Fusion of sparse reconstruction algorithms for multiple measurement vectors

Abstract

We consider the recovery of sparse signals that share a common support from multiple measurement vectors. The performance of several algorithms developed for this task depends on parameters like dimension of the sparse signal, dimension of measurement vector, sparsity level, measurement noise. We propose a fusion framework, where several multiple measurement vector reconstruction algorithms participate and the final signal estimate is obtained by combining the signal estimates of the participating algorithms. We present the conditions for achieving a better reconstruction performance than the participating algorithms. Numerical simulations demonstrate that the proposed fusion algorithm often performs better than the participating algorithms .

Appendices

Appendix

Proof of lemma 1

The proof is inspired by proposition 3.5 by Needell and Tropp [16].

Define set S as the convex combination of all matrices, which are $R+K$ sparse and have unit Frobenius norm.

$$\begin{aligned} S=\hbox {conv}\left\{ {\mathbf {X}}: \left\| {\mathbf {X}}\right\| _0 \le R+K, \left\| {\mathbf {X}}\right\| _F=1 \right\} . \end{aligned}$$

Using the relation $\left\| {\mathbf {AX}}\right\| _F \le \sqrt{1+\delta _{R+K}} \left\| {\mathbf {X}}\right\| _F$, we get

$$\begin{aligned} {\mathop {{{\mathrm{arg\,max}}}}\limits _{{\mathbf {X}} \in S}} \left\| {\mathbf {AX}}\right\| _F \le&\sqrt{1+\delta _{R+K}}. \end{aligned}$$

Define

$$\begin{aligned} Q=\left\{ {\mathbf {X}}: \left\| {\mathbf {X}}\right\| _F + \frac{1}{\sqrt{R+K}} \left\| {\mathbf {X}}\right\| _{2,1} \le 1 \right\} . \end{aligned}$$

The lemma essentially claims that

$$\begin{aligned} {\mathop {{{\mathrm{arg\,max}}}}\limits _{{{\mathbf {X}} \in Q}}} \left\| {\mathbf {AX}}\right\| _F \le {\mathop {{{\mathrm{arg\,max}}}}\limits _{{{\mathbf {X}} \in S}}} \left\| {\mathbf {AX}}\right\| _F. \end{aligned}$$

To prove this, it is sufficient to ensure that $Q \subset S$.

Consider a matrix ${\mathbf {X}} \in Q$. Partition the support of ${\mathbf {X}}$ into sets of size $R+K$. Let set ${\mathcal {I}}_0$ contain the indices of the $R+K$ rows of ${\mathbf {X}}$, which have largest row $\ell _2$-norm, breaking ties lexicographically. Let set ${\mathcal {I}}_1$ contain the indices of the next largest (row $\ell _2$-norm) $R+K$ rows and so on. The final block ${\mathcal {I}}_J$ may have lesser than $R+K$ components. This partition gives rise to the following decomposition:

$$\begin{aligned} {\mathbf {X}}={\mathbf {X}}|_{I_0}+\sum _{j=1}^{J}{\mathbf {X}}|_{I_j} = \lambda _0 {\mathbf {Y}}_0 + \sum _{j=1}^{J} \lambda _j {\mathbf {Y}}_j, \end{aligned}$$

where $\displaystyle \lambda _j =\left\| {\mathbf {X}}|_{I_j}\right\| _F \quad \text {and} \quad {\mathbf {Y}}_j= \lambda _j^{-1} {\mathbf {X}}|_{I_j}.$

By construction, each matrix ${\mathbf {Y}}_j$ belongs to S because it is $R+K$ sparse and has unit Frobenius norm. We will show that $\sum _{j} \lambda _j \le 1$. This implies that ${\mathbf {X}}$ can be written as a convex combination of matrices from the set S. As a result ${\mathbf {X}} \in S$. Therefore, $Q\subset S$.

Fix some j in the range $\{1,2, \ldots , J\}$. Then, ${\mathcal {I}}_j$ contains at most $R+K$ elements and ${\mathcal {I}}_{j-1}$ contains exactly $R+K$ elements. Therefore,

$$\begin{aligned} \lambda _j = \left\| {\mathbf {X}}|_{I_j}\right\| _F \le \sqrt{R+K} \left\| {\mathbf {X}}|_{I_j}\right\| _{2,\infty } \le \sqrt{R+K} \cdot \frac{1}{R+K} \left\| {\mathbf {X}}|_{I_{j-1}}\right\| _{2,1}. \end{aligned}$$

The last inequality holds because the row $\ell _2$-norm of ${\mathbf {X}}$ on the set $I_{j-1}$ dominates its largest row $\ell _2$-norm in ${\mathcal {I}}_j$. Summing these relations, we get

$$\begin{aligned} \sum _{j=1}^{J} \lambda _j \le \frac{1}{\sqrt{R+K}} \sum _{j=1}^J \left\| {\mathbf {X}}|_{I_{j-1}}\right\| _{2,1} \le \frac{1}{\sqrt{R+K}} \left\| {\mathbf {X}}\right\| _{2,1}. \end{aligned}$$

Also, we have $\lambda _0 = \left\| {\mathbf {X}}|_{I_0}\right\| _F \le \left\| {\mathbf {X}}\right\| _F$. Since ${\mathbf {X}} \in Q$, we conclude that

$$\begin{aligned} \sum _{j=0}^{J} \lambda _j \le \left[ \left\| {\mathbf {X}}\right\| _F + \frac{1}{\sqrt{R+K}} \left\| {\mathbf {X}}\right\| _{2,1}\right] \le 1. \end{aligned}$$

$\square$

Proof of lemma 2

Let $\mathbf {y}^{(i)}$ denote the $i\text {th}$ column of matrix ${\mathbf {Y}}$ and $\mathbf {r}^{(i)}$ denote the $i\text {th}$ column of matrix ${\mathbf {R}}$, $i=1,2,\ldots L$. Then, we have from lemma 2 of [22]

$$\begin{aligned} \left( 1-\frac{\delta _{|{\mathcal {T}}_1|+|{\mathcal {T}}_2|}}{1-\delta _{|{\mathcal {T}}_1|+|{\mathcal {T}}_2|}} \right) \left\| {\mathbf {y}}^{(i)}\right\| _2&\le \left\| \mathbf {r}^{(i)}\right\| _2 \le \left\| {\mathbf {y}}^{(i)}\right\| _2. \end{aligned}$$

Summing the above relation, we obtain

$$\begin{aligned} \left( 1-\frac{\delta _{|{\mathcal {T}}_1|+|{\mathcal {T}}_2|}}{1-\delta _{|{\mathcal {T}}_1|+|{\mathcal {T}}_2|}} \right) \sum _{i=1}^{L}\left\| {\mathbf {y}}^{(i)}\right\| _2^2&\le \sum _{i=1}^L\left\| \mathbf {r}^{(i)}\right\| _2^2 \le \sum _{i=1}^L\left\| {\mathbf {y}}^{(i)}\right\| _2^2. \end{aligned}$$

Equivalently, we have

$$\begin{aligned} \left( 1-\frac{\delta _{|{\mathcal {T}}_1|+|{\mathcal {T}}_2|}}{1-\delta _{|{\mathcal {T}}_1|+|{\mathcal {T}}_2|}} \right) \left\| {\mathbf {Y}}\right\| _F&\le \left\| {\mathbf {R}}\right\| _F \le \left\| {\mathbf {Y}}\right\| _F. \end{aligned}$$

$\square$