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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.