Let we have an endomorphism $f:\mathbb{R^3\to\mathbb{R^3}}$ and the base $\beta=${$(-1,1,0),(1,0,0),(0,1,1)$}. Let's suppose that the associated matrix for that base is$$
M_{\beta}(f)=\begin{pmatrix}
1 & 1 & 0\\
0 & 1 & 0\\
0 & 0 &1
\end{pmatrix}
$$
If $f^*$ is the dual aplication of $f$, which is the associated matrix with $f^*$ for the canonic dual base? I mean, which is $ M_{{{\beta}_k}^*}(f^*)$?
First of all I don't know if I do this is ok:
$M_{{\beta_{k}}^*}(f^*)=(M_{\beta_{k}}(f))^t$ Is this true?
If that's ok I would calculate $M_{\beta_{k}}(f)=M_{\beta}^{\beta_{k}}M_{\beta}(f)M_{\beta_{k}}^{\beta}$