= Solution
Use the basis of <1/b/solution> for the two doublets, treating the barred left-handed fields as the components to which the printed <representation> assignment applies. Write $x=(\bar\psi_L^1,\bar\psi_L^2)^T$ and $y=(\psi_R^1,\psi_R^2)^T$. The doublet contribution to the <Yukawa interaction> is $H x^T L y$, so <invariant tensors> obey
$$
D_1(Q)^TLD_1(Q)=L,\qquad P^TLP=L.
$$
The first equation forces $L_{11}=L_{22}=0$, since their factors are $\alpha^{\pm2}\ne1$. The second sets $L_{12}=L_{21}$. Thus the doublet <invariant tensor> is $x_1y_2+x_2y_1$. The product of either doublet with $R_1$ has no <trivial representation>, whereas $R_1\otimes R_1=R_0$. Therefore
$$
\boxed{\lambda^{\mathrm{(i)}}=
\begin{pmatrix}0&y&0\\y&0&0\\0&0&y_3\end{pmatrix},\qquad y,y_3\in\mathbb C.}
$$
If instead the unbarred left-handed doublet is chosen to transform by $D_1$, its barred components transform by $D_1^*$; in that equivalent basis the upper block is $yI_2$. Exchanging the two left components relates these conventions. The <singular value decomposition> gives the same physical masses in either basis.
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