices, we have: $$\det(2\,1\!\mathrm{l}-B)\stackrel{(25)}{=}2^2-\det(B)-2\mathrm{Tr}(B)=\det(B),$$

so, for every $2$-by-$2$ matrices, the solutions of the equation $A=\tfrac{1}{2}\{A,B\}$ must satisfy the relation: $$2-\mathrm{Tr}(B)=\det(B).$$ Of course, $B=1\!\mathrm{l}$ is a solution. Anyway, the general solution is: $$B={b_{11}\ b_{12}\choose b_{21}\ b_{22}}:\quad 2-b_{11}-b_{22}=b_{11}b_{22}-b_{21}b_{12}.$$ For example, $$B={0\ -1\choose 2\quad 0}$$ is a solution different from the identity. The case for higher dimensions matrices is more complicated, but I guess that similar results to the case $n=2$ will hold; hence, there always exist solutions for $A=\frac{1}{2}\{B,A\}$ different from the identity!


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