We are in $M_{n\times n}(\mathbb{R})$. Define $<A,B>=tr(AB^t)$. First I had to prove that this was indeed an inner product and that part is easy.
The second question is, given a fixed matrix $C$, define $f_C(A)=CA-AC$. Find the adjoint of this operator.
I was trying something along the lines of finding $M_C$, the matrix representation of $f_C$, which would be an $n^2\times n^2$ matrix, and then taking the conjugate transpose of this matrix, and obtaining $M^*_C$ which would be the matrix representation for $f_C$, but this route seems a bit too messy.
I also tried playing with the definition of adjoint, that is, the unique linear operator $f^*_C$ such that $<f_C(A),B>=<A,f^*_C(B)>$ for all $A,B$ but this was not fruitful either.
Does anyone know how to tackle it?
Thanks!
Best Answer
You are thinking too hard.
$\langle Y, f_C(X) \rangle = \mathbb{tr}( Y (CX-XC)^T) = \mathbb{tr}( Y X^TC^T) - \mathbb{tr}( Y C^T X^T)$, using linearity. Using the cyclic property of the trace, we have $\mathbb{tr}( Y X^TC^T) = \mathbb{tr}( C^TY X^T)$, from which we get $\langle Y, f_C(X) \rangle = \mathbb{tr}( (C^TY-Y C^T) X^T) = \langle f_{C^T}(Y),X \rangle = \langle f_{C}^*(Y),X \rangle$.
Hence $f_C^* = f_{C^T}$.