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Let $A\in \mathbb{R}^{m\times n}$ be a matrix and let us denote by $A_S$ the submatrix of $A$ with the columns restricted to a set $S\subset [n]:=\{1,2,\cdots, \ n\}$. Then one says that the matrix $A$ has restricted isometry property of order $K$ with constant $\delta_K$ if the matrix $A$ has the property $$\delta_K=\max_{S\subset[n]:|S|\le K}\|A_S^TA_S-I\|_{2\to 2}$$

There is a very nice Cauchy-Scwartz like inequality concerning this constant which can be stated as below

If $x,y\in \mathbb{R}^n$ with $supp(x)\subset S_1,\ supp(y)\subset S_2,\ S_1\cap S_2=\emptyset$, then, $$|\langle Ax,Ay\rangle |< \delta_{|S_1|+|S_2|} \|x\|_2\|y\|_2$$

Emmanuel Candes proved this inequality here. The proof is rather simple.

I am trying to find a similar inequality, for the quantity in the LHS, but now a lower bound and also with the constraint $S_1\cap S_2\ne \emptyset$. I do not know if it is trivially $0$, but I hope it is not. I have still not succeeded to find it. I have searched for such an inequality in the literature but failed to get one.

So it will be really helpful if anyone can direct me either towards some relevant literature or giving some helpful tips to come up with a non-trivial lower bound. Thanks in advance.

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  • $\begingroup$ What is the restricted isometry property? You have defined $\delta_K$ but not given the condition on it for $A$ to have the property you are talking about. $\endgroup$ – Jason Jan 31 '15 at 6:54
  • $\begingroup$ Any $A$ should have a $\delta_K$ since we are searching over a finite set. $\endgroup$ – Samrat Mukhopadhyay Jan 31 '15 at 14:09
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    $\begingroup$ Right, but what does it mean for a matrix $A$ to have the restricted isometry property? $\endgroup$ – Jason Feb 1 '15 at 0:59
  • $\begingroup$ If $m<<n$, a matrix $A\in \mathbb{R}^{m\times n}$ to have a RIC $\delta_K$ means $A$ is approximately an isometry with an error of $\delta_K\|x\|_2$ $\endgroup$ – Samrat Mukhopadhyay Feb 1 '15 at 8:14
  • $\begingroup$ Maybe you mean supp $x$ $\cap$ supp $y$ $\neq \emptyset$ as a constraint $\endgroup$ – user66081 Feb 3 '15 at 1:48
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I don't know if this is sufficiently interesting, but it may be the kind of thing that you are driving at when you say that the supports should overlap.

For $x, y$ write $\Delta(x, y) := \delta_{\#(\text{supp} x \cup \text{supp} y)} \|x\|_2 \|y\|_2$.

Take $x, y$. Let $\bar{x}, \bar{y}$ denote their restrictions to the common support $S_c$, set $\hat{x} := x - \bar{x}$ and $\hat{y} := y - \bar{y}$.

Then $$ \langle A x, A y \rangle = \langle A \bar{x}, A \bar{y} \rangle + \langle A \hat{x}, A \hat{y} \rangle + \langle A \bar{x}, A \hat{y} \rangle + \langle A \hat{x}, A \bar{y} \rangle \geq \langle A \bar{x}, A \bar{y} \rangle - \Delta_3, $$ where $\Delta_3(x, y) := \Delta(\hat{x}, \hat{y}) + \Delta(\bar{x}, \hat{y}) + \Delta(\hat{x}, \bar{y})$; note that those 3 pairs have disjoint support.

Using $\|A_{S_c}^T A_{S_c} - I\| \leq \delta_{|S_c|}$ then $$ \langle A x, A y \rangle \geq \langle \bar{x}, \bar{y} \rangle - \Delta_4(x,y), $$ where $\Delta_4(x,y) := \Delta_3(x,y) + \Delta(\bar{x}, \bar{y})$.

It is clear that you need something with $\langle \bar{x}, \bar{y} \rangle$, because for the 2x2 identity matrix and $x = (1, -1)$, $y = (1, 1)$, the quantity you are interested in is zero, despite nontrivial common support.

Addendum:

If you start with $\text{supp} x = \text{supp} y$ then the following can be said (this, however, has nothing to do with compressed sensing). Let us suppose $S_c = [n]$ for ease of notation, otherwise everything has to be restricted to $S_c$. Since $A^T A$ is a symmetric positive (possibly semi-)definite matrix, there is an orthonormal basis of its eigenvectors for $\mathbb{R}^n$, say $v_1, \ldots, v_n$ with corresponding eigenvalues $\lambda_i \geq 0$. Expand $x = \sum_i x_i v_i$ and $y = \sum_j y_j v_j$. Then $$ \langle A x, A y \rangle = x^T A^T A y = \sum_{i,j} x_i y_j v_i^T A^T A v_j = \sum_{i,j} x_i y_j \lambda_j v_i^T v_j = \sum_{i,j} x_i y_j \lambda_j v_i^T v_j \stackrel{{!}}{\color{red}\geq} (\min_k \lambda_k) \sum_{i,j} x_i y_j v_i^T v_j = m \langle x, y \rangle $$ with $m := \min_k \lambda_k$. The marked inequality is not valid in general, as the signs of the summands can be different (OP pointed this out). But on the other hand such a bound cannot hold in general: take $A = \text{diag}(1.2, 1)$ and $x = (1 / \sqrt{1.2}, 1)$, $y = (-1 / \sqrt{1.2}, 1)$. Then $\langle A x, A y \rangle = 0$ while $\langle x, y \rangle > 0$.

From the inequality $\|A_{S_c}^T A_{S_c} - I\| \leq \delta_{|S_c|}$ we get $|\lambda_k - 1| \leq \delta_{|S_c|}$ for each $k$, as can be checked by assuming the contrary. Therefore $m \geq 1 - \delta_{|S_c|}$. This is positive if $\delta_{|S_c|} < 1$, which usually the case in the context of the RIP.

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  • $\begingroup$ How do you show that $\|A_{S_c}^TA_{S_c}-I\|\le \delta_{S_c}\implies \langle A\bar{x},A\bar{y}\rangle \ge (1-\delta_{S_c})\langle \bar{x},\bar{y}\rangle $? $\endgroup$ – Samrat Mukhopadhyay Feb 3 '15 at 6:35
  • $\begingroup$ @SamratMukhopadhyay: $\langle Ax, Ay \rangle = \langle A^T A x, y \rangle = \langle x, y \rangle + \langle (A^T A -I) x, y \rangle \geq \langle x, y \rangle - |\langle (A^T A -I) x, y \rangle|$ $\endgroup$ – user66081 Feb 3 '15 at 11:37
  • $\begingroup$ Sorry, I misinterpreted the inequality as $\langle Ax, Ay \rangle\ge (1-\delta)\langle x,y \rangle$ and was suprised whereas it should be really $\langle Ax, Ay \rangle\ge \langle x,y \rangle-\delta \|x\|_2\|y\|_2$ which I can understand. $\endgroup$ – Samrat Mukhopadhyay Feb 3 '15 at 12:11
  • $\begingroup$ I appreciate your effort. I think the core question is what can we say about a nontrivial lower bound of $|\langle Ax, Ay\rangle|$ with $supp(x)=supp(y)$. As in the last comment, we can easily find some lower bound, but I want it to be of the form $m\langle x,y\rangle$ with some constant $m$ and also the lower bound should be positive. Thanks for the effort anyway. $\endgroup$ – Samrat Mukhopadhyay Feb 3 '15 at 12:20
  • $\begingroup$ @SamratMukhopadhyay: ok, see addendum. $\endgroup$ – user66081 Feb 3 '15 at 15:14

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