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A real two-mode Bogoliubov transformation preserving the full canonical anticommutation relations. The self anticommutators become $\{a,a^{†}\}=\{b,b^{†}\}=u^2+v^2$, while $\{a,b\}=-uv+uv=0$ and every other cross anticommutator vanishes. Thus its matrix is an orthogonal matrix. For the two-orbital fermionic pairing Hamiltonian, let $E=\sqrt{{\epsilon }^2+{\Delta }^2}$, take $u=\sqrt{(E+\epsilon )/(2E)}$ and $v=\sqrt{(E-\epsilon )/(2E)}$, and obtain $H=E(c^{†}c+d^{†}d)+\epsilon -E$. Inverting gives $c=ua-v{b}^{†}$, $d=v{a}^{†}+ub$.

Rotate the standard basis through $30^{\circ }$ using the rotation matrix whose columns are the new basis vectors:Here $u\cdot v=0$ and $u\cdot u=v\cdot v=1$, so $Q$ is an orthogonal matrix with $Q^{T}Q=I$, where $T$ denotes transpose and $I$ is the identity matrix. For the same vector $p=(2,1{)}^T$, its coordinate vector in this Cartesian basis isThus $\alpha \approx 2.232$ and $\beta \approx -0.134$: travel forwards along $u$ and a little backwards along $v$ to reconstruct $p=\alpha u+\beta v$. The orthogonal projections onto the two axes give these signed coordinates because the measuring vectors are both perpendicular and of unit length.

The squared norm remainsThis is useful when choosing axes aligned with a geometric feature: the coordinates change but distances and dot products retain the same sum-of-products formula. The two directions of the change of basis must be kept straight: $Q$ reconstructs the vector in old coordinates, $p=Q[p{]}_{\mathcal{U}}$, and $Q^T$ extracts its new coordinates. Applying $Q$ to the old coordinate vector $p$ instead would actively rotate the vector.

A short proof explains both formulas. If $p=Qc$, multiplication by $Q^T$ gives $c=Q^{T}p$. Then $\Vert p{\Vert }^2=c^T{Q}^{T}Qc=c^{T}c$. Perpendicular unit-length columns make the matrix inverse equal the transpose; this is the computational convenience supplied by a Cartesian basis.

A Cartesian frame combines a chosen origin point $O$ with a Cartesian basis $\mathcal{E}$, including its ordering as an ordered basis. A point $P$ has coordinate vector $c$ when $\overrightarrow{OP}={\sum }_i{c}_i{e}_i$. The basis supplies measuring directions and unit lengths; the coordinate vector is the resulting tuple of numbers; the Cartesian frame additionally fixes where measurement starts. A vector already belongs to a vector space with a zero vector, so expressing it in a basis requires no additional choice of spatial origin.

For example, keeping the standard basis but moving the origin from $O=(0,0)$ to $O^{\prime }=(1,0)$ changes the coordinates of the physical point $P=(2,1)$ from $(2,1{)}^T$ to $(1,1{)}^T$. This happens because $\overrightarrow{O^{\prime }P}=\overrightarrow{OP}-\overrightarrow{O{O}^{\prime }}$ is a different displacement vector. By contrast, the two diagrams above keep $O$ fixed and change only the basis. In a fixed reference Cartesian frame, a new frame with origin column $o$ and basis-column matrix $Q$ describes points by $p=o+Qc$; the new coordinates are $c=Q^T(p-o)$ when $Q$ is an orthogonal matrix.

For a real skew-symmetric matrix $A$, $I-A$ is invertible: $(I-A)x=0$ would imply $\Vert x{\Vert }^2=x^{T}Ax=0$. Its Cayley transform is an orthogonal matrix, since transposition gives $Q^T=(I+A{)}^{-1}(I-A)$ and the commuting factors cancel in $Q^{T}Q$. Also $\det (I+A)=\det ((I+A{)}^T)=\det (I-A)$, so $\det Q=1$ in every dimension. If $Av=\lambda v$, then $Qv=(1+\lambda )(1-\lambda {)}^{-1}v$. For $\lambda =it$, the transformed eigenvalue is $e^{2i\arctan t}$. In odd dimension $\det A=\det (-A)=-\det A$, so $A$ has a nonzero kernel and $Q$ fixes that direction.

For a real three-dimensional linear map $A$, preservation of every surface area is equivalent to being an orthogonal matrix. For necessity, diagonalize the symmetric matrix $A^{T}A$ in an orthonormal basis, with nonnegative eigenvalues ${\lambda }_i$. The three unit coordinate parallelograms have image areas $\sqrt{{\lambda }_1{\lambda }_2}$, $\sqrt{{\lambda }_1{\lambda }_3}$, and $\sqrt{{\lambda }_2{\lambda }_3}$. Setting all three equal to $1$ forces every ${\lambda }_i=1$. Sufficiency follows from preservation of dot products and $\Vert \mathbf{u}\times \mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2\Vert \mathbf{v}{\Vert }^2-(\mathbf{u}\cdot \mathbf{v}{)}^2$. This specifically uses dimension three; in dimension two, preservation of planar area alone only requires $|\det A|=1$.

For a second-degree equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$, the number ${\Delta }_2=B^2-4AC$ is the discriminant of a binary quadratic form associated with its quadratic part. For a nonempty, nondegenerate real conic section, a negative value gives an ellipse, zero a parabola, and a positive value a hyperbola. It is invariant under a rotation of orthonormal coordinates.^[ref]

A short explanation uses the symmetric matrixAn orthogonal matrix $R$ changes this matrix to $R^{\mathsf{T}}HR$, whose determinant equals $\det H$. The spectral theorem for real symmetric matrices then removes the mixed term. Equal-sign, opposite-sign and one-zero eigenvalues give the three types after the remaining linear terms and offsets are considered.

The qualification matters. The equations $x^2+y^2=0$, $y^2=0$ and $xy=0$ describe a point, a repeated straight line, and two crossing straight lines, respectively. Their quadratic discriminants have the signs expected of ellipse, parabola and hyperbola types, but all three are degenerate conics. Also, $x^2+y^2=-1$ has an empty set of real solutions despite its negative quadratic discriminant.

A contracting Euclidean similarity multiplies every Euclidean distance by a constant $c<1$; equivalently it has the form $cRx+b$ for an orthogonal matrix $R$ and translation vector $b$. Finite families of such maps generate compact sets with self-similarity.

If an orthogonal matrix $Q$ satisfies $\dot{Q}=WQ$ for the spin tensor $W^T=-W$, the Jaumann derivative becomes an ordinary derivative in that frame:This follows by differentiating all three factors and using ${\dot{Q}}^T=-Q^{T}W$. It converts a tensor relaxation law using the Jaumann derivative into a componentwise ordinary differential equation.

If a matrix with orthonormal columns is divided into two row blocks, each having at least as many rows as there are columns, it admits simultaneous thin factorizations $A_1=U_{1}C{V}^T$, $A_2=U_{2}S{V}^T$, with orthonormal $U_1,U_2$, an orthogonal matrix $V$, and nonnegative diagonal matrices $C,S$ satisfying $C^2+S^2=I$. To prove this, apply the spectral theorem to $A_1^T{A}_1$; the same orthonormal eigenbasis diagonalizes $A_2^T{A}_2=I-A_1^T{A}_1$. Normalize the nonzero columns of $A_{i}V$ and complete the zero columns to orthonormal sets. The entries of $C,S$ can be written as cosines and sines of angles in $[0,\pi /2]$.

For real rectangular matrices, expanding $(A+H{)}^t(A+H)$ leaves the quadratic remainder $H^{t}H$, whose matrix norm is $O(\Vert H{\Vert }^2)$. Thus the displayed Fréchet derivative is a linear map in the increment $H$. At an orthogonal matrix, its kernel consists of increments $H=A\Omega$ with $\Omega$ skew-symmetric. This is the tangent-space condition obtained by differentiating $A^{t}A=I$.

For real orthogonal matrices in odd dimension $d$, multiplying by their determinant defines a group homomorphism from $O(d)$ onto $SO(d)$. Scalar factors commute, so $\pi (AB)=\pi (A)\pi (B)$, while $\det \pi (A)=(\det A{)}^{d+1}=1$. The map fixes $SO(d)$ and has kernel $\{I,-I\}$. Since $-I$ is central and has determinant $-1$ in odd dimension, $O(d)\cong SO(d)\times C_2$. In even dimension the same expression does not map every orthogonal matrix into the special orthogonal group.

A finite-dimensional linear operator is diagonalizable over a specified field when it has an eigenbasis. Equivalently, its matrix $A$ admitswhere $P$ is an invertible matrix whose columns are eigenvectors, and $D$ is a diagonal matrix of the corresponding eigenvalues. This change of basis reveals a linear map as independent one-dimensional scalings. The matrix may look coupled in the original coordinates, but each coordinate in the eigenbasis evolves on its own.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ computed in eigenvector, takeThen $AP=PD$, hence $P^{-1}AP=D$. The orthonormal basis version is $Q=P/\sqrt{2}$, an orthogonal matrix satisfying $Q^{T}AQ=D$. Every real symmetric matrix has an orthonormal eigenbasis by the spectral theorem; a general diagonalizable matrix need not have orthogonal eigenvectors.

The immediate payoff is repeated application: adjacent factors $P^{-1}P$ cancel, giving $A^n=P{D}^n{P}^{-1}$. In this example,The component in $E_3$ grows while the component in $E_1$ stays fixed. After division by its Euclidean norm, this particular sequence approaches $(1,1{)}^T/\sqrt{2}$. This is the mechanism behind the power method; an initial vector entirely in $E_1$ would stay there and never find the dominant eigenvalue $3$.

Two different obstructions deserve comparison. The planar rotation $R$ in eigenvector is diagonalizable over $\mathbb{C}$ but not over $\mathbb{R}$, because its eigenvalues are $i$ and $-i$. By contrast, the shear mappinghas only $\lambda =1$, of algebraic multiplicity two, and $E_1=\mathrm{span}\{(1,0{)}^T\}$, of geometric multiplicity one. There is no second independent eigenvector, even over $\mathbb{C}$. This is a defective matrix, already a two-dimensional Jordan block. Its powers areThus even though both eigenvalues are one, repeated application can produce growth. Eigenvalues alone miss the coupling recorded by a Jordan block. A repeated eigenvalue is harmless when it has enough independent eigenvectors, as $I$ demonstrates. More generally, a matrix is diagonalizable exactly when its characteristic polynomial splits over the chosen field and every eigenvalue has equal geometric multiplicity and algebraic multiplicity. The proof outline is to join bases of the distinct eigenspaces; their direct sum fills the space exactly under these conditions.

A physical motivation is the separation of normal modes. Consider two equal masses $m$ between three identical springs of stiffness $k$, with the outer ends fixed. By Hooke's law and Newton's second law, the displacement vector $q=(q_1,q_2{)}^T$ obeysThe eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $1$ and $3$: the masses move together in the first normal mode and oppositely in the second. Writing $q=a(t)(1,1{)}^T+b(t)(1,-1{)}^T$ separates the coupled equations into two harmonic oscillators,Their angular frequencies are $\sqrt{k/m}$ and $\sqrt{3k/m}$; the eigenvalues give squared angular frequencies after multiplication by $k/m$. This is the beauty of diagonalization: finding the right directions turns interacting motions into independent patterns. For a physical treatment, see Yen-Jie Lee's "Coupled Oscillators, Normal Modes" (MIT 8.03SC, 2016), and the local article normal modes of two equal masses between three springs.^[ref]

For a differentiable material frame of a rod, there is a unique vector $\mathbf{u}$ such that ${\mathbf{d}}_i^{\prime }=\mathbf{u}\times {\mathbf{d}}_i$. To see this, form the orthogonal matrix $R=({\mathbf{d}}_1,{\mathbf{d}}_2,{\mathbf{d}}_3)$. Differentiation of $R{R}^T=I$ shows that $R^{\prime }{R}^T$ is a skew-symmetric matrix; every three-dimensional skew-symmetric matrix has the unique form $\mathbf{v}\mapsto \mathbf{u}\times \mathbf{v}$. Primes may denote differentiation with respect to reference or current arc length, but that choice must be stated. In a planar bend about ${\mathbf{d}}_2$, reference differentiation gives ${\mathbf{d}}_3^{\prime }=u{\mathbf{d}}_1$ and ${\mathbf{d}}_1^{\prime }=-u{\mathbf{d}}_3$.

Let ${\varphi }_0,\dots ,{\varphi }_n$ be orthonormal polynomials for a positive weight, and let $x_i,w_i$ define its $(n+1)$-point Gaussian quadrature. Put $A_{ij}={\varphi }_{j-1}(x_{i-1})$ and $W=\mathrm{diag}(w_0,\dots ,w_n)$. Exactness through polynomial degree $2n+1$ gives $A^{T}WA=I$. Positivity of the quadrature weights makes $W^{1/2}A$ a square orthogonal matrix, so $A{A}^T=W^{-1}$ also. Thus the unweighted row inner products of the evaluation matrix are ${\delta }_{ij}/w_{i-1}$.

A discrete wavelet transform computes multiscale approximation and detail coefficients from sampled data. For the Haar wavelet, replace each adjacent pair $(u,v)$ byThe approximation coefficient $a$ is $\sqrt{2}$ times the pair's arithmetic mean; it is not the ordinary mean. The detail coefficient $d$ records their signed contrast. Repeat on the approximation coefficients until only one remains, keeping every detail coefficient.

For a worked example, start with the vector $(2,4,6,8)$. The first pass gives approximation coefficients $(3\sqrt{2},7\sqrt{2})$ and detail coefficients $(-\sqrt{2},-\sqrt{2})$. Transforming the remaining pair gives the coarse coefficient $10$ and the half-to-half detail $-4$. In the order coarse value, coarse detail, then fine details, the transform isThis is an orthogonal matrix transformation. Each pair preserves $u^2+v^2=a^2+d^2$, so the full transform preserves the squared Euclidean norm:The inverse function for each pair is $u=(a+d)/\sqrt{2}$, $v=(a-d)/\sqrt{2}$. Reversing the passes therefore recovers every original value. This squared-norm bookkeeping is often called signal energy; it need not represent physical energy. ^[ref]

Continuous and discrete normalizations differ. If $n=2^J$ data values are interpreted as the heights of a step function on $n$ equal cells of $[0,1]$, its normalized L2 inner product coefficients are the discrete coefficients divided by $\sqrt{n}$. For our four values they are $(5,-2,-1/\sqrt{2},-1/\sqrt{2})$, as the Haar projection example shows. The same normalization is used by the wavelet regression estimator on $[0,1]$.

Under independent Gaussian noise with a common variance, orthogonality preserves that coefficient noise variance and independence. It does not make every arbitrary correlated signal independent: this statistical conclusion uses the Gaussian model and its equal-variance independent inputs.

The Euclidean group $E(n)=O(n)⋉{\mathbb{R}}^n$ is the group of all isometries $x\mapsto Ax+b$ of Euclidean space, with $A$ an orthogonal matrix. Its orientation-preserving subgroup is $SE(n)=SO(n)⋉{\mathbb{R}}^n$, the semidirect product of the special orthogonal group with translations. The Euclidean motion of Euclidean three-space and the proper Euclidean motion of Euclidean three-space are the respective three-dimensional instances.

Every self-adjoint operator on a finite-dimensional real or complex inner product space has an orthonormal eigenbasis, and all its eigenvalues are real. In orthonormal coordinates, a real symmetric matrix has the form $A=Q\Lambda Q^T$, with $Q$ an orthogonal matrix and $\Lambda$ a real diagonal matrix. A complex Hermitian matrix instead has $A=U\Lambda U^{\ast }$, with $U$ a unitary matrix and $U^{\ast }$ its conjugate transpose. These are the self-adjoint cases of the spectral theorem; Axler gives the real and complex finite-dimensional statements in Section 7B.^[ref]

An orthonormal eigenbasis turns a seemingly coupled linear map into independent scalar multiplications while preserving lengths and angles. If $x={\sum }_j{c}_j{e}_j$ and $A{e}_j={\lambda }_j{e}_j$, thenThe last identity removes every cross term from the quadratic form. The eigenvalues tell us whether it is positive, negative, or mixed in sign, and the eigenvectors identify the corresponding directions. This connects positive-definite matrices, quadratic forms, and the Hermitian Rayleigh quotient range.

For a fully computed example, takeDirect multiplication gives $A{e}_{+}=3e_{+}$ and $A{e}_{-}=e_{-}$. Their dot product is zero and each has norm one. ThusWriting $x=c{e}_{+}+d{e}_{-}$ changes $x^{T}Ax=2x_1^2+2x_1{x}_2+2x_2^2$ into $3c^2+d^2$. Its level set $x^{T}Ax=1$ is therefore an ellipse with semiaxis lengths $1/\sqrt{3}$ and $1$, along $e_{+}$ and $e_{-}$. The coordinates change, but the ellipse does not: orthogonal diagonalization reveals its existing geometry.

For distinct eigenvalues, orthogonality follows at once from self-adjointness: $\lambda ⟨u,v⟩=⟨Au,v⟩=⟨u,Av⟩=\mu ⟨u,v⟩$. Within a repeated eigenspace, choose an orthonormal basis; arbitrarily chosen eigenvectors of the same eigenvalue need not be orthogonal. One outline of the full argument finds an eigenvector, observes that its orthogonal complement is invariant, and repeats in that smaller space. Merely having real eigenvalues is weaker: $\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has two real eigenvalues, but its eigenvectors $(1,0{)}^T$ and $(-1,1{)}^T$ are not orthogonal.

Take four congruent octagonal regular polygons with four alternate sides labelled $a,b,c,d$. The intervening sides have Neumann boundary conditions. Pair equal face parameters in the positive reference boundary direction. The complete gluing data areAll unpaired labelled faces not in a Dirichlet column also have Neumann boundary conditions. Encode these rules by signed permutation matrices, using a positive swap for a paired interface and diagonal $-1$ or $+1$ for an unpaired Dirichlet boundary condition or Neumann boundary condition. The four-by-four Hadamard matrix $C$ above satisfies $C{M}_s=N_{s}C$ for every label. Explicitly, if its columns are $h_0,h_1,h_2,h_3$, applying $N_a,N_b,N_c,N_d$ gives respectively $(h_0,h_1,h_3,h_2)$, $(h_0,h_2,h_1,h_3)$, $(h_1,h_0,-h_2,-h_3)$, and $(h_0,-h_1,h_2,-h_3)$, which are exactly the columns of the corresponding $C{M}_s$. On every intervening face the face matrix is the identity.

Since $C^{T}C=4I$, the orthogonal matrix $C/2$ preserves the $L^2$ inner product and the sum of Dirichlet energies. Its intertwining relations map matching Sobolev traces and zero Dirichlet traces bijectively to those of the other assembly. It therefore preserves the quadratic form domains and the forms themselves. The associated self-adjoint operators are unitarily equivalent, proving equality of all eigenvalues with their multiplicities for the specified mixed boundary conditions.

The topology of disk tiles glued along disjoint boundary arcs shows that $D$, whose gluing graph is $0,1,2,3$ in a path, is a closed disc and is simply connected and orientable. The gluing graph of $B$ has edges $13,12,02,01,23$; it is connected with two independent cycles, so $\chi (B)=-1$ and ${\pi }_1(B)\cong F_2$. Its odd cycle $1,2,3,1$ makes it a nonorientable surface. Each assembled surface has nonempty polygonal boundary. This explicitly realizes isospectrality between a simply-connected orientable bordered surface and a nonorientable, non-simply-connected one. The boundary corners and mixed boundary conditions are essential qualifications; it does not contradict the Euler characteristic from the heat trace of a bordered surface for smooth boundaries and pure boundary conditions.

A standard real Gaussian random matrix has independent and identically distributed standard normal entries. Each column is an orthogonally invariant random vector: its probability density function is proportional to $\exp (-\Vert z{\Vert }_2^2/2)$. Consequently multiplying on either side by a fixed orthogonal matrix preserves the joint distribution. Its singular values are random; the expressions $\sqrt{m}\pm \sqrt{n}$ for an $m\times n$ matrix describe useful spectral scales, not deterministic endpoints that hold for every realization.

A Givens rotation eliminates one component of a vector by rotating just two coordinates, preserving its Euclidean norm. For example,The length remains five: elimination moves the component into the retained coordinate rather than discarding it. Applied to a matrix, the same operation mixes two rows; QR decomposition by Givens rotations uses these local operations to build an upper triangular matrix.

For indices $i<j$, the matrix $G_{ij}(c,s)$ is the identity matrix except for its block on coordinates $i,j$:This article uses left matrix multiplication on column vectors. If $c=\cos \theta$ and $s=\sin \theta$, the block is a rotation matrix through angle $-\theta$: the minus sign belongs in the lower left. Its transpose reverses the rotation. Direct multiplication gives $G^{T}G=I$ and determinant one, so it is an orthogonal matrix; consequently $\Vert Gx{\Vert }_2^2=x^T{G}^{T}Gx=\Vert x{\Vert }_2^2$. Only two coordinates change, but this Euclidean norm identity holds for the entire vector.