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A Cartesian basis is an ordered basis of unit vectors along the mutually perpendicular axes of a Cartesian coordinate system. It is an orthonormal basis of a real Euclidean space: $e_i\cdot e_j={\delta }_{ij}$, with ${\delta }_{ij}$ the Kronecker delta. In the plane, the familiar standard basis is $e_1=(1,0{)}^T$, $e_2=(0,1{)}^T$; the vector $p=(2,1{)}^T$ is $2e_1+e_2$. A rotated pair of perpendicular unit vectors is also a Cartesian basis, even though its basis vectors have different standard basis entries.

The advantage is that coordinates can be measured separately by dot products, and the squared norm is the sum of their squares:These are the real Euclidean space instances of the orthonormal basis expansion.^[ref]

An orthogonal basis only requires perpendicular basis vectors; their lengths need not be one. For example, $(2,0{)}^T,(0,3{)}^T$ are orthogonal, but $p=(2,1{)}^T$ has coordinates $(1,1/3{)}^T$, while its dot products with them are $4,3$. For an orthogonal basis $(a_i)$, the coordinate along $a_i$ is $(p\cdot a_i)/(a_i\cdot a_i)$. Dividing each nonzero basis vector by its norm makes it a unit vector without changing its direction. Unit length alone does not imply orthogonality: $(1,0{)}^T$ and $(1,1{)}^T/\sqrt{2}$ are both unit vectors but have dot product $1/\sqrt{2}$.

The examples below show why these distinctions matter: rotating a Cartesian basis changes the coordinate vector while keeping lengths easy to compute; using nonperpendicular basis vectors still gives unique coordinates, but requires the dual basis or a matrix inverse to extract them.

Starting observations at a finite time, subtract the best linear predictor based on the available earlier observations. The resulting residual is orthogonal to their linear span. The unit-triangular relation between observations and residuals makes the residuals an orthogonal basis of the same finite observation space. These differ from infinite-past innovations until the initialization effect disappears.

For the fundamental representation of SU(N), take traceless Hermitian generators with $\mathrm{Tr}(T^a{T}^b)={\delta }^{ab}/2$. Along with the identity matrix, they form an orthogonal basis of all complex matrices. Expanding a matrix in this basis and subtracting its identity component proves the displayed completeness relation. It separates exchanged-index and unchanged-index contributions to color amplitudes.

An orthogonal basis of an inner-product vector space consists of nonzero mutually orthogonal vectors that span the space. The orthogonal projection of $v$ onto a finite-dimensional span is ${\sum }_i⟨v,u_i⟩u_i/⟨u_i,u_i⟩$. Dividing each vector by its norm produces an orthonormal basis. For centered random variables, the inner product is covariance; an orthogonal basis therefore consists of uncorrelated variables, without requiring unit variance.

A finite-dimensional linear subspace $W$ of an inner product space has a unique closest vector to any given vector $x$. This is its orthogonal projection $P_{W}x$. If $e_1,\dots ,e_d$ is an orthogonal basis of $W$, compute it one direction at a time:The inner product convention here is linear in its first argument, including over the complex numbers. For an orthonormal basis, each denominator is one. When $W$ consists only of the zero vector, the empty sum gives $P_{W}x=0$. The finite-dimensional formula is described in Francis Narcowich, Inner Products and Norms: minimization problems.

Take $W=\mathrm{span}\{(1,2{)}^T\}$ and $x=(3,1{)}^T$, using the dot product. With $e=(1,2{)}^T$,The orthogonal projection residual has $r\cdot e=2-2=0$. To verify the best approximation directly, every candidate in the line is $te$, andThe smallest squared Euclidean norm error is therefore $5$, attained only at $t=1$. The diagram connects the perpendicular residual to this one-variable quadratic polynomial.

The general mechanism is the Pythagorean theorem in an inner-product space. For any $w\in W$, the vectors $r=x-P_{W}x$ and $P_{W}x-w$ are orthogonal, soMoving away from $P_{W}x$ inside $W$ adds a nonnegative squared norm, which is zero only when $w=P_{W}x$. This proves both optimality and uniqueness.

The useful feature is separation of preparation from evaluation. Construct an orthonormal basis once with the Gram-Schmidt process, then obtain each new best approximation using inner products instead of a fresh optimization problem. When $W$ is the column space of a matrix, this gives linear least squares; when $W$ is spanned by orthogonal polynomials, it gives a least-squares polynomial in an orthogonal-polynomial basis. The ambient inner product space need not be a Hilbert space for this finite-dimensional construction. In an infinite-dimensional target linear subspace, existence requires additional hypotheses, supplied by orthogonal decomposition by a closed subspace.

For positive integer indices, distinct members are orthogonal and each has nonzero squared norm $\pi$. Thus $\{\sin (nx):n\ge 1\}$ is an orthogonal basis in the Hilbert space of odd L2 functions on $(-\pi ,\pi )$. The zero-index member must be omitted. More precisely this is a complete Fourier sine basis, with generally infinite L2 norm convergence, rather than a Hamel basis with finite linear combinations. A claim about arbitrary real functions requires suitable integrability and a specified sense of convergence.

After discarding $f_0=0$, the preceding inner products show that distinct sine functions are orthogonal; dividing each by $\sqrt{\pi }$ makes them orthonormal. More precisely, their infinite expansions form an orthogonal basis of the odd part of $L^2(-\pi ,\pi )$ in the Hilbert-space sense. This is not a claim that every arbitrary odd function is a finite linear combination of sines.

Take the inner product of $x={\sum }_{n\ge 1}{a}_n\sin (nx)$ with $\sin (mx)$. Orthogonality givesTherefore the Fourier series of a periodic sawtooth, scaled to this interval, isIts periodic extension is piecewise smooth, so the Dirichlet convergence theorem for Fourier series gives pointwise convergence to $x$ inside the interval. At the jump endpoints of that periodic extension, the series instead converges to zero, the midpoint of the two one-sided limits.

For admissible initial data, for example a square-integrable function on the interval, expand in the sine eigenfunctions of the Dirichlet Laplacian:The expansion follows either from separation of variables or the complete sine orthogonal basis. If $\kappa <{\pi }^2$, every growth rate is negative and Parseval's identity givesThus all finite-energy data decay. The convergence is also uniform in $x$ after any fixed positive time: apply Cauchy-Schwarz inequality to the sine series, using the summable factors $e^{-2(j^2-1){\pi }^{2}t}$, to bound its supremum by a constant times $e^{-({\pi }^2-\kappa )t}$ for $t\ge t_0>0$.

Conversely choose $u(x,0)=\sin \pi x$. Then $u(x,t)=e^{(\kappa -{\pi }^2)t}\sin \pi x$, which is stationary at equality and grows when $\kappa >{\pi }^2$. All initial conditions decay if and only if $\kappa <{\pi }^2$. This is the continuous part of the reaction-diffusion spectral decay threshold.

Write $h=\Delta x=1/(M+1)$. The semidiscrete matrix is $B=h^{-2}\mathrm{tridiag}(1,-2,1)+\kappa I$. For $j=1,\dots ,M$, set $v_m^{(j)}=\sin (j\pi m/(M+1))$. The endpoint values vanish and the sine addition formula givesThese form an orthogonal basis of eigenvectors. Every initial vector decays precisely when every eigenvalue is negative. Since the largest is ${\lambda }_1$, the necessary and sufficient condition isNecessity follows by taking the first eigenvector as the initial data; at equality it remains unchanged. Sufficiency follows by decomposing any vector into the displayed decaying modes. The discrete reaction-diffusion spectral decay threshold is below ${\pi }^2$, since $\sin x<x$ for $x>0$, and equals ${\pi }^2-{\pi }^4{h}^2/12+O(h^4)$ as the mesh is refined. Therefore a coarse semidiscretization can spuriously grow for some parameters for which the continuous equation still decays.

A vector space over the real numbers is a set $V$ with an addition making it an abelian group and a scalar multiplication by $\mathbb{R}$ satisfying, for scalars $a,b$ and vectors $u,v$, the identities $a(u+v)=au+av$, $(a+b)v=av+bv$, $a(bv)=(ab)v$ and $1v=v$. A vector subspace is a nonempty subset closed under addition and scalar multiplication; a proper vector subspace is one unequal to $V$. A spanning set is a set whose finite linear combinations give every vector of $V$. A basis is a linearly independent spanning set. The dimension is the number of vectors in a basis, or its cardinality when the basis is infinite.

The sum of vector subspaces and their intersection of vector subspaces areBoth are vector subspaces: addition and scalar multiplication preserve the displayed forms or simultaneous membership. The intersection of vector subspaces is never empty because $0\in U\cap W$.

For the two given hyperplanes, adding and subtracting their defining equations gives $x_1=x_2$ and $x_3=x_4$. HenceThese two nonzero vectors are orthogonal, and consequently form an orthogonal basis of the intersection of vector subspaces. Put $u=(1,-1,-1,1)$ and $w=(1,-1,1,-1)$. Direct substitution gives $u\in U$, $w\in W$. Their dot products with $b_1,b_2$ vanish, and $u\cdot w=0$. Each defining equation of $U$ or $W$ is one nonzero linear constraint in four variables, so each vector subspace has dimension three. Thus each space has the indicated orthogonal basis:The last four vectors are nonzero and mutually orthogonal, so they are a basis of ${\mathbb{R}}^4$. They lie in $U+W$, proving $U+W=V$. The dimension identity is therefore verified numerically:

Assume the consistency regularity implicit in the requested rate, for example $u\in C^4$ on the closed square. Bounded fourth derivatives and Taylor's theorem give a uniform residual ${\tau }_{ij}={\Delta }_{h}u(ih,jh)-f(ih,jh)$ with $|{\tau }_{ij}|\le C_0{h}^2$. Subtracting the exact grid values from the numerical equations gives ${\Delta }_{h}e=-\tau$, with zero boundary error.

Let $A=-h^2{\Delta }_h$ be the unscaled positive Dirichlet matrix. Its eigenvectors are products of discrete sine vectors, which form an orthogonal basis; direct substitution gives eigenvaluesSince $\sin (\pi h/2)\ge h$ for $0<h\le 1$, ${\lambda }_{11}\ge 8h^2$. Therefore $\Vert A^{-1}{\Vert }_2\le 1/(8h^2)$. The error equation is $Ae=h^2\tau$, and its residual has norm $\Vert \tau {\Vert }_2\le M{C}_0{h}^2\le C_{0}h$. Combining the bounds yieldsThis is the unweighted grid error for the five-point Poisson formula. The unweighted norm loses one power of $h$ because the grid has order $h^{-2}$ points; the area-weighted norm is second-order accurate. A regularity assumption is needed for the uniform truncation estimate and hence for this rate; the question does not specify it explicitly.

The same Weyl integration formula for SU2 gives $⟨{\chi }_n,{\chi }_m⟩={\delta }_{nm}$. More strongly, these characters are complete in the space of square-integrable class functions: multiplication by $\sin \theta$ identifies that weighted space with $L^2(0,\pi )$, and sends ${\chi }_n$ to $\sin ((n+1)\theta )$. The sine functions form a complete orthogonal basis of $L^2(0,\pi )$, as follows from the Fourier series theorem applied to odd extensions.

Suppose a finite-dimensional continuous irreducible group representation were not isomorphic to any $V_n$. Character orthogonality would make its character orthogonal to every ${\chi }_n$. Completeness forces its character to vanish in $L^2$, and continuity then forces it to vanish everywhere. This is impossible at the identity, where the character equals the positive dimension of the representation. Hence $\boxed{\text{every such irreducible representation is some }{V}_n}$. Since the dimensions $n+1$ differ, the index is unique.

Take the boundary-fitted grid $h=1/(M+1)$ and assume the standard consistency of a numerical method hypothesis $u\in C^4(\overline{\Omega })$ with bounded fourth partial derivatives. Taylor expansion along each coordinate, including a bounded fourth-order remainder, gives the five-point Laplacian truncation errorThe numerical equation and exact grid values therefore imply $A_h\mathbf{e}=\mathbf{\tau }$, where $A_h=-{\Delta }_h$ with zero boundary values. Its discrete sine transform eigenvectors are $\sin (\pi rih)\sin (\pi sjh)$, $1\le r,s\le M$, and its eigenvalues areThe one-dimensional sine vectors form an orthogonal basis, hence so do their tensor products. The least eigenvalue occurs at $r=s=1$. Since $\sin (\pi h/2)\ge h$ for $0<h\le 1$, ${\lambda }_{11}\ge 8$, so the Euclidean operator norm of $A_h^{-1}$ is at most $1/8$. There are $M^2$ residual components, and $Mh<1$, givingThe factor $M$ matters: pointwise second-order consistency of a numerical method becomes first order in the unweighted Euclidean grid norm.

The PDF does not explicitly impose regularity on the exact solution. Without such a hypothesis its unrestricted error claim is false. Here is a counterexample, with a classical solution smooth in the open square and continuous with zero boundary values:The sampled numerical solution separates as $v_i\sin (\pi jh)$. Its error $e_i=v_i-g(ih)$ solves $Ae=\tau$, where $A=-{\Delta }_{h,x}+{\lambda }_{y}I$ and ${\lambda }_y=4h^{-2}{\sin }^2(\pi h/2)$. The one-dimensional residual is ${\tau }_i={\Delta }_{h,x}g(ih)-g^{\prime \prime }(ih)+({\pi }^2-{\lambda }_y)g(ih)$. At the first node,The leading coefficient is strictly negative. Moreover $g^{\prime \prime \prime \prime }<0$ on $(0,1)$, so the triangular-average formula for the second difference and concavity of $g^{\prime \prime }$ give ${\Delta }_{h,x}g\le g^{\prime \prime }$. Thus the positive part of every residual is at most $C{h}^2$. The matrix $A$ has a nonnegative inverse, and its first diagonal inverse entry is at least $1/(2h^{-2}+{\lambda }_y)$, as follows from the nonnegative Neumann expansion of its tridiagonal matrix. The discrete comparison principle bounds the inverse image of the residual's positive part by $C{h}^2$, while the negative first residual gives $e_1\le -c\sqrt{h}+C{h}^2$. Hence $|e_1|\ge c^{\prime }\sqrt{h}$ for small $h$. Since ${\sum }_{j=1}^M{\sin }^2(\pi jh)=1/(2h)$, the full Euclidean error is bounded below by a positive constant. It cannot be $O(h)$. Thus the boxed estimate holds for the intended smooth solution class; a regularity condition is genuinely necessary.

The autocovariance of an MA(1) process comes directly from shared noise terms:Work with the zero-mean white noise convention and ${\sigma }^2>0$. Put ${\mathcal{H}}_{t-1}=\mathrm{span}(X_1,\dots ,X_{t-1})$. Suppose the preceding finite-past innovations $U_1,\dots ,U_{t-1}$ are mutually orthogonal. Their triangular relation to the observations means that they span the same finite past. For $s\le t-2$, $U_s$ is a linear combination of $X_1,\dots ,X_s$, so $\mathrm{Cov}(X_t,U_s)=0$. Moreover,Projection onto this orthogonal basis therefore has only its last coefficient nonzero:The residual is orthogonal to the entire preceding span, proving the induction. Expanding its variance givesThe initialization is $U_1=X_1$, ${\phi }_1={\gamma }_0$, ${\lambda }_1=0$ and $U_0=0$. Each finite covariance matrix is a positive-definite matrix: the last observation in any nonzero finite linear combination contains a noise term absent from earlier observations. Thus all the projection denominators are positive. These are the finite-sample innovations of an MA(1) process, requiring no Gaussian assumption.

Substituting the covariances into the recursion givesThe denominator has a positive limit when ${\lambda }_{t-1}\to \lambda \in [-1,1]$. Taking limits gives $(\lambda -\theta )(\theta \lambda -1)=0$. The permitted interval selectsFor $\theta =0$, every coefficient is zero; at $\theta =\pm 1$, the roots coincide. This is the limiting MA(1) innovations coefficient. The limiting innovation variance is correspondingly ${\sigma }^2\max (1,{\theta }^2)$.

For the observed numerical case, ${\gamma }_0=2$ and ${\gamma }_1=-1$. The exact innovation calculations areTherefore the predictor of $X_3$ based on the two observations and its error variance areFor $X_4$, the available information is still only ${\mathcal{H}}_2$. Since its covariances with both $X_1$ and $X_2$ are zero, its orthogonal projection onto that span is zero:This is two-step prediction for an MA(1) process. Applying the next one-step recursion would instead use an observed $X_3$, which is not available here.

With the standard real normalization, the Fourier series of a period-$2L$ function iswhereFor the partial sum $S_N$, Fourier orthogonality eliminates all cross terms. The constant term contributes $2L(a_0/2{)}^2$, while every squared sine or cosine integrates to $L$, givingFor a square-integrable function, completeness of the trigonometric orthogonal basis in the Hilbert space $L^2[-L,L]$ makes $S_N\to f$ in norm. Passing to the limit proves Parseval's identity:Square integrability is the needed hypothesis here; periodicity alone is not an integrability or convergence condition.

Take the dot product of the orthogonal basis expansion with ${\mathbf{u}}_j$. All terms except the $j$th vanish by orthogonality, leavingThereforeThe basis is not assumed normalized, so its squared lengths must remain in the denominators. Applying $S$ and using $S{\mathbf{u}}_i={\lambda }_i{\mathbf{u}}_i$ gives

Definition and equations. A normal mode is a motion in which every component of a linear system has the same time dependence with fixed amplitude ratios and phases. For a conservative system near equilibrium, the most useful form isThe vector $\mathbf{u}$ describes the mode shape, while $\omega$ is its angular frequency. Linear combinations of modes describe arbitrary sufficiently small motions when the modes form a complete basis.

Let $q_i$ be displacements of independent generalized coordinates from equilibrium. Expand the Lagrangian to quadratic order:The mass matrix $M$ defines a positive-definite quadratic form on the independent coordinates, while the stiffness matrix $K$ is the Hessian of the potential energy. Substituting a mode gives the generalized eigenvalue problem for small oscillationsA practical calculation identifies the equilibrium, chooses independent coordinates, computes the quadratic kinetic and potential energies, solves for the eigenvalues and then obtains their eigenvectors. One can equivalently diagonalize the real symmetric matrix $M^{-1/2}K{M}^{-1/2}$. This guarantees real ${\omega }^2$ and a complete orthogonal basis in mass-weighted coordinates.

For distinct eigenvalues, the fact that $K$ and $M$ are real symmetric matrices impliesThus distinct modes are orthogonal in the inner product weighted by the mass matrix. Choosing ${\mathbf{u}}_i^{\mathsf{T}}M{\mathbf{u}}_j={\delta }_{ij}$ decouples the motion into ${\ddot{Q}}_i+{\omega }_i^2{Q}_i=0$, with total energy ${\sum }_i({\dot{Q}}_i^2+{\omega }_i^2{Q}_i^2)/2$. A positive-definite matrix $K$ gives positive eigenvalues and oscillatory stability. Zero modes can describe rigid translations or rotations without restoring force; a negative eigenvalue gives exponential instability rather than an oscillatory mode.

Symmetry. A symmetry preserving both quadratic energies maps modes into modes of the same frequency. Coordinates adapted to symmetry divide the eigenvalue problem into smaller independent blocks. For two identical subsystems, exchanging them gives symmetric and antisymmetric modes. The degeneracy of normal modes means that independent modes share one frequency. Higher-dimensional group representations of a symmetry can enforce such degeneracies; symmetry does not require every pair of modes to be degenerate, and accidental degeneracies can also occur. Within a degenerate eigenspace, orthogonal linear combinations are equally valid mode shapes.

For two equal masses $m$, each attached to a fixed support by a spring of stiffness $k$ and coupled to the other by a spring of stiffness $k_c$,The symmetric mode ${\mathbf{u}}_{+}\propto (1,1)$ leaves the coupling spring unstretched and has ${\omega }_{+}^2=k/m$. The antisymmetric mode ${\mathbf{u}}_{-}\propto (1,-1)$ stretches it and has ${\omega }_{-}^2=(k+2k_c)/m$. Superposing the two gives beats when their frequencies are close.

Continuous systems. The same idea replaces matrices by differential operators and mode vectors by functions satisfying boundary conditions. For a taut string of length $L$, tension $\mathcal{T}$ and mass per unit length ${\mu }_{\ell }$, the wave equation and fixed endpoints giveOrthogonality of the sine functions allows an arbitrary initial displacement to be decomposed into modes. Changing the endpoints changes the spectrum: boundary conditions are part of the physical mode problem.

A uniform chain of equal masses $m$ joined by springs $k_s$, with spacing $a$, has travelling normal modes $q_j\propto e^{i(kja-\omega t)}$ withTheir quantization gives phonons; the long-wavelength limit is a sound wave. Rectangular electromagnetic cavity modes satisfy Maxwell's equations and conductor boundary conditions, with $\omega =c|\mathbf{k}|$ in vacuum. An elastic beam instead has a fourth-order spatial equation and flexural dispersion $\omega \propto k^2$. These examples show how restoring forces and geometry determine the frequency spectrum.

Real systems. Small-amplitude motion is often approximately linear even when the full equations are nonlinear. Large amplitudes produce frequency shifts, mode coupling and transfer of energy between modes. Friction broadens resonances and makes free oscillations decay. A damping matrix proportional to $M$ or $K$ can retain the undamped mode shapes; general damping need not, and the appropriate eigenfrequencies and modes can be complex. External periodic forcing predominantly excites modes near resonance, with amplitude set by coupling to the forcing and dissipation. Imperfections split ideal degeneracies, while radiation or coupling to a surrounding fluid can remove energy from the chosen subsystem. Normal modes remain useful when linearization, the boundary conditions and the treatment of damping accurately describe the motion being measured.

The identity and the three Pauli matrices form an orthogonal basis for the complex vector space of two-by-two matrices, with Hilbert-Schmidt inner product $\mathrm{Tr}(P^{†}Q)=2{\delta }_{PQ}$. Thus every operator has the expansion $A=\frac{1}{2}{\sum }_{P\in \{I,X,Y,Z\}}\mathrm{Tr}(PA)P$. For a density matrix, the identity coefficient is $1/2$ and the remaining real coefficients are half the components of its Bloch vector. This converts unitary conjugation into transformations of three real components.