Denote orthonormal basis in $\mathbb{R}^2$: $(\epsilon_1,\epsilon_2)=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $A=\begin{pmatrix}1&0&\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}\\0&1&\frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2}\end{pmatrix}$.
Now let frames $(e_1,e_2,e_3,e_4)=(\epsilon_1,\epsilon_2)A$ and its dual $(\tilde{e}_1,\tilde{e}_2,\tilde{e}_3,\tilde{e}_4)=(\epsilon_1,\epsilon_2)B$. For $x=(1,1)^T$, one can decompose and reconstruct $x$ through $\{e_j\}$ and $\{\tilde{e}_j\}$: $x=\sum_{j}\langle x,\tilde{e}_j\rangle e_j=AB^Tx=\sum_j\alpha_j e_j$ The problem is to determine $B$ such that: 1) $\|\alpha\|_2$ is minimized; 2) $\alpha$ has the least non-zero entries.
The first one can be solved by Moore–Penrose pseudoinverse, but I have no idea about P2. Intuitively it appears to be some analogues of Moore–Penrose pseudoinverse but minimizing $L^1$ norm. Can anyone give hints? Thank you.