Q:

How to calculate dual frames under constraints? 

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. 

A:

OK I think I've got it. 
By SVD, $$A=U\Sigma V^T=
\begin{pmatrix}0&1\\1&0\end{pmatrix}
\cdot
\begin{pmatrix}\sqrt{2}&0&0&0\\0&\sqrt{2}&0&0\end{pmatrix}
\cdot
\begin{pmatrix}0 & \frac{1}{\sqrt{2}} & \frac{1}{2} & -\frac{1}{2} \\
 \frac{1}{\sqrt{2}} & 0 & \frac{1}{2} & \frac{1}{2} \\
 -\frac{1}{2} & \frac{1}{2} & 0 & \frac{1}{\sqrt{2}} \\
 -\frac{1}{2} & -\frac{1}{2} & \frac{1}{\sqrt{2}} & 0\end{pmatrix}$$
Let $\beta=V^T\alpha$, then
$$\begin{pmatrix}\sqrt{2}&0&0&0\\0&\sqrt{2}&0&0\end{pmatrix}
\cdot\begin{pmatrix}\beta_1\\\beta_2\\\beta_3\\\beta_4\end{pmatrix}
=\begin{pmatrix}1\\1\end{pmatrix}\qquad\cdots\cdots(*)$$
Therefore $\beta_1=\beta_2=\frac{1}{\sqrt{2}}$. Further,
$$\alpha=V\beta=\left(\frac{1}{2}-\frac{\beta _3}{2}-\frac{\beta _4}{2},\frac{1}{2}+\frac{\beta _3}{2}-\frac{\beta _4}{2},\frac{1}{\sqrt{2}}+\frac{\beta _4}{\sqrt{2}},\frac{\beta _3}{\sqrt{2}}\right)^T$$
To achieve sparsity, it must hold that $\beta_3=0$ and $\beta_4=1$, yielding $\alpha=(0,0,\frac{1}{\sqrt{2}},0)^T$
Now from $(*)$ and $AB^T=I$, we have
$$\begin{pmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\\1\end{pmatrix}
=\begin{pmatrix}\frac{1}{\sqrt{2}}&0\\0&\frac{1}{\sqrt{2}}\\0&0\\0&1\end{pmatrix}
\cdot
\begin{pmatrix}1\\1\end{pmatrix}=\tilde{B}^T\begin{pmatrix}1\\1\end{pmatrix}$$
The entries in lower part of $\tilde{B}^T$ are chosen out of simplicity. 
Finally, $$B=\left(V\tilde{B}^TU^T\right)^T=\begin{pmatrix}
 0 & -\frac{1}{2} & \frac{3}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} \\
 0 & \frac{1}{2} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}}
\end{pmatrix}$$

