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

The Thomas algorithm is Gaussian elimination specialized to a tridiagonal matrix: each row couples an unknown only to its immediate neighbours. Eliminating one unknown changes just the next diagonal entry and right-hand side. Provided all pivot elements are nonzero, a forward sweep followed by back substitution solves an $n$-unknown linear system in $O(n)$ time and storage, using Big O notation. Treating the same entries as a full matrix wastes the zeros: classical dense LU factorization and solve costs grow cubically for factorization and quadratically for storage.

A concrete reason to care is the Dirichlet Poisson solution by tridiagonal elimination: a three-point finite difference for the second derivative couples adjacent points on a mesh, producing precisely this tridiagonal matrix. The same structure occurs in one-dimensional Crank-Nicolson method calculations. NPTEL's archived Lecture 13 introduces the Thomas algorithm from that time-stepping problem and labels the procedure “Thomas algorithm (1949)”.^[ref]

A linear map $T:V\to W$ has a trivial kernel when $\ker T=\{0_V\}$: the zero vector is the only input producing the zero output. The kernel is a singleton set, not the empty set, because every linear map sends zero to zero. Equivalently, its nullity is zero and $T$ is injective. This holds even without a finite-dimensional hypothesis; see Linear Algebra Done Right by Sheldon Axler (4th edition).

The reason this zero-output test controls all outputs is linearity:With a trivial kernel, equal outputs force equal inputs. Therefore every equation $T(x)=b$ has at most one solution. Existence is a separate question: $b$ must belong to the image of a linear map. In the fiber description under nullity, $x_0+\{0\}$ is just the single point $x_0$.

For example, the linear map $J:{\mathbb{R}}^2\to {\mathbb{R}}^3$, $J(x,y)=(x,y,0)$, has trivial kernel and full column rank:The output $(3,4,0)$ comes from exactly $(3,4)$, while $(3,4,1)$ is unreachable. Conversely, $P:{\mathbb{R}}^3\to {\mathbb{R}}^2$, $P(x,y,z)=(x,y)$, is surjective and has full row rank, but its kernel is $\mathrm{span}\{(0,0,1)\}$: every output $(a,b)$ comes from the solution straight line $(a,b,s)$.

For an $m\times n$ matrix, a trivial kernel is equivalent to matrix rank $n$ and linear independence of its columns. It can therefore occur only if $n\le m$. If $m=n$, the rank-nullity theorem also forces surjectivity, giving an invertible matrix; if $m>n$, the example $J$ shows why unique recovery of reachable inputs does not imply that every output is reachable. Finally, a trivial kernel guarantees exact uniqueness without guaranteeing numerical stability: the singular matrix example $A(t)$ has trivial kernel for every $t\ne 0$, even as its condition number grows without bound when $t$ approaches zero.

Two separated short-range scatterers are coupled because the outgoing wave from either centre supplies part of the regular incoming field at the other. Applying the Bethe-Peierls boundary condition at both centres gives the displayed two-by-two linear system, where $s_j$ is the incident plane-wave phase. matrix inversion sums repeated scattering between the centres. Near a small determinant, effective-range corrections may matter even if the nominal microscopic range is short.

Consider the real matrixThe Cayley-Hamilton theorem gives $A^2=3A-I$. ConsequentlyThusAll four entries come from just two scalar coefficients. If $A^n=a_{n}A+b_{n}I$, multiplying once by $A$ and replacing $A^2$ gives the linear recurrence relationEquivalently $a_{n+2}=3a_{n+1}-a_n$, with $a_0=0,a_1=1$. The appearance of Fibonacci numbers is explained by $A=Q^2$, where $Q=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ advances the Fibonacci recurrence. No eigenvectors need to be computed.

The same identity computes an inverse. Rearrange it as $A(3I-A)=I$, givingThis works because the constant term of this characteristic polynomial is nonzero and the coefficients lie in a field. Generally, when $c_0\ne 0$ over a field, rearranging the Cayley-Hamilton theorem expresses $A^{-1}$ as a polynomial evaluated at a matrix. Over a general commutative ring, a nonzero constant need not be a unit; invertibility requires that extra condition.

This doubly stochastic matrix replaces two values $a,b$ by $ta+(1-t)b$ and $(1-t)a+tb$, preserving their sum. It fixes all other coordinates when embedded in a larger matrix. A finite product of such averaging matrices and permutation matrices is again doubly stochastic.

The two-factor tensor rank is the smallest number of pure tensors whose sum equals $t$, with zero allowed as the empty sum. For two finite-dimensional vector spaces over the same field, it equals the matrix rank of the tensor-product coefficient matrix. A sum of $r$ outer products has column space contained in the span of their $r$ first factors, so its matrix rank is at most $r$. Conversely, choose a basis $u_1,\dots ,u_r$ for the column space, express each column in that basis, and collect the coefficients into row vectors $w_k^T$. This yields $C={\sum }_{k=1}^r{u}_k{w}_k^T$, hence a sum of $r$ pure tensors.

The pure tensor $(2,1)\otimes (-1,3)$ therefore has two-factor tensor rank one, whereas $e_1\otimes f_1+e_2\otimes f_2$ has two-factor tensor rank two. Both have two factors: the number of factors and the minimum number of product summands are different counts. Over the real numbers, the singular value decomposition supplies a useful decomposition $C={\sum }_k{\sigma }_k{u}_k{v}_k^T$, with one nonzero term per matrix rank. The two-factor rank equality does not extend to three or more factors by merely choosing one matrix layout.

For distinct fermionic orbitals obeying the full canonical anticommutation relations, fix the Fock state order $|11⟩=a^{†}{b}^{†}|0⟩$. With real $\epsilon ,\Delta >0$, the even-fermion parity matrix is $\left(\begin{array}{cc}0& -\Delta \\ -\Delta & 2\epsilon \end{array}\right)$, since $ab|11⟩=-|0⟩$. The two odd-fermion parity energy eigenvalues are both $\epsilon$. The even-fermion parity energy eigenvalues are $\epsilon \pm \sqrt{{\epsilon }^2+{\Delta }^2}$. This finite model isolates the relative sign and constant shift required in a fermionic Bogoliubov transformation without any momentum sum or particle-hole doubling factor.

A two-qubit parity phase gate multiplies odd-parity computational vectors by $e^{i\theta }$ and fixes even-parity ones. With $R_{\theta }=\mathrm{diag}(1,e^{i\theta })$,The first controlled-NOT gate computes $x\oplus y$ in the second wire, the phase gate multiplies by $e^{i\theta (x\oplus y)}$, and the last controlled-NOT gate restores the original bits. This compute-phase-uncompute construction leaves the data unchanged except for the desired phases. The more general Pauli-string phase by parity computation uses the same mechanism. In terms of the Pauli Z gate, $G_{\theta }=e^{i\theta /2}\exp [-i\theta (Z\otimes Z)/2]$; the scalar factor ensures equality as a matrix, rather than only up to global phase.

For $-{\psi }^{\prime \prime }+\cos (\pi x)\psi =\lambda \psi$ on $(0,1)$ with zero Dirichlet boundary conditions, use $\psi =\sin (\pi x)+\alpha \sin (2\pi x)$. Its Rayleigh quotient is $({\pi }^2+\alpha +4{\pi }^2{\alpha }^2)/(1+{\alpha }^2)$. In the normalized sine basis the restricted operator has matrix $\left(\begin{array}{cc}{\pi }^2& 1/2\\ 1/2& 4{\pi }^2\end{array}\right)$. Its smaller eigenvalue gives the displayed upper bound, attained within this trial space at $\alpha =3{\pi }^2-\sqrt{9{\pi }^4+1}$. The other stationary parameter gives the larger restricted eigenvalue. This is a finite-dimensional Rayleigh-Ritz method for a regular Sturm-Liouville problem with a sign-changing potential.

After a computational-basis projective measurement, an observer who lacks the result assigns a diagonal density matrix. If the observer's preparation prior has mean $q=\mathbb{E}|\alpha {|}^2$, the assignment is the displayed matrix. No outcome record and no stated prior do not logically force $q=1/2$; that value follows from a symmetric or maximum-entropy ignorance convention. An observer who knows the preparation instead uses $q=|\alpha {|}^2$.

For a uniformly bounded two-level Fourier companion matrix $C(\theta )$ with distinct roots $g_1,g_2$ of modulus at most one, its powers satisfyA positive frequency-uniform gap therefore gives a uniform bound on every power. The Parseval identity then supplies spatial L2 norm stability. When the gap closes at a repeated unit root, a Jordan block can produce growth proportional to the number of steps despite both roots having modulus one.

A unimodular matrix is a square integer matrix whose determinant is $1$ or $-1$. The decisive property is that its inverse matrix also has integer entries: multiplication by $U$ gives a bijection of the integer lattice ${\mathbb{Z}}^n$ onto itself. Thus an invertible linear map can preserve not only real coordinates but the requirement that every coordinate is an integer. These matrices form the integral general linear group.^[ref]

For an original two-dimensional example, takeIts determinant is one, and it maps $(a,b)$ to $(a+b,b)$. Conversely, the integer point $(x,y)$ comes from $(x-y,y)$. For instance, $(3,2)$ has new coordinates $(1,2)$ relative to the columns $(1,0)$ and $(1,1)$. No integer point disappears, although the basis vectors and their fundamental cell change shape.

This is useful whenever coordinates must remain integral. A unimodular change of lattice basis can give a shorter or more convenient description without changing the objects represented. Smith normal form and Hermite normal form use controlled integer operations to simplify integer matrices; ordinary Gaussian elimination may divide by numbers that destroy integrality. The geometric freedom is substantial: a sheared basis may look completely different while describing exactly the same infinite set of integer points.

A determinant of $-1$ is allowed too. Swapping two basis columns reverses orientation while preserving the integer lattice. The condition is therefore preservation of the complete integer structure, rather than a requirement to preserve orientation.

A square matrix $U$ is unipotent when $U-I$ is nilpotent. Over an algebraically closed field, this is equivalent to all eigenvalues being one. If $j$ belongs to a nilpotent ideal of an operator algebra, $I+j$ is a unipotent matrix. This realizes the radical factor in the Levi decomposition of a quiver automorphism group.

Let $A$ be a real positive-definite matrix and minimize the displayed quadratic function over all of ${\mathbb{R}}^n$. Its gradient is $Ax-b$, so its only stationary point solves the linear system $A{x}_{\ast }=b$. The matrix is an invertible matrix, making this solution exist and be unique. Completing the square around that solution proves the stronger global statement:The quadratic function is strictly convex, and its level sets above the minimum are translated ellipsoids. Positivity therefore turns solving a linear system into locating the bottom of a bowl.^[ref]

For the opening matrix $A=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)$, take $b=(1,1{)}^T$ and $c=0$. Then $A^{-1}=\frac{1}{8}\left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)$, $x_{\ast }=(1/4,1/4{)}^T$, and $f(x_{\ast })=-1/4$. Every displacement $h$ raises the value by $\frac{1}{2}{h}^{T}Ah$; along the unit eigenvectors $e_{+}$ and $e_{-}$ the increases are respectively $2s^2$ and $s^2$ for displacement $s{e}_{+}$ or $s{e}_{-}$.

For a singular positive semidefinite matrix, a solution of $Ax=b$ is still a global minimizer, but adding any kernel vector gives another. If $b$ is outside the image of a linear map defined by $A$, some kernel vector $v$ has $b^{T}v\ne 0$, and $f(sv)=c-s{b}^{T}v$ is unbounded below. For an indefinite quadratic form, a negative eigenvalue gives a direction along which the quadratic term tends to $-\infty$ regardless of $b$. Thus a stationary point alone does not establish a minimum. The same energy-difference identity extends to the quadratic variational principle for a symmetric positive operator, with additional existence hypotheses in infinite dimension.

Here is a short proof that uses only the row space. For a vector subspace $W\subseteq {\mathbb{F}}^n$, let $d_j$ be the dimension of its projection map onto the first $j$ coordinates, with $d_0=0$. In any reduced row echelon form with row space $W$, $d_j$ counts the pivot columns among those first $j$ columns: later-pivot rows project to zero, while earlier-pivot rows remain linearly independent. Thus the positions where $d_j-d_{j-1}=1$ determine all pivot positions from $W$ alone.

If those positions are $p_1,\dots ,p_r$, the map sending $w\in W$ to $(w_{p_1},\dots ,w_{p_r})$ is an isomorphism $W\to {\mathbb{F}}^r$: the reduced nonzero rows map to the standard basis. Its $i$th row is therefore the unique vector of $W$ with pivot-coordinate values equal to the $i$th standard basis vector. Any other reduced row echelon form with that row space has identical nonzero rows in identical order. At a fixed matrix size, the remaining rows must be zero. Together with existence by Gaussian elimination, this proves uniqueness.

A complex matrix $Z=-Z^T$ can be put by unitary congruence into two-by-two skew blocks and zero blocks; the magnitudes $|z_r|$ are its singular values, each repeated twice. One proof uses the antilinear map $A(v)=Z\bar{v}$. Because $Z^{†}=-\bar{Z}$, $A^2=-Z{Z}^{†}$. On a nonzero eigenspace of $Z{Z}^{†}$ with eigenvalue $s^2$, any unit vector $v$ has orthogonal partner $A(v)/s$, and $A(A(v)/s)=-sv$. These orthonormal pairs give the blocks; the orthogonal complement is invariant, so the procedure repeats. Phases of the paired vectors set the block phases.

For a number field $K$ and an algebraic integer $\alpha \in {\mathcal{O}}_K$, the criterion isHere the absolute field norm means the field norm down to $\mathbb{Q}$; its absolute value is one. Necessity follows from $N(\alpha )N({\alpha }^{-1})=1$ with both factors integers. Conversely, multiplication by $\alpha$ on an integral basis is a matrix with integer entries and determinant $\pm 1$. Its inverse also has integer entries, and applying that inverse to $1$ gives ${\alpha }^{-1}\in {\mathcal{O}}_K$. The integrality hypothesis matters: $\alpha =(3+\sqrt{2})/(3-\sqrt{2})=(11+6\sqrt{2})/7$ has field norm one, but is not in $\mathbb{Z}[\sqrt{2}]$.

In $\mathbb{Z}[\sqrt{2}]$, the criterion becomes $a^2-2b^2=\pm 1$. Thus $(1+\sqrt{2}{)}^{-1}=\sqrt{2}-1$ and $(3+2\sqrt{2}{)}^{-1}=3-2\sqrt{2}$ are again algebraic integers. These units produce infinite families of solutions to the Pell equation.

The unitriangular group ${\mathrm{UT}}_n(F)$ consists of upper triangular matrices over a field $F$ with every diagonal entry equal to one. Every such matrix is $I_n+N$ for a strictly upper triangular matrix $N$. Products have the same form, and $N^n=0$ gives the inverseThus these matrices form a subgroup of the general linear group. The case ${\mathrm{UT}}_3({\mathbb{F}}_3)$ is noncommutative but every element has cube equal to the identity matrix: $(I_3+N{)}^3=I_3$ because $N^3=0$ and $3=0$ in the finite field ${\mathbb{F}}_3$.

For a three-dimensional strictly upper triangular matrix $A$, the binomial theorem stops after $A^2$. Over a finite field of odd prime characteristic $p$, both coefficients $p$ and $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ vanish, so every nonidentity matrix in $U{T}_3({\mathbb{F}}_p)$ has order $p$. In characteristic two, a matrix with nonzero $A^2$ instead has order four.