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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

Use the natural empty-sum convention ${\mu }_1(0)=0$ for cumulative coherence. This is needed for the printed case $s=1$. For $x=0$ the conclusion is immediate. Otherwise let $S$ be its support of a vector, with $k=|S|\le s$. The Gram matrix $G=A_S^{\ast }{A}_S$ has diagonal entries one because the columns have unit Euclidean norm. Each off-diagonal row sum is bounded byThe last inequality uses monotonicity of cumulative coherence: enlarging an index set only adds nonnegative summands. There are enough indices to enlarge it because $s\le N$.

The Gershgorin circle theorem places every eigenvalue of $G$ within distance ${\mu }_1(s-1)$ of one. Since $G$ is a Hermitian matrix, its eigenvalues are real, and the finite-dimensional spectral theorem givesThis is the cumulative coherence bound for restricted isometry. The lower bound remains valid when ${\mu }_1(s-1)>1$, although it is then nonpositive. No assumption that the matrix is already a near isometry is required. The distortion is bounded by ${\mu }_1(s-1)$ on every order-$s$ sparse vector.

The relevant fixed coordinate subspace is $\{(u,0):u\in {\mathbb{R}}^k\}$. Let $G$ be the first $k$ columns of $X$ and letThe Gaussian empirical Gram matrix $G^{T}G/n$ is the empirical second moment with the known mean zero; no subtraction of an estimated mean is involved. For $\theta =(u,0)$, the ratio under consideration is $|u^{T}Du|/\Vert u{\Vert }_2^2$. Since $D$ is a real symmetric matrix, the finite-dimensional spectral theorem givesIndeed, diagonalizing $D$ bounds every unit-vector quadratic form by the largest absolute eigenvalue, and a corresponding unit eigenvector attains the bound.

We first construct a unit sphere net from ball covering. Enlarge the numerical covering constant, if necessary, to $A_0\ge 1$. Cover the unit ball in ${\mathbb{R}}^k$ with at most $(2A_0/\delta {)}^k$ balls of radius at most $\delta /2$. For each such ball meeting the unit sphere, choose a point of the sphere in it and discard the others. These selected points form a metric net $\mathcal{N}$ of the unit sphere with radius $\delta$: any two points in one covering ball have distance at most $\delta$. This argument ensures that the net points have unit length even if the original covering centers did not.

Take $\delta =1/4$ and put $B=8A_0$, so $|\mathcal{N}|\le B^k$. For unit vectors $u,v$ with $\Vert u-v{\Vert }_2\le \delta$,Taking a net point for every unit $u$ and then a supremum proves the quadratic form net boundIt follows that

Fix $v\in \mathcal{N}$. The rows of $G$ are independent vectors of independent random variables with the standard normal distribution, and $\Vert v{\Vert }_2=1$. Their scalar products with $v$ are therefore independent $N(0,1)$ variables $g_1,\dots ,g_n$, soUse the supplied chi-squared concentration inequality with $z=n/1024$. Its threshold isConsequently $P(|v^{T}Dv|>1/4)\le 2e^{-n/1024}$. The union bound gives the stronger fixed-subspace estimateThis is Gaussian Gram matrix concentration on a fixed subspace.

Since $1\le k<p$, we have $p\ge 2$. Under $n\ge Ck\log p$,Choose the numerical constant $C\ge 1024(1+\log B/\log 2)$. Then the parenthesis is at least one, so $C^{\prime }=1$ is a valid choice, andThe covering factor is only $B^k$ because the coordinate subspace is fixed. The ambient dimension enters through the assumed sample-size bound.

The linear-system residual at an approximate solution $x_m$ is $r_m=b-A{x}_m$. For symmetric positive definite $A$, the Conjugate gradient method has $x_m\in x_0+K_m(A,r_0)$ and minimizes the squared $A$-norm of the error over that affine space; its residual is orthogonal to $K_m(A,r_0)$.

If $A$ has $t$ distinct eigenvalues, interpolate $q(\lambda )=1/\lambda$ on them with a polynomial of degree at most $t-1$. The finite-dimensional spectral theorem gives $A^{-1}=q(A)$, so the exact correction $x_{\ast }-x_0=A^{-1}{r}_0$ lies in $K_t(A,r_0)$. The minimum possible error at step $t$ is therefore zero. Thus $\boxed{\text{CG terminates in at most }t\text{ steps}}$ in exact arithmetic, possibly earlier if the initial residual has fewer eigencomponents. If $r_0=0$, it terminates immediately.

A symmetric positive definite matrix preconditioner $M$ changes the system to $M^{-1/2}A{M}^{-1/2}z=M^{-1/2}b$, with $x=M^{-1/2}z$; practical preconditioned CG applies solves with $M$. It can accelerate convergence by reducing the spectral condition number of a positive-definite matrix, which improves the usual bound involving $(\sqrt{\kappa }-1)/(\sqrt{\kappa }+1)$, and by clustering eigenvalues, which allows low-degree residual polynomials small on the spectrum even when the extreme-eigenvalue bound is pessimistic.

Choosing $\boxed{M=A}$ makes the transformed matrix $I$, so one step suffices. But applying $M^{-1}$ already requires solving the original problem, and constructing an exact factorization generally incurs its original computational cost. It is therefore not a useful shortcut for a single unsolved system, though an available factorization can be reused for multiple right-hand sides. Effective matrix preconditioning balances spectral improvement against inexpensive application.

A self-adjoint operator $T$ satisfies $⟨Tx,y⟩=⟨x,Ty⟩$. If $Tx=\lambda x$, then $\lambda ⟨x,x⟩=⟨Tx,x⟩=⟨x,Tx⟩=\overline{\lambda }⟨x,x⟩$, so $\lambda \in \mathbb{R}$. An eigenvector exists over $\mathbb{C}$; its orthogonal complement is $T$-invariant. Induction on dimension produces an orthonormal basis of eigenvectors, proving the finite-dimensional spectral theorem.

A self-adjoint $T$ is positive definite exactly when every eigenvalue is positive, since $⟨Tx,x⟩={\sum }_i{\lambda }_i|x_i{|}^2$ in an eigenbasis. The quadratic forms of two positive-definite operators add, so their sum is positive definite.

The same eigenbasis shows that $\mathrm{spec}T\subset [a,b]$ exactly when $T-\lambda I$ is positive definite for every $\lambda <a$ and negative definite for every $\lambda >b$. Finally, if $v$ is a unit eigenvector of $\alpha +\beta$ with eigenvalue $\gamma$, the Rayleigh quotient bounds giveThus every eigenvalue of $\alpha +\beta$ lies in $[a+c,b+d]$.

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.

Put $p=⟨\psi |M_1^2|\psi ⟩\ge 1-\epsilon$. Since $M_1$ is a positive contraction, the successful post-measurement state is the pure state represented by $|{\psi }^{\prime }⟩=M_1|\psi ⟩/\sqrt{p}$. Its squared quantum fidelity with the input isThe finite-dimensional spectral theorem and $0\le M_1\le I$ give $M_1^2\le M_1$, so $⟨M_1⟩\ge p$. Therefore the pure-state gentle measurement bound isThe positive-operator assumption matters: an arbitrary unitary measurement operator can have success probability one while rotating the state to an orthogonal vector.

Put $H=A-B$. This is a nonzero Hermitian matrix. By the finite-dimensional spectral theorem, it has a unit eigenvector $v$ with a nonzero real eigenvalue $lambda$. Define the rank-one orthogonal projection matrixThe cyclic property of the matrix trace givesHence this $P$ distinguishes $A$ and $B$ through the required traces.

A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.

Because $A$ is a real symmetric positive-definite matrix, the finite-dimensional spectral theorem supplies an orthonormal eigenbasis $u_1,\dots ,u_n$ withIts eigendecomposition is also its singular value decomposition. If $e=y^{(\delta )}-y$, thensoThe operator norms satisfy $\Vert A{\Vert }_2={\lambda }_n$ and $\Vert A^{-1}{\Vert }_2=1/{\lambda }_1$. Therefore the worst-case relative perturbation bound isThe ratio ${\kappa }_2(A)$ is the spectral condition number of a positive-definite matrix. A large ratio means that data noise aligned with an eigenvector for the smallest eigenvalue is strongly amplified, so the inverse problem is ill conditioned.