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At elimination column $k$, let $r=m-k$ be the size of the remaining trailing block. Computing its multipliers costs $r$ divisions; updating all $r^2$ entries costs $2r^2$ floating-point operations, one multiplication and one subtraction per entry. Thus the dominant count isUpdating the right side and back substitution add only $O(m^2)$ floating-point operations. Pivot searches involve comparisons and row exchanges, without changing the leading arithmetic count. Hence Gaussian elimination with pivoting has the requested asymptotic cost.

Solving the $m$ unrelated shifted systems by separate Gaussian elimination costs $\boxed{\frac{2}{3}{m}^4+O(m^3)}$ floating-point operations.

Instead compute $H=Q^{T}AQ$ once in $O(m^3)$ work. For each shift, solveThe shift changes only the diagonal, so $H-{\lambda }_{j}I$ remains an upper Hessenberg matrix. In each elimination column only the entry immediately below the pivot needs removal. With partial pivoting the only possible nonzero candidate rows are the current row and the next, and each update touches $O(m)$ entries. Factorization therefore costs $O(m^2)$ per shift; the triangular solve and both orthogonal-vector transformations also cost $O(m^2)$ each.

Thus $m$ shifted solves plus the initial Hessenberg reduction cost $\boxed{O(m^3)}$ in total. Pivoting is still needed in general: nonsingularity excludes a singular system but does not by itself guarantee that every unpivoted elimination pivot is nonzero.

The standard conjugate gradient convergence bound isHere $\kappa \le 5$, so the factor in parentheses is at most $q=(3-\sqrt{5})/2\simeq 0.381966$. Since $2q^{25}<10^{-10}$, 25 iterations suffice to reduce the energy-norm error by a factor exceeding $10^{10}$.

There are at most $m/20$ nonzero entries of $t$, and each contributes to at most two off-diagonals. Thus $A-I$ has at most $m^2/10$ nonzero entries. Initialize a matrix-vector product with the diagonal contribution $p$, which costs no arithmetic because the diagonal is one. Each off-diagonal term needs one multiplication and one addition, giving at most $m^2/5$ floating-point operations per product.

A standard conjugate gradient step adds two inner products, three vector updates and two scalar divisions, at most $10m$ further operations. Starting from $x_0=0$ makes $r_0=b$ and avoids an initial matrix-vector product; its squared norm needs $2m-1$ operations. The total for 25 steps is at mostThis supplies an explicit meaning of the paper's large-$m$ qualification. The quoted conjugate gradient convergence bound is the usual exact-arithmetic result.