Given is the $\mathbf{R}$-linear map $L:\mathbf{R}^2\longrightarrow\mathbf{C}^2:(a,b)\longmapsto(a+bi,a-bi)$. I am asked to find (a basis for) the subspace of $(\mathbf{C}^2)^\star$ which is mapped to $0$ under the pullback by $L$.
Since $\dim_{\mathbf{R}}(\mathbf{C}^2)=4$, we know that the dimension of its dual space is also 4. I defined the linear functionals $\psi_1,\ldots,\psi_4$ by $\psi_1(u,v)\mapsto \operatorname{Re}u$, $\psi_2(u,v)\mapsto \operatorname{Re}v$, $\psi_3(u,v)\mapsto \operatorname{Im}u$, $\psi_4(u,v)\mapsto \operatorname{Im}v$ and proved that these are indeed linearly independent and therefore form a basis for $(\mathbf{C}^2)^\star$.
Now we have that $L^\star (\lambda_1 \psi_1+\cdots+\lambda_4 \psi_4)(a,b)=(\lambda_1+\lambda_2)a+(\lambda_3-\lambda_4)b$. This gives the conditions $\lambda_1=-\lambda_2$ and $\lambda_3=\lambda_4$. My question is: how do we deduce from this a basis for the asked subspace?
Just note a linear form in the kernel of $L^*$ can be written as $$\psi=\lambda_1(\psi_1-\psi_2)+\lambda_3(\psi_3+\psi_4),$$ so a basis is $$\bigl\{\psi_1-\psi_2,\,\psi_3+\psi_4\bigr\}=\bigl\{\operatorname{Re} u-\operatorname{Re} v,\,\operatorname{Im} u+\operatorname{Im} v\bigr\}.$$