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For sine collocation of square modified Helmholtz global relations, each opposite-side mode has a scaled matrixThe assembled system has strict diagonal dominance, and its spectral condition number of a positive-definite matrix is below $(1+e^{-\pi })/(1-e^{-\pi })<1.091$. The conclusion concerns the explicit paired sine rows, not arbitrary uncombined complex rows.

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 particularly simple constructive choice uses paired collocation points for a global relation. For $n=1,\dots ,N$, putFour suitable sets, one for each side, areThe braces in each set range over all $n$. For large $p_n/k$, compute $r_n^{-1}=k/({\omega }_n+p_n)$ to avoid cancellation. The corresponding $(A,B)$ pairs are, respectively,Each pair has the same exponentially growing normal direction and opposite tangential frequencies. Take the phase-weighted difference of the two collocated global relations for a linear boundary value problem. For example, the bottom pair gives the adjoint testThe other three phase-weighted differences giveThus the four spectral sets supply four real sine-tested equations per mode. We retain these $4N$ equations; requiring every uncombined complex equation as well would instead form an overdetermined finite approximation. For real traces each retained equation is the appropriate real combination of its conjugate pair.

The key advantage is that $v_{Bn},v_{Tn}$ vanish on the vertical sides, while $v_{Ln},v_{Rn}$ vanish on the horizontal sides. Orthogonality then eliminates all off-mode unknowns. Define the known right-hand sidesThe retained global relations for a linear boundary value problem areFor example, with $f_{jn}={\int }_{-1}^1{f}_j^{(M)}(s){\varphi }_n(s)ds$, the bottom and top right-hand sides are explicitlyFor completeness the left and right right-hand sides areAll these integrals use known polynomial approximations or the exact Dirichlet boundary condition.

Scale each row by $e^{-{\omega }_n}$. Every block becomesThere is just one off-diagonal entry in each row. This proves diagonal dominance of paired square global-relation collocation. Therefore the assembled $4N\times 4N$ system is strictly diagonally dominant, with row margin $1-{\delta }_n>1-e^{-\pi }$. This proof applies to the explicitly paired and scaled collocation system, rather than assuming arbitrary raw spectral rows are diagonally dominant. Each block is a positive-definite matrix and have spectral condition number of a positive-definite matrixFor instance, setting ${\rho }_{jn}=e^{-{\omega }_n}{R}_{jn}$ gives the concise answerwith the identical formula for the opposite vertical sides. If the known-data integrals use exact $f_j$, these are exactly the first $N$ sine coefficients of the true normal traces: no discarded unknown mode contributes to a retained row. With a finite Legendre polynomial approximation, the error is solely that of the known-data approximation and the final normal-trace truncation. These square modified Helmholtz Dirichlet-to-Neumann coefficients determine the unknown boundary normal derivative; Green's third identity or the separated-variable construction can then recover the interior solution.

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.

Forthe gradient is $\nabla F(x)=Ax-b=-r$. The steepest descent method with exact line search for a positive-definite quadratic therefore uses

For $A=\mathrm{diag}(1,\gamma )$ and $b=0$, the minimizer is $x^{\ast }=0$. PutStarting from $x^{(0)}=(\gamma ,1{)}^T$, direct substitution gives ${\alpha }_k=2/(\gamma +1)$ and, inductively,The Euclidean norm is unchanged by the alternating sign, soFor a positive-definite matrix, the spectral condition number of a positive-definite matrix isHere ${\kappa }_2(A)=\gamma$, and hence

The Conjugate gradient method starts with $p_0=r_0$ and updatesBy finite termination of the conjugate gradient method, it reaches the exact solution in at most the number of distinct eigenvalues. This matrix has two, so at mostare required in exact arithmetic.