A useless non-RIP Gaussian matrix

Recently, for some unrelated reasons, I discovered that it is actually very easy to generate a Gaussian matrix $ \Phi$ that does not respect the restricted isometry property (RIP) [1]. I recall that such a matrix is RIP if  there exists a (restricted isometry) constant $ 0<\delta<1$ such that, for any $ K$-sparse vector $ w\in \mathbb R^N$,

$ (1-\delta)\|w\|^2\leq \|\Phi w\|^2 \leq (1+\delta)\|w\|^2$.

This is maybe obvious and it probably serves to nothing but here is anyway the argument.

Take a $ K$-sparse vector $ x$ in $ \mathbb R$ and generate randomly two Gaussian matrices $ U$ and $ V$ in $ \mathbb R^{M\times N}$ with i.i.d. entries drawn from $ \mathcal N(0,1)$. From the vectors $ a=Ux$ and $ b=Vx$, you can form two new vectors $ c$ and $ s$ such that $ c_i = a_i / \sqrt{a_i^2 + b_i^2}$ and $ s_i = b_i / \sqrt{a_i^2 + b_i^2}$. Then, it is easy to show that matrix $ \Phi = {\rm diag}(c) V - {\rm diag}(s)U\qquad(1)&s=2$ is actually Gaussian except in the direction of $ x$ (where it vanishes to 0). This can be seen more clearly in the case where $ x = (1,0,\,\cdots,0)^T$. Then the first column of $ \Phi$ is $ 0$ and the rest of the matrix is independent of $ {\rm diag}(c)$ and $ {\rm diag}(s)$. Conditionally to the value of these two diagonal matrices, this part of $ \Phi$ is therefore Gaussian with each $ ij$ entry ($ j\geq 2$) of variance $ c_i^2 + s_i^2 = 1$. Then, the condition can be removed by expectation rule to lead to the cdf of $ \Phi$, and then to the pdf by differentiation, recovering the Gaussian distribution in the orthogonal space of $ x$. However, $ \Phi$ cannot be RIP. First, obviously since $ \Phi x = 0$ which shows, by construction, that at least one $ K$-sparse vector (that is, $ x$) is in the null space of $ \Phi$. Second, by taking vectors in $ x + \Sigma_K = x + \{u: \|u\|_0 \leq K \}$, we clearly have $ \|\Phi (x + u)\| = \|\Phi u\|$ for any $ u \in \Sigma_K$. Therefore, we can alway take the norm of $ u$ sufficiently small so that $ \|\Phi (x + u)\|$ is far from $ \|x + u\|$. Of course, for the space of $ K$-sparse vectors orthogonal to $ x$, the matrix is still RIP. It is easy to follow the argument above for proving that $ \Phi$ is Gaussian in this space and then to use classical RIP proof [2]. All this is also very close from the “Cancel-then-Recover” strategy developed in [3]. The only purpose of this post to prove the (useless) result that combining two Gaussian matrices as in (1) leads to a non-RIP matrix. References: [1] Candes, Emmanuel J., and Terence Tao. “Decoding by linear programming.” Information Theory, IEEE Transactions on 51.12 (2005): 4203-4215. [2] Baraniuk, Richard, et al. “A simple proof of the restricted isometry property for random matrices.” Constructive Approximation 28.3 (2008): 253-263. [3] Davenport, Mark A., et al. “Signal processing with compressive measurements.” Selected Topics in Signal Processing, IEEE Journal of 4.2 (2010): 445-460.

Laurent Jacques
Laurent Jacques
FNRS Senior Research Associate and Professor