I am trying to find the dual of the matrix Lie algebra $\mathfrak{g}$ with basis $$X=\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right),Y=\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0 & 0 & 1 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{array}\right), Z=\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right)$$ with resepct to the bilinear form $<A,B>=trace(AB)$. We can show that the dual to the Lie algerba $\mathfrak{g}^*$ is isomorphic to $M_3/\mathfrak{g}^{\perp}$, where $\mathfrak{g}^{\perp}=\{A \in M_n: trace(A\xi)=0, \forall \xi \in \mathfrak{g}\}$.
My question is how can I practically calculate this?
My attempt:
Let $\xi=xX+yY+zZ \in \mathfrak{g}$ and $M \in M_n$ then $$\xi=\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0 & z & x + y \\ 0 & 0 & -y + z \\ 0 & 0 & z \end{array}\right), M=\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ m_{31} & m_{32} & m_{33} \end{array}\right)$$ and $$trace(M\xi)=\newcommand{\Bold}[1]{\mathbf{#1}}m_{31} {\left(x + y\right)} - m_{32} {\left(y - z\right)} + m_{21} z + m_{33} z$$
Then we can show that $m_{31}=m_{32}=0, m_{21}=-m_{33}$
but I am not sure what to do next.