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For $m=n$, $U$ and $V$ are square orthogonal matrices. Multiply the linear system successively by $U^T$, ${\Sigma }^{-1}$ and $V$ to obtainWith the singular value decomposition already supplied, compute $c=U^{T}b$, divide $c_j$ by ${\sigma }_j$, then form $x=Vc$. The two dense matrix-vector products each cost $O(n^2)$ arithmetic operations in Big O notation, and the diagonal scaling costs $O(n)$. ThusIf the singular value decomposition first has to be computed, its usual dense cost is $O(n^3)$; that is a separate setup cost and can be reused for further right-hand sides.

Use banded Gaussian elimination followed by back substitution. To distinguish the right-hand side from the upper-diagonal coefficients $b_i$, write it as $r_i$. If the modified pivot elements are nonzero, the tridiagonal elimination recurrences areThey eliminate just the one entry below each pivot element. Solve the resulting upper bidiagonal matrix system byThere is a bounded number of arithmetic operations per row in each sweep, giving $O(n)$ arithmetic operations and $O(n)$ storage. The unpivoted recurrences cost $8n-7$ arithmetic operations under the usual scalar-operation count.

For a general tridiagonal matrix, nonsingularity alone does not guarantee these pivot elements. If a zero or unsafe pivot element occurs, use partial pivoting between the current row and the next row. These are the only possible nonzero candidates in the current column; adjacent row swaps introduce at most a second upper diagonal. Tracking that fill-in keeps both Gaussian elimination and back substitution at $O(n)$ arithmetic operations. For example, $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ is invertible but requires a row swap. If the input is singular, report a zero final pivot element; a unique solution cannot be promised, and compatibility or free-variable analysis is a separate issue. No nonsingularity or safe-pivot assumption is silently inferred from the word tridiagonal.

Evaluate the recurrence relation successively, retaining only the preceding two values. Each of the $n-1$ updates uses three multiplications and one subtraction. Hence $4(n-1)=O(n)$ arithmetic operations and $O(1)$ auxiliary storage suffice, excluding the stored input. Naively branching into two recursive calls would repeat work; the sequential computation is essential. This is an arithmetic-operation bound, not a bit-complexity bound for arbitrarily large exact integers.

Sparse Gaussian elimination solves a linear system by Gaussian elimination while exploiting its sparse matrix structure to avoid storing zeros and performing unnecessary arithmetic operations. The catch is that zeros in the input need not remain zero: eliminating a shared unknown can couple two equations that previously had no direct coupling. These new positions are fill-in, and their number can depend strongly on the order of elimination.

For a real symmetric positive-definite matrix, the usual direct method is a Cholesky decomposition. Choose an ordering, use symbolic factorization to allocate the predicted factor entries, compute their values, then solve two triangular systems. Compressed sparse storage makes the reduced storage useful. General LU decomposition may also need numerical partial pivoting, so an ordering chosen for sparsity can change during factorization.^[ref]

For a fixed ordering, let $P$ be the permutation matrix that puts the unknowns in that order. Permute both rows and columns:Here $C$ is the lower triangular matrix in the Cholesky decomposition of $B$; $z=Px$. Forward substitution and back substitution give the original solution. The simultaneous permutation preserves the symmetric positive-definite matrix property because $v^{T}Bv=(P^{T}v{)}^{T}A(P^{T}v)>0$ for every nonzero vector $v$. A row permutation alone generally loses the symmetric matrix structure.

The local mechanism is a Schur complement. If the active pivot element is $a_{kk}$, then for remaining indices $i,j$,Even when $a_{ij}=0$, two nonzero couplings through $k$ make the second term nonzero. Thus the elimination graph removes $k$ and joins its remaining neighbors into a clique. The fill-path criterion for symmetric elimination extends this local rule across several steps. The elimination tree organizes the resulting dependencies; the minimum degree algorithm and nested dissection choose orderings to control them.

This connection between graphs and linear systems was already explicit in Donald J. Rose, R. Endre Tarjan and George S. Lueker's 1976 study of fill-in and perfect elimination orderings.^[ref]

Evaluate numeric formula dependencies using depth-first search and memoization, or a topological ordering. Marking active evaluations detects cycles; repeated numerical sweeps need not terminate or determine a unique result.

A spreadsheet dependency graph turns a page of spreadsheet formulas into a plan: compute each prerequisite before the spreadsheet cell that consumes it. This section uses a deliberately small model with explicit cell references, fixed spreadsheet formulas, and deterministic computation from numerical inputs. There are no external data changes or formulas that change their references while running. Within these assumptions, a topological order gives a reproducible answer without repeatedly sweeping the visible grid.

Take two input spreadsheet cells and five spreadsheet formulas:

Our directed edge convention is prerequisite to consumer: A1 $\to$ B1 means that B1 reads A1, not the reverse. The seven directed edges are A1 $\to$ B1, A2 $\to$ B1, A1 $\to$ B2, B1 $\to$ C1, B2 $\to$ C1, C1 $\to$ D1, and A2 $\to$ E1. This spreadsheet dependency graph is a Directed acyclic graph. One topological order is A1,A2,B1,B2,C1,D1,E1; E1 could instead be computed immediately after A2. A topological order constrains prerequisites, not the physical positions of spreadsheet cells.

In that order, the initial arithmetic operations give $B1=3+4=7$, $B2=2\cdot 3=6$, $C1=7+6=13$, $D1=13/2$, and $E1=4^2=16$. The spreadsheet formulas define one value per spreadsheet cell: an outline by induction along any topological order shows that every prerequisite already has its uniquely determined value when its consumer is computed. Different permitted orders therefore agree in this model. The useful idea is that information flows through the Directed acyclic graph even when the visible grid does not display that flow.

A concrete implementation history illustrates this separation. Microsoft's documentation distinguishes the per-worksheet calculation chains used in Excel 2002 and earlier from calculation using multiple software threads introduced in Excel 2007 for independent chain sections. This is a documented implementation example, not a definition of spreadsheet dependency evaluation.^[ref]

For the displayed unpivoted Thomas algorithm with one right-hand side, count each scalar addition, subtraction, multiplication or division as one arithmetic operation. Forming $m_i,p_i,g_i$ takes five arithmetic operations per forward step. The last unknown takes one division; each remaining back substitution step takes three arithmetic operations. ThusThis is an operation-count model, excluding assignments, indexing, pivot checks and any special simplifications of the coefficients. It is not a timing benchmark. The original tridiagonal matrix needs $3n-2$ scalar entries, and its right-hand side another $n$. Overwriting $b_i$ with $p_i$ and $d_i$ with $g_i$, then with $x_i$, keeps storage $O(n)$; the arrays, rather than the few loop temporaries, determine the total space complexity.

For repeated right-hand sides, retain the LU decomposition: computing $m_i,p_i$ once costs $3(n-1)$ arithmetic operations, and each subsequent pair of forward substitution and back substitution sweeps costs $5n-4$. Hence $r$ right-hand sides cost $3(n-1)+r(5n-4)$ arithmetic operations. This is the banded counterpart of Reusing LU factors for multiple right-hand sides.