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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!

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  • $\begingroup$ $\displaystyle\large \left[C,A\right]^{\dagger} =\left[A^{\dagger},C^{\dagger}\right]$ $\endgroup$ – Felix Marin Mar 17 '14 at 20:24
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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 $f_C^* = f_{C^T}$.

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  • $\begingroup$ Perfect! I tried something like that, but I forgot I could move the things inside one of the traces cyclically, so I couldnt quite make it work! Thank you! $\endgroup$ – Daniel Montealegre Aug 29 '12 at 4:53

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