Triangular matrix

An upper or lower triangular matrix has all entries on one side of the main diagonal equal to zero. Its eigenvalues, with algebraic multiplicity, are its diagonal entries.

An invertible lower triangular matrix has a lower triangular matrix inverse. To see this, solve $Ax=e_j$ for column $j$ of the inverse. The nonzero diagonal and forward substitution force $x_i=0$ for $i<j$. Applying the result to the transpose proves the upper-triangular case as well. Invertibility is essential; a singular triangular matrix has no ordinary matrix inverse.

An upper triangular matrix system with nonzero diagonal is solved from the last row upward. Once $x_{i+1},\dots ,x_n$ are known, use $x_i=(b_i-{\sum }_{j>i}{a}_{ij}{x}_j)/a_{ii}$. The nonzero diagonal guarantees a unique solution; this is the final solve after a QR factorization.

An upper triangular matrix has zero entries below its main diagonal. Its determinant is the product of its diagonal entries, and an invertible upper triangular linear system is solved by backward substitution.

A square matrix is strictly upper triangular if every entry on or below its diagonal is zero. A product of $n$ such $n$-by-$n$ matrices vanishes, so they form a nilpotent Lie algebra under the commutator. Multiplication by an upper triangular matrix preserves strict upper triangularity and zero trace.

A lower triangular matrix has zero entries above its main diagonal. Its determinant is the product of its diagonal entries, and an invertible lower triangular linear system is solved by forward substitution.

A lower triangular matrix with every diagonal entry equal to one. Its determinant is one, and its inverse is again unit lower triangular. Such matrices store the elimination multipliers in an LU decomposition.

For an invertible lower triangular matrix, the scalar pivot $a$ and the remaining triangular block $C$ are both invertible. The displayed formula follows by matrix multiplication: the lower-left block in the product is $b/a-C(C^{-1}b/a)=0$, and the diagonal blocks are identities. Recursively invert $C$, compute $-C^{-1}b/a$, and assemble the inverse matrix; the one-dimensional base case is $a^{-1}$. This recursion reduces the size by one at each step.

The algebra $T_n(k)$ consists of the lower triangular $n$ by $n$ matrices over $k$. Its Jacobson radical $N$ consists of the strictly lower triangular matrices, andThe $n$ simple left modules $S_r$ are one-dimensional, with a matrix acting through its $r$th diagonal entry. For the left regular module,

A triangular linear system is solved by forward substitution for a lower triangular coefficient matrix or backward substitution for an upper triangular coefficient matrix, using $O(n^2)$ operations for a dense $n$ by $n$ matrix.

Forward substitution solves a lower-triangular system from its first row downward, substituting each newly determined variable into the rows below.

Backward substitution solves an upper-triangular system from its last row upward.

A triangular system is a linear system whose coefficient matrix is a triangular matrix. Nonzero diagonal entries give a unique solution.

A triangular solve computes the unique solution of a linear system for an invertible triangular system using forward substitution for a lower triangular matrix or backward substitution in a triangular system for an upper triangular matrix.