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A Euclidean vector is a vector in a finite-dimensional real inner product space. Choosing an orthonormal basis identifies it with its real vector components in ${\mathbb{R}}^n$. A geometric arrow represents a Euclidean vector by a length and a direction; translating the arrow without changing either represents the same Euclidean vector. The zero vector has length zero and no distinguished direction.^[ref]

For points $P=(1,1)$ and $Q=(4,5)$ in a Cartesian coordinate system, the displacement is $Q-P=(3,4)$. The arrow from $(-2,0)$ to $(1,4)$ represents the same Euclidean vector, even though its endpoints differ. An endpoint is a position; their difference is a vector. In an orthonormal basis $({\mathbf{e}}_1,{\mathbf{e}}_2)$,The Euclidean norm comes from the inner product: $\Vert \mathbf{v}{\Vert }_2^2=⟨\mathbf{v},\mathbf{v}⟩$. The sum of squared vector components is valid in an orthonormal basis; an arbitrary basis requires the appropriate Gram matrix. This distinction lets the geometric vector stay the same while its coordinate vector changes.^[ref]

Vector addition is componentwise. With $\mathbf{u}=(2,1)$ and $\mathbf{w}=(1,3)$,Geometrically, translate the second arrow so its tail starts at the first arrow's head. The vector sum joins the starting point to the final endpoint. Reversing the order gives the other route around the same parallelogram, so vector addition is commutative. This is the useful bookkeeping behind adding successive displacements, velocities, or forces: direction information survives the calculation. For a force, the Euclidean vector alone does not specify its point of application, which also affects the torque.^[ref]

The magnitude of a vector sum is generally different from the sum of the magnitudes. Here $\Vert \mathbf{u}+\mathbf{w}{\Vert }_2=5$, whereas $\Vert \mathbf{u}{\Vert }_2+\Vert \mathbf{w}{\Vert }_2=\sqrt{5}+\sqrt{10}>5$. The triangle inequality explains why the direct displacement cannot exceed the length of this two-segment route. A return displacement $(-3,-4)$ makes the net vector sum zero, although the lengths of the three travelled segments sum to a positive number.

The dot product exposes direction quantitatively: for nonzero Euclidean vectors, $⟨\mathbf{u},\mathbf{w}⟩=\Vert \mathbf{u}{\Vert }_2\Vert \mathbf{w}{\Vert }_2\cos \theta$. For the displayed pair, the dot product is $2\cdot 1+1\cdot 3=5$, so $\cos \theta =1/\sqrt{2}$ and the angle is $45^{\circ }$. For orthogonal vectors, that dot product is zero. Dividing by magnitudes separates direction from size, which is the purpose of vector normalization.

Draw a $k\times D$ random matrix $G$ whose entries are independent random variables with the standard normal distribution, and setThe factor $1/\sqrt{k}$ makes the expected value of the squared L2 norm equal to the original squared L2 norm. This Gaussian Johnson–Lindenstrauss transform is a linear map, generally not an orthogonal projection: its rows are not required to be orthogonal vectors. ^[ref]

For the fixed finite set $S$, let $0<\delta <1$ be an allowed failure probability, and defineWith $\log$ denoting the natural logarithm, a sufficient integer target dimension isThen one draw of this specified random matrix satisfies all the squared Euclidean distance bounds simultaneously with probability at least $1-\delta$. Since $c(\epsilon )\ge {\epsilon }^2/12$, the simpler sufficient bound $k\ge \lceil 12{\epsilon }^{-2}\log (n(n-1)/\delta )\rceil$ also works. Both are sufficient bounds for this argument; they do not claim the smallest possible target dimension.

This is a probability statement about a map drawn independently of the fixed finite set. It implies existence because the success probability is positive, but permits failed draws. The set cannot be chosen after inspecting the map with this same guarantee. In particular, no single draw preserves every possible finite set in the ambient Euclidean space when $k<D$.

A finite collection of nonzero pairwise orthogonal vectors is linearly independent. Given ${\sum }_i{a}_i{q}_i=0$, take the inner product with $q_j$ to obtain $a_j\Vert q_j{\Vert }^2=0$. Positive definiteness and $q_j\ne 0$ force $a_j=0$ for every index. The nonzero condition matters: adjoining a zero vector preserves pairwise orthogonality but destroys linear independence.

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.

Take eigenvectors $v_i,v_j$ for distinct eigenvalues ${\lambda }_i,{\lambda }_j$. Since the eigenvalues are real and $A$ is a Hermitian matrix,Hence $({\lambda }_j-{\lambda }_i)v_i^{†}{v}_j=0$, so eigenvectors for distinct eigenvalues are orthogonal vectors. They are therefore linearly independent; after normalization the given $n$ eigenvectors form an orthonormal basis of ${\mathbb{C}}^n$.

Partition the design matrix accordingly as $X=(X_1\cdots X_k)$. The blocks ${\beta }_i,{\beta }_j$ are orthogonal when $X_i^{\mathsf{T}}{X}_j=0$, meaning that their column spaces are orthogonal vectors. Mutual orthogonality means that this holds for every pair of distinct blocks, so $X^{\mathsf{T}}X$ is block diagonal.

For the three scalar coefficients, the corresponding columns are $1,x_1,x_2$. The necessary and sufficient conditions areUnder the assumed full rank, neither explanatory column is zero. The maximum-likelihood estimator is $\hat{\beta }=(X^{\mathsf{T}}X{)}^{-1}{X}^{\mathsf{T}}Y$, and the normal distribution of $Y$ impliesThus the three coefficient estimators are independent normal random variables, not merely uncorrelated. This is a sampling-distribution statement with the true ${\sigma }^2$ fixed.

Differentiate the position vector with respect to each of the spherical polar coordinates. The derivatives areTheir lengths give the scale factors of orthogonal coordinates:Dividing the derivatives by these lengths yieldsEach has unit length. For example, ${\mathbf{e}}_r\cdot {\mathbf{e}}_{\theta }=\sin \theta \cos \theta ({\cos }^2\varphi +{\sin }^2\varphi )-\cos \theta \sin \theta =0$; the other two dot products vanish because their first two terms cancel. Thus these are mutually orthogonal vectors, and the total differential has the required form. The coordinates degenerate at $r=0$ and on the polar axis, but this does not affect an area integral.

On the parametrized surface $\theta =\alpha$, use $r,\varphi$ as parameters. The surface integral area element isFor $0\le \alpha \le \pi$, its lateral area, with no closing disk, is thereforeAt $\alpha =0,\pi$ the surface collapses to a line and its area is zero.

If $A\mathbf{u}=\lambda \mathbf{u}$ and $A\mathbf{v}=\mu \mathbf{v}$, symmetry givesFor $\lambda \ne \mu$, ${\mathbf{u}}^T\mathbf{v}=0$, so the eigenvectors are orthogonal.

Let $H$ be the measured wave scattering response> linear operator, with direct background propagation removed if present. Its time reversal operator is the positive operator $T=H^{\ast }H$. For a normalized emitted signal $h$,The maximizing signal is a right singular vector, equivalently an eigenvector of $T$ with its largest eigenvalue. This is a precise intensity statement independent of a scatterer model.

Under the Born approximation for scalar wave scattering, neglect interactions between point scatterers and writewhere $b_j$ is the receiver response to scatterer $j$, $c_j^{\ast }h$ is its illumination by the source array, and ${\alpha }_j$ is its scattering wave amplitude. Well-resolved point scatterers have approximately orthogonal vectors $b_j$ and $c_j$. In the ideal orthogonal limit,Thus the most reflective, geometrically weighted scatterer gives the largest eigenvalue, and its normalized steering vector $c_j/\Vert c_j\Vert$ gives the corresponding wave focusing signal. If illumination and reception factors are equal for all scatterers, this is precisely the scatterer with largest $|{\alpha }_j{|}^2$. Geometric size by itself is not the quantity being ranked.

Repeated adjoint time reversal acoustics is power iteration on $T$. If its top eigenvalue is simple and the initial signal has a nonzero component along its eigenvector, thenA tied largest eigenvalue leaves a combination in the leading eigenspace, and an initial signal orthogonal to that eigenspace cannot excite it. This is the basis of the DORT method. Mere physical separation is insufficient if the array cannot resolve the scatterers; coherent steering-vector overlap or multiple scattering can mix the modes. Also, if $H$ includes unrestricted homogeneous transmission, its largest eigenvalue need not identify any individual scatterer. The correspondence concerns the resolved wave scattering response>. The ideal correspondence and selective wave focusing are analyzed by Prada and Fink.

Write $a=(2,1{)}^T$, so the data map is $a^{T}u$. The Gaussian likelihood and prior distribution giveCompleting the square, or using Gaussian conjugacy for a normal linear model, identifies a multivariate normal distribution with precision matrix $C_{\delta }^{-1}=I_2+{\delta }^{-2}a{a}^T$. Since $a^{T}a=5$, its posterior mean and covariance matrix areThe parameters areFor example, $(I_2+{\delta }^{-2}a{a}^T)C_{\delta }=I_2$, which checks the square completion.

The orthogonal vectorsare eigenvectors of the covariance matrix. Their respective variances areThey are independent under the posterior distribution, by independence of uncorrelated jointly normal variables. Thus uncertainty is different in the two directions even for positive noise: the data constrain $e_{\parallel }$, whereas $e_{\perp }$ spans the null space of the data map.

As $\delta \to 0$, the posterior distribution converges to the law ofThis is supported on $2u_1+u_2=m$, and its posterior mean is the minimum-norm least-squares solution. The prior distribution still determines the distribution along the unobserved null space; it has not disappeared. This example of a Gaussian posterior in the zero-noise limit distinguishes exact recovery of an observed component from recovery of the whole unknown.

Let ${\theta }_k=k\pi h=k\pi /(m+1)$ and define $q_j^{(k)}=\sin (j{\theta }_k)$. The artificial boundary values are $q_0^{(k)}=q_{m+1}^{(k)}=0$. For the symmetric tridiagonal Toeplitz matrix $H=[\beta ,\alpha ,\beta ]$, the sine addition formula givesThus all such matrices have the same discrete sine transform eigenvectors, with eigenvaluesThe sine vectors are mutually orthogonal vectors and nonzero, so the matrix $Q=(\sin (ij\pi h){)}_{i,j=1}^m$ is invertible. If $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$, then $HQ=QD$ and therefore

A matrix is Hermitian when $A^{†}=A$, and unitary when $U^{†}U=U{U}^{†}=I$.

If $A\mathbf{u}=\lambda \mathbf{u}$ with $\mathbf{u}\ne 0$, thenso every eigenvalue is real. If $A\mathbf{u}=\lambda \mathbf{u}$ and $A\mathbf{v}=\mu \mathbf{v}$, thenDistinct eigenvalues therefore have orthogonal eigenvectors.

The normalized eigenvectors form an orthonormal basis, so the matrix $U=({\mathbf{u}}_1\cdots {\mathbf{u}}_n)$ satisfies $U^{†}U=I$ and is unitary. Withthe equations $A{\mathbf{u}}_j={\lambda }_j{\mathbf{u}}_j$ say $AU=UD$, hence $A=UD{U}^{†}$. This is the finite-dimensional unitary diagonalization of a normal matrix.

A unitary conjugate of an arbitrary diagonal matrix need not be Hermitian: $U=(1)$ and $D=(i)$ give $UD{U}^{†}=(i)$. It is Hermitian exactly when all diagonal entries of $D$ are real.

For the displayed matrix, one valid decomposition isThe columns are orthonormal eigenvectors, so direct multiplication gives $UD{U}^{†}=A$.

The supports of ${\phi }_1$ and ${\phi }_2$ are disjoint, so they are orthogonal vectors. Substitution in the covariance formula givesandTherefore the associated eigenvalues are

There are uncountably many physically distinct one-qubit unitaries. Part (ii) would require an orthogonal program state for every one, whereas a $K$-qubit register has dimension only $2^K$. This contradiction is the no-programming theorem, so no deterministic gate satisfying (PROG) exists.

Use $f,g$ from i. Their inner products with $x^2$ areThus $h=x^2-2/3$. Its values at $-1,0,1$ are $1/3,-2/3,1/3$, soBy ii the three polynomials are orthogonal vectors, and normalizing $h$ givesThis orthonormal set is a basis of the three-dimensional vector space $V$. It is the result of the Gram-Schmidt process applied to $1,x,x^2$.

The relevant inner products areHence $1$ and $x$ are orthogonal vectors. Dividing by their norms gives the orthonormal basisof their linear span $U$.

At a point $T$ on a circle with centre $O$, the tangent line is perpendicular to the radius $OT$. For example, differentiating $|\mathbf{r}(s)-O{|}^2=R^2$ along a parametric curve on the circle gives $(\mathbf{r}-O)\cdot {\mathbf{r}}^{\prime }=0$, showing that the radial vector and tangent direction are orthogonal vectors. Conversely, the perpendicular straight line through $T$ has only that point in common with the circle: for a nonzero displacement $v$ along it, $|T+v-O{|}^2=R^2+|v{|}^2>R^2$.