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For a nonzero normal vector $\mathbf{n}$ and the plane equation $\mathbf{n}\cdot \mathbf{x}=d$, the closest point on a plane to $\mathbf{p}$ isIt subtracts only the normal displacement. Substituting gives $\mathbf{n}\cdot \mathbf{q}=d$, so $\mathbf{q}$ lies on the plane; the difference $\mathbf{p}-\mathbf{q}$ is a scalar multiple of its normal vector. These two checks identify the orthogonal projection onto a plane without solving a three-variable minimization problem.

For $\mathbf{n}=(2,-1,2)$, $d=3$, and $\mathbf{p}=(4,1,2)$, the residual $8$ givesIndeed, $(40-17+4)/9=3$, and $\Vert \mathbf{p}-\mathbf{q}\Vert =(8/9)\cdot 3=8/3$. Choosing the plane point $(1,1,1)$ instead gives distance $\sqrt{10}>8/3$: being on the plane does not make a point the nearest one.

As a function of $\mathbf{p}$, this projection isIt is an affine map, becoming a linear map when $d=0$. Projecting a second time changes nothing. It preserves displacements along the plane and removes the normal displacement, so it is a concrete example of an idempotent map.

Let $A,B$ be unital associative algebras over a field, each a simple ring, and $J\ne 0$ a two-sided ideal of their tensor product. Choose $0\ne x={\sum }_{i=1}^n{a}_i\otimes b_i\in J$ of minimal two-factor tensor rank. Both factor families are linearly independent. Since $A{a}_{1}A=A$, choose ${\sum }_j{u}_j{a}_1{v}_j=1$ and replace $x$ by ${\sum }_j(u_j\otimes 1)x(v_j\otimes 1)$. The result has first coefficient $1$, is nonzero by the linear independence of the $b_i$, and retains minimal two-factor tensor rank; its first factor family is therefore still linearly independent. Now apply the same construction in $B$ to make the first second-factor coefficient $1$. The result is again nonzero of minimal two-factor tensor rank, and both its coefficient families are linearly independent.

If $Z(A)=k$, commutation with $a\otimes 1$ gives a tensor in $J$ expressible with at most $n-1$ pure tensors, hence zero by minimality. Independence forces every $c_i$ to be central, hence a scalar multiple of $1$. Independence of $1,c_2,\dots ,c_n$ then forces $n=1$. Thus $1\otimes 1\in J$, proving simplicity of the tensor product without a splitting-field argument.

A nonzero vector $d$ is a direction vector of a line if the line has a parametrization $r=p+td$, $t\in \mathbb{R}$, for some point $p$. Replacing $d$ by a nonzero scalar multiple describes the same line; dividing by its Euclidean norm gives a unit vector in one of the two orientations. A cross product of two nonparallel normal vectors supplies a direction vector for the intersection of their planes.

A nonzero vector $v$ is an eigenvector of a linear operator $A$ when $Av=\lambda v$ for some eigenvalue $\lambda$. For a real vector space and $\lambda \ne 0$, the line through $v$ is preserved: the image can point forward or backward, but cannot turn away from that line. When $\lambda =0$, the image is zero and has no direction. The zero vector is excluded because it satisfies the equation for every scalar and identifies no special direction.

Every nonzero scalar multiple of an eigenvector is an eigenvector for the same eigenvalue, since $A(cv)=cAv=\lambda (cv)$. Thus there is usually no preferred length or sign; choosing a unit vector is a convenient normalization. Eigenvectors for a fixed eigenvalue, together with zero, form its eigenspace.

For a fully computed example, consider the real symmetric matrixIts characteristic polynomial isFor $\lambda =3$, $(A-3I)v=0$ gives $-x+y=0$, so $v_3=(1,1{)}^T$ is an eigenvector. For $\lambda =1$, $(A-I)v=0$ gives $x+y=0$, so $v_1=(1,-1{)}^T$ is an eigenvector. Direct multiplication checks both:The map stretches the line $y=x$ threefold and fixes the line $y=-x$ pointwise. These eigenvectors are orthogonal because $v_3^T{v}_1=0$; dividing each by $\sqrt{2}$ gives an orthonormal basis. A general vector turns because its components along these two lines receive different scalings. For example, $(1,0{)}^T$ maps to $(2,1{)}^T$ and is not an eigenvector.

A contrasting example is the planar rotation through $90$ degrees:Every nonzero real vector turns to a perpendicular direction, so there are no real eigenvectors. Its characteristic polynomial is ${\lambda }^2+1$, which has no real roots. Over the complex numbers, however, $R(1,-i{)}^T=i(1,-i{)}^T$ and $R(1,i{)}^T=-i(1,i{)}^T$. The same matrix therefore has an eigenbasis in ${\mathbb{C}}^2$ even though it has none in ${\mathbb{R}}^2$: the choice of field matters.

Equality in the Heisenberg uncertainty relation requires the centred vectors $x^{\prime }\psi$ and $p^{\prime }\psi$ to be imaginary scalar multiples, where $x^{\prime }=x-x_{0}I$, $p^{\prime }=p-p_{0}I$ and $x_0,p_0$ are the expectation values. Thus, for some $s>0$,In the position representation, using the momentum operator turns this into a first-order differential equation, whose normalizable solutions areHence the quantum states saturating the Heisenberg uncertainty relation are these Gaussian wave packets, with arbitrary position and momentum means and global phase. Their width is ${\sigma }^2=\hslash /(2s)$. A Gaussian wave packet with an additional quadratic quantum phase need not saturate this product bound.

For $1<p<\infty$ and nonzero elements $f,g$ of the same real or complex Lp space, equality in the Minkowski inequality holds if and only ifEquivalently, $g=cf$ almost everywhere for a single positive real number $c$. The multiplier must be constant across $X$; matching signs at each input alone is insufficient. For complex numbers, an arbitrary complex scalar multiple is also insufficient: $g=if$ gives $\Vert f+g{\Vert }_p=\sqrt{2}\Vert f{\Vert }_p<2\Vert f{\Vert }_p$ for nonzero $f$. If either element is zero in Lp space, equality holds automatically, even if a representative has nonzero values on a null set.

The reason is strictly convex normed space geometry. Normalize $f,g$ to unit Lp norm and use weights $\Vert f{\Vert }_p/(\Vert f{\Vert }_p+\Vert g{\Vert }_p)$ and $\Vert g{\Vert }_p/(\Vert f{\Vert }_p+\Vert g{\Vert }_p)$. Distinct normalized functions give a convex combination of Lp norm strictly below one. This follows by integrating the strict convexity of $z\mapsto |z{|}^p$ on the real numbers or complex numbers; equality therefore forces the normalized functions to agree almost everywhere.

At $p=1$, equality instead means $|f+g|=|f|+|g|$ almost everywhere: the nonnegative difference has zero integral. For real functions, this is $fg\ge 0$ almost everywhere. For complex functions, their nonzero values must have the same argument wherever both are nonzero, but their magnitude ratio may vary. The disjoint-coordinate vectors $(1,0)$ and $(0,1)$ illustrate equality without constant proportionality.

At $p=\infty$, no such constant proportionality condition is necessary: $(1,0)$ and $(1,1)$ have L-infinity norms one and a sum of L-infinity norm two. Nor must an essential supremum be attained: under Lebesgue measure on $(0,1)$, $f(x)=g(x)=x$ give equality $2=1+1$, although neither function takes the value one. The endpoint Lp spaces need not be strictly convex normed spaces.

Every real $2\times 2$ antisymmetric matrix has the form $A=sJ$, where $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Thus $A^2=-s^{2}I$, $\mathrm{tr}A=0$, and $\det A=s^2$. In particular,A convergent scalar power series applied to $A$ can therefore be reduced into an even scalar multiple of $I$ plus an odd scalar multiple of $A$. For real coefficients and $s\ne 0$, this is $f(A)=\mathrm{Re}f(is)I+[\mathrm{Im}f(is)/s]A$, provided the series converges at $\pm is$. Absolute convergence justifies separating even and odd terms. This converts many small matrix series problems into ordinary scalar series, without computing eigenvectors.

The linear span of a subset $S$ of a vector space is the set of all finite linear combinations of elements of $S$. It is the smallest vector subspace containing $S$: every vector subspace containing the original vectors must contain their sums and scalar multiples.

In ${\mathbb{R}}^3$, the linear span of $(1,0,0)$ is a line, the linear span of $(1,0,0),(0,1,0)$ is the plane $z=0$, and adding $(0,0,1)$ gives all of ${\mathbb{R}}^3$. Adding $(2,0,0)$ to the first set changes none of its attainable vectors, because it is already a scalar multiple of $(1,0,0)$. Thus a spanning set can contain redundant generators; linear independence identifies when this redundancy is absent. In an infinite-dimensional space, an algebraic linear span still uses finite sums; a closed linear span can also include limits in a specified norm.

A normal mode, also called an eigenmode, is a pattern of free oscillation in a linear system whose relative displacements stay fixed while a single temporal factor changes their size and sign. For an undamped stable vibrational mode,Here $\varphi$ is the mode shape, $A$ sets the amplitude, $\omega$ is the eigenfrequency expressed as an angular frequency, and $\phi$ is the oscillation phase. Every snapshot is a scalar multiple of the same mode shape; neighbouring regions may move in opposite directions. Normal modes of a double pendulum show that normal modes also describe coupled discrete systems, beyond strings. Their shapes are eigenvectors, rather than spatial eigenfunctions. ^[ref]

For the ideal wave equation on a string, take a uniform string of length $L$ with constant tension $\tau$ and linear mass density $\mu$. Assume small transverse slope, negligible bending stiffness, no damping, and fixed endpoints. Its wave speed is $c=\sqrt{\tau /\mu }$, and the nonzero normal modes areThe angular frequency ${\omega }_n$ is in radians per second; the frequency $f_n$ is in hertz. The wavelength is $2L/n$. The mode shape has $n$ loops separated by nodes of a standing wave, including both fixed endpoints. ^[ref]

As a worked example, choose $L=1\mathrm{m}$, $\tau =1\mathrm{N}$, and $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$. Then $c=1\mathrm{m}{\mathrm{s}}^{-1}$ and $f_1=0.5\mathrm{Hz}$, $f_2=1\mathrm{Hz}$, $f_3=1.5\mathrm{Hz}$. Their oscillation periods are $T_1=2\mathrm{s}$, $T_2=1\mathrm{s}$, and $T_3=2/3\mathrm{s}$. These are deliberately simple model parameters, rather than measurements of a musical instrument. The plot divides displacement by $A$ so that the three mode shapes can be compared, and uses each mode's own oscillation period for its snapshots.

For $\phi =0$, at $t=T_n/4$ every point has zero displacement, but most points have nonzero velocity. Only zeros of the mode shape are permanent nodes of a standing wave. At $t=T_n/2$ the original mode shape has reversed sign, and at $t=T_n$ it returns. The synchronized conversion between kinetic energy and potential energy is shown in normal-mode energy exchange.

The short derivation uses separation of variables: substituting $u=\varphi q$ into the wave equation gives ${\varphi }^{\prime \prime }+k^2\varphi =0$ and $q^{\prime \prime }+c^2{k}^{2}q=0$. The fixed-end boundary conditions remove the cosine spatial term and require $\sin (kL)=0$, hence $k=n\pi /L$ and ${\omega }_n=ck$. Different geometries and boundary conditions give different eigenfunctions; a sine mode shape is a property of this uniform fixed string, not a universal form of an eigenmode.

The usefulness is that a complicated vibration becomes a collection of harmonic oscillators. For sufficiently regular fixed-end initial data, the Fourier sine series givesThe initial displacement and velocity determine $a_n$ and $b_n$. This is the superposition principle in a Fourier sine basis, and modal energy distribution measures how strongly each normal mode is excited. The striking simplification is that each component keeps its identity while their sum can continually change shape. ^[ref]

These independent normal modes belong to the linear model. At larger slopes, changing tension introduces nonlinear effects and can couple the modes. A linearly damped string still admits a Fourier sine basis, but its temporal factors decay and need not oscillate; its energy dissipation identity for a linearly damped string replaces conservation of energy. More general damping need not leave undamped mode shapes independent.

A null direction of a linear map $T$ is represented by a nonzero vector $v$ in its kernel. Every scalar multiple $tv$ is an undetectable displacement: $T(x+tv)=T(x)$. The vector subspace $\mathrm{span}\{v\}$ describes the direction independently of the chosen nonzero representative. More generally, a basis of the kernel supplies all independent null directions of a linear map; their number is the nullity. The zero vector belongs to the kernel but supplies no direction.

In part (i), set every $b_i=1$. Its squared norm is $n$, givingThe equality condition is $a_1=\cdots =a_n$, because $a$ must be a scalar multiple of $(1,\dots ,1)$.

Keep the nonzero $z$ of minimal two-factor tensor rank constructed in part (i). For every $a\in A$, its commutator with $a\otimes 1$ lies in $J$ and equalsIf this tensor were nonzero, its two-factor tensor rank would be at most $n-1$, contradicting minimality. Thus it is zero. The linear independence of $d_2,\dots ,d_n$ forces $a{c}_i=c_{i}a$ for every $a$ and $i\ge 2$. Since $Z(A)=k$, each $c_i$ is a scalar multiple of $1$.

But $1,c_2,\dots ,c_n$ are linearly independent. Hence $n=1$, so $z=1\otimes 1\in J$. A two-sided ideal containing the multiplicative identity is the entire ring; therefore $J=A{\otimes }_{k}B$.

We have proved the stronger result that $A{\otimes }_{k}B$ is a simple ring. It has finite dimension $({\dim }_{k}A)({\dim }_{k}B)$, so part (b) provesPart (c)(ii) also gives $Z(A{\otimes }_{k}B)=k$, so this proves that a tensor product of central simple algebras is again a central simple algebra.

Let $S=S{L}_n(F)$ and let $Z$ be its scalar center. Its action on the lines of $F^n$ has kernel exactly $Z$: a linear map fixing every line is diagonal in a basis, and fixing the lines spanned by $e_i+e_j$ makes all its diagonal entries equal. The action is two-transitive. Indeed an ordered pair of distinct lines can be represented by two independent vectors and extended to a basis. A linear map between two such bases can be adjusted to determinant one by scaling one target basis vector; this preserves both target lines. Therefore $S/Z$ has a faithful primitive group action.

Fix $L=Fv$. The mapsform an abelian subgroup: all products of the rank-one terms vanish. It is normal in the line stabilizer, since conjugation replaces $v$ by a scalar multiple and transports $f$. Its conjugates contain every transvection, and hence generate $S$ by 4(b). The same statements hold for its image in $S/Z$.

It remains to prove perfectness rather than assume it. Use the group commutator convention $[x,y]=xy{x}^{-1}{y}^{-1}$. If $n\ge 3$, distinct $i,j,k$ giveEvery elementary generator is consequently in $S^{\prime }$. If $n=2$ and $|F|>3$, choose $a\in F^{\ast }$ with $a^2\ne 1$. There are at most two roots of $a^2=1$, so such a choice exists. For $h=\mathrm{diag}(a,a^{-1})$,Varying $b$ gives every upper transvection; conjugating by $w(1)$ gives every lower one. Again $S^{\prime }=S$, and its projective quotient is a perfect group.

Apply 4(a) to this faithful primitive projective action. Every nontrivial normal subgroup contains the commutator subgroup, which is the whole quotient. ThusThe restriction is real: $PS{L}_2(2)\cong S_3$ and $PS{L}_2(3)\cong A_4$ are not simple.

The orthogonal projection has range $\mathcal{S}$, not all of $C[0,1]$: the latter is an infinite-dimensional vector space whereas $\mathcal{S}$ is a finite-dimensional vector space. For a concrete counterexample, $e^x$ cannot belong to any finite-degree piecewise polynomial space, since all its derivatives are nonzero on every open interval. For $k\ge 2$, the B-splines are continuous, so the orthogonal projection is a map $C[0,1]\to \mathcal{S}\subset C[0,1]$. For order one it instead has a generally discontinuous piecewise constant range. For example, a single interval indicator on $[1/4,3/4)$ is the projection of the constant function one onto its span, and is not continuous.

Write $s^{\ast }={\sum }_j{a}_j{N}_j$. Since $M_i$ is a positive scalar multiple of $N_i$, the operator normal equations $⟨f-s^{\ast },N_i⟩=0$ are equivalent toThe mixed-normalization spline Gram matrix is invertible: $G=DH$ where $D_{ii}=k/(t_{i+k}-t_i)>0$ and $H_{ij}=⟨N_i,N_j⟩$ is the ordinary positive-definite matrix of inner products of a linearly independent basis. The positivity and unit-integral normalization of a B-spline give $|b_i|\le \Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines givesHere the operator norm of a matrix on ${\ell }^{\infty }$ is its maximum absolute row sum. Consequently,This proves the maximum-norm bound for spline projection.

Inner-product preservation between programs for $U$ and $V$ givesfor all $|\alpha ⟩,|\beta ⟩$. If $⟨P_U|P_V⟩\ne 0$, this identity forces every matrix element of $U^{†}V$ to be the same scalar multiple of the corresponding identity matrix element. Hence $U^{†}V=e^{i\theta }I$, contrary to physical distinctness. Therefore

An elementary row operation is one of three reversible changes: interchange two rows; multiply one row by a nonzero scalar; or add a scalar multiple of one row to a different row. Their inverses are, respectively, the same interchange, multiplication by the reciprocal, and addition of the negative scalar multiple.

Applied to an augmented matrix, these changes interchange linear equations, rescale an equation by a nonzero factor, or add a multiple of one equation to another. Every old solution satisfies the new equations. Applying the inverse elementary row operation gives the converse, so the solution set is unchanged at every step. Equivalently, all the operations together givewhere $E$ is an invertible matrix, a product of elementary matrices.

LetThe vector triple product identity $(\mathbf{a}\wedge \mathbf{b})\wedge \mathbf{a}=|\mathbf{a}{|}^2\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{a}$ shows that ${\mathbf{r}}_0\wedge \mathbf{a}=\mathbf{b}$. Moreover, $(\mathbf{r}-{\mathbf{r}}_0)\wedge \mathbf{a}=0$ exactly when $\mathbf{r}-{\mathbf{r}}_0$ is a scalar multiple of $\mathbf{a}$. Thus the given line isAlso $\mathbf{a}\wedge {\mathbf{r}}_0=-\mathbf{b}$, so $\mathbf{N}=\mathbf{a}\wedge (\mathbf{y}-{\mathbf{r}}_0)$. The assumption $\mathbf{N}\ne 0$ says precisely that $\mathbf{y}$ is outside $L$. Consequently $L$ and $\mathbf{y}$ determine a unique plane, with normal vector $\mathbf{N}$.

To identify its equation, use cyclic invariance of the scalar triple product:For $\mathbf{r}\in L$, this equals $\mathbf{y}\cdot \mathbf{b}$, while $\mathbf{r}\cdot \mathbf{b}=\mathbf{r}\cdot (\mathbf{r}\wedge \mathbf{a})=0$. Hence $\mathbf{r}\cdot \mathbf{N}=\mathbf{y}\cdot \mathbf{b}$ throughout $L$. The point $\mathbf{y}$ also satisfies that equation because $\mathbf{y}\cdot (\mathbf{a}\wedge \mathbf{y})=0$. Since $\mathbf{N}\ne 0$, the required plane is therefore exactlyThis also proves the reusable plane through a point and a line given by a cross product construction.

The hypothesis implies $v_1,v_2\ne 0$, since the zero vector is a scalar multiple of every vector. SetSince $v_1$ is not a scalar multiple of $v_2$, $w\ne 0$. Positive definiteness and bilinearity of the real inner product giveMultiplying by $\Vert v_2{\Vert }^2>0$ and taking square roots proves the strict Cauchy-Schwarz inequalitySince every real number $s$ satisfies $s\le |s|$, this proves, in particular,The argument derives the required inequality directly from the inner product axioms.

The three elementary row operations on an $m\times n$ real matrix are:

Each operation changes whole rows, so it acts on every column by the same rule. Each can be undone by an elementary row operation.

Every $v\in V$ has a unique basis expansion $v={\sum }_{i=1}^n{a}_i{v}_i$. Define the linear functionalUniqueness of the coordinates makes this definition well defined. The coordinates of a sum or a scalar multiple are the corresponding sums or scalar multiples of the coordinates, so this formula defines a linear map. It has $T(v_i)={\mu }_i$. Conversely, any linear map with those values must satisfy the same formula by linearity, proving uniqueness.

If the finite field $\mathbb{F}$ has $q$ elements, each of the $n$ values ${\mu }_i$ can be chosen independently in $q$ ways. The number of linear maps $V\to \mathbb{F}$ is therefore $\boxed{q^n}$, including the zero linear map.