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For a matrix with at least two columns of unit Euclidean norm, its mutual coherence is the largest magnitude of an inner product between distinct columns. It measures the strongest pairwise ambiguity of linear measurements. The Cauchy-Schwarz inequality gives $0\le \mu \le 1$. The order-two restricted isometry constant equals the mutual coherence, because every two-column Gram matrix has eigenvalues $1\pm |⟨a_i,a_j⟩|$.

For normalized columns and $1\le s\le N$, the restricted isometry constant is bounded by the cumulative coherence at $s-1$. A restricted Gram matrix has diagonal one and every off-diagonal row sum at most ${\mu }_1(s-1)$. The Gershgorin circle theorem and the finite-dimensional spectral theorem bound its eigenvalues between $1-{\mu }_1(s-1)$ and $1+{\mu }_1(s-1)$. At $s=1$ the distortion is zero, and at $s=2$ it is exactly the mutual coherence.

Work in ${\mathbb{C}}^N$, the domain of the matrix $A$; the printed ${\mathbb{C}}^n$ in the introduction is a dimension typo. For a fixed nonempty support of a vector $S$, write $u$ for the coordinates of $x$ in $S$. ThenThe Gram matrix $A_S^{\ast }{A}_S$ is a Hermitian matrix, so its difference from the identity matrix is also a Hermitian matrix. By the finite-dimensional spectral theorem, its matrix 2-norm is the largest absolute eigenvalue, equivalentlyIndeed, an expansion in an orthonormal eigenbasis bounds every Rayleigh quotient by the largest absolute eigenvalue, and an appropriate eigenvector attains that bound. The restricted isometry constant must bound this quantity for every $S$ of size at most $s$, and the maximum of these quantities suffices for all sparse vectors of order $s$. There are finitely many sets, so the maximum exists. The empty set contributes zero. Therefore

For one-column Gram matrices, unit Euclidean norm of the columns gives $A_S^{\ast }{A}_S=1$. The restricted isometry constant formula therefore gives ${\delta }_1=0$. For a two-column Gram matrix, put $c=⟨a_i,a_j⟩$. The difference from the identity matrix has the formup to the convention for the complex inner product. Its characteristic polynomial is ${\lambda }^2-|c{|}^2$, so the eigenvalues are $\pm |c|$ and its matrix 2-norm is $|c|$. Maximizing over pairs gives ${\delta }_2=\mu$, where $\mu$ is the mutual coherence.

Part (b), together with the minimality defining the restricted isometry constant, gives ${\delta }_s\le {\mu }_1(s-1)$. Every summand defining cumulative coherence is at most the mutual coherence, so ${\mu }_1(s-1)\le (s-1)\mu$. Thus, for normalized columns and $N\ge 2$,The upper range $s\le N$ matters because the printed definition of cumulative coherence stops at $N-1$. For $N=1$, only ${\delta }_1=0$ is needed; the maximum over pairs defining mutual coherence is otherwise empty.

The ordinary column Gram matrix is $G=X^{T}X$. Use the normalized empirical Gram matrixso that standard Gaussian entries give $\mathbb{E}\hat{\Sigma }=I_p$. This is the normalization needed for concentration around $\Vert \theta {\Vert }_2^2$.

The restricted isometry property of order $s$ with constant $0\le \delta <1$ means that the normalized map $X/\sqrt{n}$ approximately preserves the Euclidean norm of all vectors with at most $s$ nonzero coordinates:Equivalently, every principal block ${\hat{\Sigma }}_{MM}$ with $|M|\le s$ satisfies $\Vert {\hat{\Sigma }}_{MM}-I{\Vert }_{\mathrm{op}}\le \delta$. The least such $\delta$ is its restricted isometry constant. In the unnormalized definition apply this property to $X/\sqrt{n}$ itself.

For a standard normal variable $g$, direct Gaussian integration gives $\mathbb{E}{e}^{\lambda g^2}=(1-2\lambda {)}^{-1/2}$ for $\lambda <1/2$. Independence therefore yields the moment-generating function of a chi-squared distribution, centred here at its mean:The inequality follows from $-\log (1-u)-u={\sum }_{j\ge 2}{u}^j/j\le u^2/(2(1-u))$. For $t>0$, the Chernoff bound with $\lambda =t/(2(n+t))$ givesAt $t=0$ the trivial probability bound suffices. This proves the requested bound, with a stronger prefactor one.

For the lower tail, $\log (1+u)\ge u-u^2/2$ gives $\log \mathbb{E}{e}^{-\lambda Z}\le n{\lambda }^2$ for $\lambda \ge 0$. Taking $\lambda =t/(2n)$ yields $\mathrm{Pr}(Z\le -t)\le e^{-t^2/(4n)}$. Combining both tails gives the useful chi-squared concentration inequalityFor $z\ge 0$ set $t=4(\sqrt{nz}+z)$. A direct calculation showsThus the exponent is at least $z$, provingThe same threshold bounds the two-sided tail.

Finally fix a deterministic $\theta$ and put $r=\Vert \theta {\Vert }_2$. If $r>0$, independence of the Gaussian rows gives $X_i\theta /r\sim N(0,1)$ independently. ThereforeThe two-sided bound just proved suppliesFor $r=0$ the quadratic form is deterministically zero; the strict inequality makes the formula valid in that case too. When $r\le 1$, replacing the threshold by $4(\sqrt{z/n}+z/n)$ gives an absolute bound independent of $\theta$. In particular every fixed such direction concentrates at rate $n^{-1/2}$ for a fixed confidence level. The restricted isometry property requires a simultaneous statement over sparse directions; this fixed-direction calculation alone is not that stronger assertion.

The printed definition of the restricted isometry constant omits the quantifier: the two inequalities must hold for every vector with at most $s$ nonzero entries. The constant is the smallest nonnegative value, or the infimum over positive values; it can be zero.

Let $T=\mathrm{supp}u\cup \mathrm{supp}v$ and $B=A_T^{\ast }{A}_T-I$. The restricted isometry property of order $s+t$ says $|z^{\ast }Bz|\le {\delta }_{s+t}\Vert z{\Vert }_2^2$ for every $z$ supported on $T$. Since $B$ is a Hermitian matrix, the finite-dimensional spectral theorem implies $\Vert B{\Vert }_{2\to 2}\le {\delta }_{s+t}$. Disjoint supports give $⟨u,v⟩=0$, henceThis argument works over $\mathbb{C}$ and avoids the loss of a factor caused by treating real and imaginary parts separately.

The printed bound is false, even for $s=2$ and zero noise. Let $P=I_7-11^T/7$ and choose $A$ with six orthonormal rows spanning $1^{\perp }$, so $A^{T}A=P$. For a vector supported on at most $s$ entries,Take $x_0=(0,0,1,1,1,1,1)$, $s=2$, and $\eta =0$. Every feasible vector is $x_0+t1$, whose ${\ell }^1$ objective is $2|t|+5|1+t|$. Its unique minimizer is $\hat{x}=(-1,-1,0,0,0,0,0)$. YetNo choice of the noise constant repairs this example. Also $s=1$ must be excluded: $A=(1,1)$ has ${\delta }_1=0$, but its two unit-coordinate vectors have identical measurements and both minimize basis pursuit.

The corrected conclusion, for $s\ge 2$, iswith both constants depending on the restricted isometry constant. The order-$s$ threshold itself is valid for $s\ge 2$; it need not be replaced by an order-$2s$ threshold. A precise sharp robust recovery theorem gives, for $\delta ={\delta }_s<1/3$, the admissible constantsThis is the standard stable recovery result, invoked here explicitly; its order restriction and constants are given in arxiv.org/pdf/1302.1236, Theorem 3.3.

For completeness, the mechanism behind a robust null space property proof is as follows. Put $h=\hat{x}-x_0$, let $S$ index the largest $s$ entries of $x_0$, and write $\sigma =\Vert x_{0,S^c}{\Vert }_1$. Feasibility and minimality give the tube and cone inequalitiesIf $A$ has the robust null space property with $0\le \rho <1$ and $\tau >0$, namely $\Vert h_T{\Vert }_2\le \rho \Vert h_{T^c}{\Vert }_1/\sqrt{s}+\tau \Vert Ah{\Vert }_2$ for every $|T|\le s$, thenApply that property again to the largest $s$ entries $T$ of $h$. Sorted coefficients satisfy $\Vert h_{T^c}{\Vert }_2\le \Vert h{\Vert }_1/\sqrt{s}$, because each tail entry is at most $\Vert h{\Vert }_1/s$. ThereforeThis proves the usual two-constant conclusion under the explicitly stated robust null space property and explains why a fixed coefficient one is not supplied by that argument. The sharp order-$s$ theorem above supplies the conclusion for the actual printed threshold, after the necessary corrections.