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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.

Let $L$ be the triangular matrix with $L_{ij}=1$ for $j\le i$ and zero otherwise. Then $(L{L}^T{)}_{ij}=\min (i,j)$ and $\det L=1$, so the determinant of $L{L}^T$ is one. More generally, for $0=t_0<t_1<\cdots <t_n$, the matrix with entries $\min (t_i,t_j)$ factors as $L\mathrm{diag}(t_k-t_{k-1})L^T$ and has determinant ${\prod }_k(t_k-t_{k-1})$. This is the finite-dimensional structure behind the Brownian covariance kernel.

Let $L_{ik}=1$ when $k\le i$ and $L_{ik}=0$ otherwise. The matrix product has entriesso $A_n=L{L}^T$. Both $L$ and its matrix transpose are triangular matrices with diagonal entries one. Multiplicativity of the determinant therefore gives $\det A_n=(\det L{)}^2=1$ for every $n\ge 1$. This determinant of the minimum-index matrix also follows by subtracting each preceding row from the next, working from the bottom upwards: the resulting matrix is upper triangular with unit diagonal.

Each QR decomposition $A_k=Q_k{R}_k$ givesInduction therefore yields the first Accumulated QR factorization identity:

We next prove the corresponding factorization of a matrix power. The case $k=0$ is just $A=Q_0{R}_0$. Ifthen the similarity formula above impliesConsequently,The product ${\overline{Q}}_k$ is orthogonal, and the product ${\overline{R}}_k$ is upper triangular. Henceis a QR decomposition.

Using Euler's formula,ThusOrdering the modes as $(-1,0,1)$, let $a=1/2+i/4$. ThenConsequentlyThis matrix is triangular, so its eigenvalues are its diagonal entries:

A square matrix is invertible precisely when its determinant is nonzero. Equivalent conditions are full rank, a trivial kernel, and zero not being an eigenvalue.

For a lower triangular matrix, the first row has only the possibly nonzero entry $A_{11}$. A Laplace expansion along that row gives $\det A=A_{11}\det A^{(1)}$, where the remaining principal submatrix is again lower triangular. Induction from the one-by-one case yieldsThis proof includes singular triangular matrices; it makes no division by diagonal entries.

For the diagonal example, put $N=n(n+1)/2$. Its determinant is $b^{1+2+\cdots +n}=b^N$, and invertibility requires $b\ne 0$. Write the complex number as $b=r{e}^{i\theta }$ with $r>0$. A nonzero purely imaginary determinant requires $\cos (N\theta )=0$. ThusEquivalently $b^N\in i\mathbb{R}\setminus \{0\}$; the angle is understood modulo $2\pi$.

For skew-Hermitian matrices, reversing products under the conjugate transpose gives $(CD{)}^{†}=D^{†}{C}^{†}=DC$. ConsequentlyThe latter condition is vanishing of the anticommutator. A universal choice for the last request isIndeed $(CD-DC{)}^{†}=DC-CD=-(CD-DC)$, so the commutator of two skew-Hermitian matrices is skew-Hermitian.

Write $A=D+L+R$, where $D$ is the diagonal matrix of diagonal entries of $A$, and $L,R$ are its strictly lower- and upper-triangular parts. Since $A$ is symmetric, $R=L^T$. Subtract the two factors in the SOR iteration:Therefore, for $\omega \ne 0$,Positive definiteness makes each $D_{ii}>0$, so the triangular matrix $B(\omega )$ is an invertible matrix for every $\omega \ne 0$. The original formula gives $C(0)=I$; the expression for $B(\omega )$ itself is undefined there. The affine term ensuring that the desired solution is a fixed point is $c=B(\omega {)}^{-1}b$.

Let $P$ be the $3$-by-$3$ evaluation matrix $P_{i,\ell +1}=p_{\ell }({\lambda }_i)$. The kernel evaluation matrix is the Gram matrix $P{P}^T$, so multiplicativity of the determinant givesBecause the polynomials are monic of consecutive degrees, their coefficient matrix relative to $1,x,x^2$ is a triangular matrix with unit diagonal. It has determinant one, so the Vandermonde determinant invariance under monic basis change givesSquaring removes the ordering sign. Thus the Gaussian unitary ensemble joint probability density function has precisely the kernel formNo orthogonality of the polynomials is needed for this step.

For spatially homogeneous perturbations write $q=1+u$ and $c=\eta +v$. The linearized reaction equations areThis triangular matrix has eigenvalues $-1$ and $-\kappa$, both negative. Hence the homogeneous steady state is linearly asymptotically stable against homogeneous perturbations. Even when $\kappa =1$, where the two eigenvalues coincide and a polynomial factor may occur in the matrix exponential, all such perturbations still decay.

A triangular matrix has determinant equal to the product of its diagonal entries. Since $L$ is a unit lower triangular matrix, $\det L=1$. ThusEquivalently, the elimination used only row additions, which leave the determinant unchanged; there were no row swaps or row scalings to account for.

For square matrices of the same size,For an upper or lower triangular matrix $T$, its determinant is the product of its diagonal entries:These formulas include singular matrices and zero diagonal entries.

Let $e_i$ be standard coordinate columns. The required elementary matrix isIt is the identity matrix with the extra entry $\alpha$ in position $(j,i)$. Left matrix multiplication implements the row operation: all other rows are unchanged. Since $j>i$, $E$ is a lower triangular matrix with every diagonal entry one. Part (a) therefore gives $\det (E)=1$ andHence adding a multiple of one row to another leaves the determinant unchanged, even if $A$ is singular.

Use $-A=(-I_n)A$. The diagonal matrix $-I_n$ has $n$ diagonal entries equal to $-1$, so the product and triangular matrix formulas from part (a) giveEquivalently, multiplying each of the $n$ rows by $-1$ supplies one factor $-1$ per row.

For $A$, the triangular matrix formula gives the characteristic polynomialThe three eigenvalues are distinct, so part (d) already proves the existence of an eigenvector basis. Explicit choices, ordered by eigenvalues $1,3,2$, areTheir column matrix has determinant one, and direct multiplication gives respectively one, three and two times the indicated vector.

For $B$, expansion along the last row givesAgain the three eigenvalues are distinct. An explicit eigenvector basis, ordered by eigenvalues $4,2,3$, isIts column matrix has determinant $-2\ne 0$. In particular, $B$'s diagonal entries are not its eigenvalues: it is not a triangular matrix.

Finally $C=I_3$, so every nonzero vector is an eigenvector with eigenvalue one. Choose the standard basis $e_1,e_2,e_3$. These are linearly independent even though their eigenvalues coincide. Distinct eigenvalues are sufficient for an eigenvector basis, and are not necessary.

Use multiplicativity of the determinant and the determinant criterion for an invertible matrix:Neither factor can therefore be zero, so both matrices are nonsingular. The determinant of a triangular matrix is the product of its diagonal entries:Each product is nonzero, forcing every diagonal factor to be nonzero. These explicitly stated determinant properties justify both conclusions.

A triangular matrix has determinant equal to the product of its diagonal entries, so $\det L=1$ and $\det U=3\cdot 4\cdot (-11)=-132$. A permutation matrix exchanging two rows has determinant $-1$. Using multiplicativity of the determinant in $PA=LU$ yields $(-1)\det A=-132$. Hence $\det A=132$.

A Hermitian matrix $A$ is a positive-definite matrix exactly when its leading principal minorsare all positive. The leading principal minor uses the first $k$ rows and the first $k$ columns in the same ordering. For the two-by-two Cholesky decomposition example, ${\Delta }_1=4$ and ${\Delta }_2=4\cdot 5-2^2=16$, so the criterion proves positive definiteness without calculating eigenvalues.

The successive pivot elements before taking square roots are ${\Delta }_k/{\Delta }_{k-1}$, with ${\Delta }_0=1$. Here they are $4$ and $16/4=4$. In general, a leading triangular matrix factor has determinant equal to the product of its diagonal entries, soConversely, positive leading principal minors make each successive Schur complement pivot element positive, allowing the Cholesky decomposition to be constructed. The identity $u^{T}Au=\Vert L^{T}u{\Vert }_2^2$ then proves positive definiteness; for complex matrices, replace the transposes by conjugate transposes. This gives a short inductive proof of the criterion.

For $B=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$, the first minor is positive but $\det B=-3$. The second Cholesky radicand would be $1-2^2=-3$, and $v=(1,-1{)}^T$ directly gives $v^{T}Bv=-2<0$. Thus positive diagonal entries are insufficient. Replacing “positive” by “nonnegative” in just the leading-minor test is also insufficient for positive semidefiniteness: $\mathrm{diag}(0,-1)$ has leading minors $0,0$, but its quadratic form is negative on $(0,1{)}^T$.

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

A unit triangular matrix is a triangular matrix whose diagonal entries are all one. It is an invertible matrix with determinant one. In the convention used for LU decomposition, $L$ is a unit lower triangular matrix, so forward substitution needs no division by its diagonal entries.

The upper unitriangular group consists of upper triangular matrices with every diagonal entry equal to one. Over ${\mathbb{F}}_p$ it has order $p^{n(n-1)/2}$ and is a Sylow subgroup of $G{L}_n({\mathbb{F}}_p)$.

Its elements form the upper triangular, diagonal-one subset of triangular matrices.