Let $T:\mathbb{C^3} \to \mathcal{M}_{2x2}$ and $T(x,y,z)=\begin{pmatrix} 3x+iy & y-iz\\ iy+z & 2x \end{pmatrix}$ with the inner product $<A, B>=tr(\bar{B}^t, A)$.
Find $T^*\begin{pmatrix} a & b\\ c & d \end{pmatrix}$
What I've been doing:
Using the definition for adjoint transformation:
$<(x, y, z), T^*\begin{pmatrix} a & b\\ c & d \end{pmatrix}>=<\begin{pmatrix} a & b\\ c & d \end{pmatrix}, T(x, y, z)>$
So using the given inner product and the standart basis:
$<\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 3x+iy & y-iz\\ iy+z & 2x \end{pmatrix}>=tr(\begin{pmatrix} 3x+iy & iy+z\\ y-iz & 2x \end{pmatrix}\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix})=3x+iy=(3, i, 0)$
$<\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 3x+iy & y-iz\\ iy+z & 2x \end{pmatrix}>=tr(\begin{pmatrix} 3x+iy & iy+z\\ y-iz & 2x \end{pmatrix}\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix})=y-iz=(0, 1, -i)$
$<\begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 3x+iy & y-iz\\ iy+z & 2x \end{pmatrix}>=tr(\begin{pmatrix} 3x+iy & iy+z\\ y-iz & 2x \end{pmatrix}\begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix})=iy+z=(0,i,1)$
$<\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 3x+iy & y-iz\\ iy+z & 2x \end{pmatrix}>=tr(\begin{pmatrix} 3x+iy & iy+z\\ y-iz & 2x \end{pmatrix}\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix})=2x=(2,0,0)$
And then I found $T^*\begin{pmatrix} a & b\\ c & d \end{pmatrix}$:
$T^*\begin{pmatrix} a & b\\ c & d \end{pmatrix}=a(3,i,0)+b(0,1,-i)+c(0,i,1)+d(2,0,0)=(3a+2d, ia+b+ic, -ib+c)$
The given solution was $T^*\begin{pmatrix} a & b\\ c & d \end{pmatrix}=(3a+2d, -ia+b-ic, ib+c)$