Prove that if $A=\frac{1}{2}\{B,A\}$ then $B=1\!\mathrm{l}$ 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? 
 A: 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 $A=B=E_{11}$.
A: 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!
