Is the Adjoint of a Commutator Negative of the Commutator?

[stumpoman]

Homework Statement

Show that
\left [ A,B \right ]^{\dagger}=-\left [A,B \right ]

Homework Equations

\left [ A,B \right ] = AB-BA\left (AB \right)^{\dagger}= B^{\dagger}A^{\dagger}

The Attempt at a Solution

\left [ A,B \right ]^{\dagger}=\left (AB-BA \right )^{\dagger}=\left (AB \right )^{\dagger}-\left (BA \right )^{\dagger}=B^{\dagger}A^{\dagger}-A^{\dagger}B^{\dagger}=-\left (A^{\dagger}B^{\dagger}-B^{\dagger}A^{\dagger} \right )=-\left [ A^{\dagger},B^{\dagger} \right ]
I can only see this working if the operators are Hermitian but the question did not specify it as such.

 

[andrewkirk]

A quick way to solve the problem of being given a potentially incorrectly specified question is to look for a counterexample. Keep it as simple as possible. In this case, you can choose two 2 x 2 real matrices that are non-symmetric - hence non-Hermitian. Then do the calc and see if the result holds.

I used the following R code to test this (the 't' function performs a transpose and %*% does matrix mult)

A=array(c(1,0,1,2),dim=c(2,2))
A=array(c(2,1,0,1),dim=c(2,2))
t(A%*%B-B%*%A)
-(A%*%B-B%*%A)
t(A%*%B-B%*%A)-(-(A%*%B-B%*%A))