Linear map

A linear map preserves vector addition and scalar multiplication.

A real-linear map is a linear map over the real numbers. For a map from the real line to itself, real linearity means multiplication by a fixed real number, its value at one. Rational-linear maps need not be real-linear without an additional regularity condition.

A rational-linear map is a linear map when its domain and codomain are regarded as vector spaces over the rational numbers. An additive map on a rational vector subspace of the real numbers is automatically rational-linear: integer homogeneity follows by repeated addition, and division by a positive integer gives rational homogeneity of an additive function.

Let $D$ be a rational vector subspace of the real numbers and let $t\notin D$. Every element of $D+\mathbb{Q}t$ has a unique expression $d+qt$. For any prescribed real number $a$, the rule $h(d+qt)=f(d)+qa$ extends a rational-linear map $f$ on $D$ and remains rational-linear. The uniqueness follows because a nonzero rational multiple of $t$ cannot lie in $D$.

An invertible linear transformation replaces one system of coordinates by another. If $x=Pu$ in a linear differential equation $\dot{x}=Ax$, then $\dot{u}=P^{-1}APu$. The similar matrices $A$ and $P^{-1}AP$ have the same characteristic polynomial and eigenvalues, although their coordinate descriptions of the eigenvectors differ. Such a change can turn a planar linear system with eigenvalues $\alpha \pm i\beta$ into $\dot{u}=\alpha u-\beta v$, $\dot{v}=\beta u+\alpha v$.

For a unit vector $u$ in a real inner product space, stretching by a factor $\alpha$ along $u$ and leaving its perpendicular subspace fixed givesDecompose $x=(u^{T}x)u+(x-(u^{T}x)u)$. The first term is the orthogonal projection onto the line; the second is perpendicular to it. The displayed map multiplies the first by $\alpha$ and fixes the second. In the plane, if $u=(\cos \theta ,\sin \theta {)}^T$, then $T=R(\theta )\mathrm{diag}(\alpha ,1)R(-\theta )$.

A linear map $T:V\to W$ between finite-dimensional vector spaces admits a linear left inverse $S:W\to V$ if and only if its kernel of a linear map is $\{0\}$, equivalently if and only if it is injective.

Necessity follows from $Tv=0⟹v=STv=0$. For sufficiency, choose a basis $v_1,\dots ,v_n$ of $V$. Triviality of the kernel of a linear map implies that $T{v}_1,\dots ,T{v}_n$ are linearly independent. By basis extension, they extend to a basis $T{v}_1,\dots ,T{v}_n,w_{n+1},\dots ,w_m$ of $W$. Define $S(T{v}_i)=v_i$ and $S(w_j)=0$, and extend by linearity. Then $ST{v}_i=v_i$ on a basis, proving $ST=I_V$.

The freedom in $S$ lies in its values on a direct-sum complement of the image of a linear map. For example, for $T(x,y)=(x,y,0)$, every linear map $S(u,v,w)=(u+aw,v+bw)$ is a left inverse, with arbitrary scalars $a,b$. A right inverse instead requires $T$ to be a surjective linear map.

For a fixed nonzero real vector $b$ in three dimensions, a real linear map commuting with every proper rotation about that axis has the displayed form, where $[b{]}_{\times }x=b\times x$ is the cross-product matrix. On the axis it acts by $a+c\Vert b{\Vert }^2$. On the perpendicular plane its matrix in an oriented orthonormal basis is $\left(\begin{array}{cc}a& -d\Vert b\Vert \\ d\Vert b\Vert & a\end{array}\right)$. Its eigenvalues are therefore $a+c\Vert b{\Vert }^2$ and $a\pm id\Vert b\Vert$, andTo see the converse, commuting with all rotations preserves their common fixed axis and its perpendicular plane; there can be no nonzero coupling vector fixed by every planar rotation. A planar matrix commuting with every rotation is a scalar identity plus a scalar right-angle rotation. Thus the block description is exhaustive. If $d\ne 0$, only the axial eigenvalue is real; if $d=0$, the perpendicular plane also consists of real eigenvectors.

For $\alpha :U\to V$, maps in $M^l(\alpha )$ factor uniquely through $V/\mathrm{im}\alpha$; maps in $M^r(\alpha )$ have image in $\ker \alpha$. Thus these vector subspaces are naturally $L(V/\mathrm{im}\alpha ,U)$ and $L(V,\ker \alpha )$, of dimensions $(\dim V-\mathrm{rank}\alpha )\dim U$ and $(\dim U-\mathrm{rank}\alpha )\dim V$. Passing to a dual map reverses composition and interchanges these two spaces. This gives a basis-free route to equality of the ranks of a map and its dual.

For endomorphisms $S,T$ of the same finite-dimensional vector space, $ST=I$ makes $S$ surjective and $T$ injective. The rank-nullity theorem makes both invertible, hence $TS=I$. The finite-dimensional and equal-domain assumptions matter; unilateral shift operators on an infinite-dimensional sequence space give one-sided inverses that are not two-sided.

The rank of a linear map is the dimension of its image of a linear map. In finite-dimensional spaces it equals the rank of a matrix representing the map in any bases.

The image of $A+B$ lies in $\mathrm{im}A+\mathrm{im}B$. The dimension formula for a sum of subspaces gives the upper bound. Applying it to $A=(A+B)-B$ and then interchanging $A,B$ gives the lower bound. Both extremes are sharp: cancelling diagonal blocks gives the difference of ranks; diagonal maps with supports covering the whole space and positive coefficients give full rank when the sum of prescribed ranks is at least the dimension.

For a nonzero vector $v$ and nonzero linear functional $f$ vanishing on $v$, this invertible map fixes the hyperplane $\ker f$ pointwise and has determinant one. Its inverse is $I-vf$, since $(vf{)}^2=0$. A symplectic transvection is obtained by taking $f=cB(-,v)$ for a nondegenerate alternating bilinear form.

Every nonidentity transvection has an adapted basis in which it is $I+E_{12}$. Thus all such maps are conjugate in the general linear group. In dimension at least three, a scaling of a third basis vector centralizes this matrix and has arbitrary nonzero determinant; adjusting a conjugator by this scaling proves conjugacy in the special linear group as well.

// Target: algebra.bigb

This matrix adds $t$ times coordinate $j$ to coordinate $i$. Left multiplication performs the corresponding row addition. Such matrices generate the special linear group over a finite field: row elimination also realizes signed interchanges and compensating diagonal scalings using products of these matrices.

Row additions reduce an invertible matrix to diagonal form without changing determinant. A determinant-one diagonal matrix is a product of diagonal two-coordinate blocks $\mathrm{diag}(a,a^{-1})$, each expressible as a product of six elementary transvection matrices. Therefore these matrices generate the special linear group over a field.

// Target: algebra.bigb

A horizontal shear fixes the horizontal axis pointwise and sends $(x,y)$ to $(x+ky,y)$. Its determinant is one and both eigenvalues are one. A nonzero shear preserves area but generally changes lengths and angles. Composing a shear with a planar rotation can produce a symmetric stretch; having determinant one alone does not make a linear map a shear.

A uniform dilation about the origin multiplies every vector by one positive real scale $s$. The linear map $\mathbf{x}\mapsto s\mathbf{x}$ multiplies lengths by $s$ and $d$-dimensional volumes by $s^d$. This differs from stretching different axes by different factors. The case $s=1$ is the identity map.

A real-linear map between complex vector spaces that commutes with multiplication by $i$. On $\mathbb{C}$, its real matrix is $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$, representing multiplication by $a+ib$. On $\mathbb{C}$, such a map with real determinant one is a rotation. In higher complex dimension, determinant one does not force the map to be a unitary operator.

The image of a linear map $T:V\to W$ is the vector subspace $\mathrm{im}T=\{T(v):v\in V\}$ of its codomain.

For a real matrix $A$ and a fixed vector $n$, every vector in the image of a linear map defined by $A$ is orthogonal to $n$ exactly when $A^{T}n=0$. Indeed, $n^{T}Ar=(A^{T}n{)}^{T}r$. If this vanishes for every $r$, choose $r=A^{T}n$ to obtain $|A^{T}n{|}^2=0$. Equivalently each column of $A$ is orthogonal to $n$. Requiring the statement only for nonzero $r$ changes nothing, since the zero input already has zero dot product. This is the orthogonal complement relation between the image of a linear map and the kernel of a linear map defined by its transpose.

If a linear functional $\ell$ vanishes on the image of a linear map $T$ but $\ell (b)\ne 0$, the equation $Tx=b$ has no solution. In Euclidean coordinates, a vector $n$ with $n^{T}A=0$ provides such an obstruction whenever $n^{T}b\ne 0$. In finite dimensions these annihilating functionals exactly characterize membership in the image: $b\in \mathrm{im}A$ precisely when it is orthogonal to every vector in $\ker A^T$.

A linear map $T:V\to W$ is surjective when its image is all of $W$. For finite-dimensional spaces this is equivalent to $\mathrm{rank}T=\dim W$.

For a linear map $T:V\to W$, the induced mapis a vector-space isomorphism.

A linear isomorphism is a bijective linear map. Its inverse is also linear.

This is a scalar-valued linear functional, so it vanishes at the origin; an affine function with a nonzero constant term is a different notion.

A real linear function on ${\mathbb{R}}^n$ has the form $x\mapsto c^{T}x$.

An affine function on a real or complex vector space has the form $x\mapsto Lx+b$, where $L$ is a linear map and $b$ is constant. Its derivative is the constant map $L$. It is linear exactly when $b=0$.

An expression of the form $a_0+{\sum }_{j=1}^m{a}_j{x}_j$, with fixed coefficients $a_j$. It contains a constant term and terms of degree one, without products of variables. It represents an affine function once the scalar domain is specified.

Linearity is the property $T(ax+by)=aT(x)+bT(y)$.

For a homogeneous linear equation, every linear combination of solutions is again a solution.

A linear operator is a linear map from a vector space to itself.

An operator on a complex Hilbert space is accretive when the displayed inequality holds throughout its domain. Its negative is a dissipative operator. For a bounded everywhere-defined operator, expansion gives $\Vert (I+T)x{\Vert }^2\ge \Vert x{\Vert }^2$. Its adjoint operator satisfies the same accretivity condition. Consequently $I+T$ has closed range and trivial kernel, and its range is dense because its adjoint has trivial kernel. Thus $I+T$ is onto and $\Vert (I+T{)}^{-1}\Vert \le 1$. Boundedness and the full domain are material here; for unbounded operators accretivity alone need not imply surjectivity.

For a bounded accretive operator, $I+T$ is invertible and the transform satisfies $\Vert S\Vert \le 1$. The identity $\Vert (I+T)x{\Vert }^2-\Vert (I-T)x{\Vert }^2=4\mathrm{Re}⟨Tx,x⟩$ proves contraction after the substitution $z=(I+T)x$. If the real part vanishes for every $x$, this identity proves norm preservation. Both $T$ and $-T$ are then accretive, making both $I+T$ and $I-T$ invertible; hence $S$ is onto and is a unitary operator.

For a linear operator on a real vector space, the identity $T^3=T$ gives an explicit eigenspace decompositionIndeed, the linear operatorssum to $I$. Their images lie respectively in the eigenspaces for $0$, $1$, and $-1$, since $T{P}_0=0$, $T{P}_{+}=P_{+}$, and $T{P}_{-}=-P_{-}$. On each of these eigenspaces, its corresponding $P$ is the identity and the other two vanish, which proves uniqueness of the decomposition. Thus a finite-dimensional tripotent linear operator admits a diagonalization with diagonal entries in $\{0,1,-1\}$. The condition $T^2\ne T$ is equivalent to the presence of a nonzero $-1$ eigenspace, and $T^2\ne I$ is equivalent to a nonzero kernel.

A linear operator is semisimple if it becomes diagonalizable over an algebraic closure; equivalently, its minimal polynomial has no repeated root over that closure. If an invertible operator $A$ has finite order $r$ prime to the characteristic of a field, its minimal polynomial divides $X^r-1$. The derivative $r{X}^{r-1}$ is coprime to this polynomial, so $A$ is semisimple. Its invariant irreducible summands over a finite field become multiplication by scalars in suitable finite field extensions.

A linear operator $T$ is locally nilpotent if for each vector $v$ there is an integer $m(v)$ such that $T^{m(v)}v=0$. In finite dimension this is equivalent to being a nilpotent operator. Differentiation on the polynomial ring is locally nilpotent in characteristic zero but has no uniform nilpotence exponent.

Over a field of characteristic zero, a locally nilpotent operator has an algebraically defined exponential, because the sum is finite for each vector. Its inverse is $\exp (-T)$. The nilpotence bound may depend on the vector. In an sl2 Lie algebra Verma module, the raising operator is locally nilpotent but the lowering operator is not, so the latter's exponential does not define an endomorphism of the ordinary algebraic module.

// Target: lie-algebra.bigb

The commutator measures failure of two operators to commute, on a common domain where the products are defined. Direct expansion gives $[AB,C]=A[B,C]+[A,C]B$. A common eigenvector of $A,B$ lies in the kernel of $[A,B]$, so an injective commutator excludes all nonzero common eigenvectors. A nonzero commutator need not itself be injective, and therefore does not alone exclude an individual common eigenvector.

For a finite almost-everywhere measurable function $f$, the multiplication operator on L2 space has maximal operator domain $D(M_f)=\{u\in L^2:fu\in L^2\}$. It is a closed operator: if $u_n\to u$ and $f{u}_n\to v$ in L2 space, subsequences converge almost everywhere, giving $v=fu$. Its domain is dense, since truncating a function to the sets $|f|\le n$ approximates it in L2 space. If $f$ is bounded, this is a bounded linear operator of norm equal to the essential supremum of $|f|$.

For a real measurable function $m$ on a sigma-finite measure space, multiplication on $D(M_m)=\{f\in L^2:mf\in L^2\}$ is an unbounded self-adjoint operator. Testing its adjoint relation on $\{|m|\le n\}$ identifies the same maximal domain. Outside the essential range, multiplication by $(m-z{)}^{-1}$ is a bounded resolvent. Inside it, normalized indicators of finite-positive-measure sets where $m$ is close to $z$ are approximate eigenvectors, so the inverse cannot be bounded.

A linear projection is an idempotent linear operator: $P^2=P$. Every vector decomposes as $x=Px+(I-P)x$, into the image of a linear map and the kernel of a linear map. Those two vector subspaces form a direct sum. An orthogonal projection additionally uses the orthogonal complement in an inner product space.

For linear projections $E,F$, expanding the square proves that their sum is a linear projection exactly when $EF=-FE$. Multiplying this equality on the left and right by $E$ gives $EF=-EFE$ and $EFE=-FE=EF$, hence $2EF=0$. If the field has characteristic different from two, this forces $EF=FE=0$, and those conditions also suffice. In characteristic two, the condition instead reduces to commutation; $E=F=I$ has sum zero but nonzero product. For three commuting linear projections, the square-minus-sum is $2(EF+EG+FG)$, so vanishing of the triple product alone is not sufficient.

Over a field of characteristic different from two, three linear projections summing to the identity annihilate one another in both orders. Each pair sum is the complementary projection to the third. The sum of two linear projections criterion therefore proves zero products. Every vector is the sum of its three images; applying each projection to a zero sum proves that these images form a direct sum. Choosing a basis in each image gives a simultaneous eigenbasis, with triples of eigenvalues $(1,0,0)$, $(0,1,0)$ or $(0,0,1)$. This is algebraic orthogonality, and does not assume that the projections are orthogonal projections for a specified inner product.

For a linear projection $P$ on a nonzero finite-dimensional vector space, $V=\ker P\oplus \mathrm{im}P$. The linear map $\alpha I+\beta P$ acts by the scalar $\alpha$ on the first summand and $\alpha +\beta$ on the second. Its minimal polynomial is the product of the distinct corresponding linear polynomials whose summands are nonzero. If $\beta =0$ this is $x-\alpha$; if $P=0$ or $P=I$ only the associated single factor remains. Otherwise, for $\beta \ne 0$, it is $(x-\alpha )(x-\alpha -\beta )$. This follows directly from an adapted basis, so it works in every characteristic.

For an idempotent linear map $P$, $I-P$ is also idempotent andFor the first identity, $P(I-P)=0$ proves one inclusion. If $Pv=0$, then $v=(I-P)v$, proving the reverse inclusion. Apply the same argument to $I-P$ for the second identity. Consequently the ambient vector space is the direct sum of the image and kernel of a linear map of $P$. Orthogonality is not required; it is a separate property of an orthogonal projection.

For $P=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the off-diagonal equations in $P^2=P$ are $b(a+d-1)=c(a+d-1)=0$. Thus either $b=c=0$ or the trace is one. The condition alone is not sufficient: $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ has trace one but is not an idempotent matrix. For a two-by-two matrix of trace one, idempotence is equivalent to zero determinant, by the Cayley-Hamilton theorem.

For an idempotent linear map $P$, induction gives $P^j=P$ for all integers $j\ge 1$. Summing the matrix exponential series therefore proves the displayed identity. The direct sum of its kernel of a linear map and image of a linear map is fixed and scaled by $e^{\alpha }$, respectively. The result requires no Hermitian matrix hypothesis and is valid for real or complex $\alpha$.

Let ${\ell }_0,{\ell }_1$ be norm-one extensions to $L^{\infty }((0,1))$ of endpoint point evaluations on $C([0,1])$. For continuous functions $e_0,e_1$ with $e_i(j)={\delta }_{ij}$, the linear projection $Pf={\ell }_0(f)e_0+{\ell }_1(f)e_1$ has the displayed operator norm, independently of which extensions are chosen.

The triangle inequality proves the upper bound. At a point $x_{\ast }$ maximizing $|e_0|+|e_1|$, choose unit complex numbers $a_j$ with $a_j{e}_j(x_{\ast })=|e_j(x_{\ast })|$. The continuous function $f(x)=(1-x)a_0+x{a}_1$ has supremum norm one and ${\ell }_j(f)=a_j$, so $(Pf)(x_{\ast })=|e_0(x_{\ast })|+|e_1(x_{\ast })|$. The L-infinity norm of the continuous output equals its supremum norm, proving equality even at an endpoint.

For a finite-dimensional real or complex linear projection $P$, every vector splits as $Px+(I-P)x$ in the image of a linear map and the kernel of a linear map, respectively. Their intersection is zero, so a basis adapted to this direct sum represents $P$ as $\mathrm{diag}(I_r,0)$. Its trace and rank both equal $r$. In particular, a projection's determinant is zero unless it is the identity. Hermiticity is unnecessary for this algebraic identity; a Hermitian matrix projection is additionally orthogonal.

Every finite-dimensional vector subspace $E$ of a Banach space $X$ is the range of a bounded linear projection with the displayed norm bound. In maximal inscribed Euclidean or Hermitian ellipsoid coordinates, write the John contact decomposition as ${\sum }_r{c}_r{u}_r{u}_r^{\ast }=I_E$. Its contact functionals ${\varphi }_r(x)=u_r^{\ast }x$ have norm one on $E$. Extend them to $X$ by the Hahn-Banach theorem and set $Px={\sum }_r{c}_r{\tilde{\varphi }}_r(x)u_r$. This restricts to the identity on $E$. The contact synthesis operator has Euclidean norm one, the weights sum to $\dim E$, and the Euclidean unit ball is contained in the norm unit ball; these facts give $\Vert Px{\Vert }_E\le \sqrt{\dim E}\Vert x{\Vert }_X$.

For an idempotent linear map $P$ on a finite-dimensional real vector space, take any positive definite inner product on that space. The displayed new inner product is positive definite since its squared norm vanishes only if both $Pu$ and $(I-P)u$ vanish. It makes the kernel of a linear map and image of a linear map orthogonal. Since $P^2=P$, one has $⟨Pu,v{⟩}_P=⟨u,Pv{⟩}_P$; thus $P$ becomes an orthogonal projection.

If $T^2=T$, every vector splits as $v=(v-Tv)+Tv$ in $\ker T\oplus \mathrm{im}T$. On these summands $T$ is respectively zero and the identity. Its matrix in an adapted basis is therefore $\mathrm{diag}(0,I_r)$, where $r$ is its rank of a linear map. Two idempotent linear maps on the same finite-dimensional vector space are similar matrices exactly when their ranks agree.

If two closed subspaces of a Hilbert space give $H=V\oplus W$, the projection onto $V$ along $W$ fixes $V$ and vanishes on $W$. It need not be an orthogonal projection. The bounded inverse theorem applied to the addition map $(v,w)\mapsto v+w$ on the complete component spaces proves boundedness of the component maps. With both complementary subspaces nonzero, its operator norm and that of $I-Q$ equal $\sec {\theta }_{V,W^{\perp }}$, with the directed subspace angle convention. Indeed, for fixed $v\in V$, the smallest possible norm of $v+w$ over $w\in W$ is $\Vert P_{W^{\perp }}v\Vert$, which proves the formula for $\Vert Q\Vert$. Equality with $\Vert I-Q\Vert$ follows from the block representation $Q=\left(\begin{array}{cc}I& B\\ 0& 0\end{array}\right)$ on $V\oplus V^{\perp }$: both squared operator norms are $1+\Vert B{\Vert }^2$. If a summand is zero, $Q$ is zero or the identity operator, and one complementary operator norm is zero; those cases must be treated directly.

For equal finite-dimensional closed subspaces of a Hilbert space $T,S$, positivity of $\cos {\theta }_{T,S}$ makes $P_S{|}_T$ invertible. The displayed reconstruction is the unique element of $T$ whose orthogonal projection onto $S$ matches that of $f$. It is the oblique projection onto $T$ along $S^{\perp }$. Its operator norm is the value of the secant function at the directed subspace angle, and its error is at most that secant function value times the best orthogonal projection error. The lower error bound follows from the Pythagorean identity. Zero-dimensional cases are handled directly.

The domain $\mathcal{D}(A)$ of an operator $A$ is the set of vectors on which $A$ is defined. For an unbounded differential operator, the domain includes regularity and boundary conditions and is part of the operator's definition.

A linear operator $T$ on a Hilbert space $H$ is densely defined when its operator domain is a dense subset of $H$. Density makes the adjoint operator unique wherever its defining inner-product identity has a solution.

The identity operator maps every vector to itself. Every nonzero vector is an eigenvector with eigenvalue one.

Given a map $\tau :X\to X$, the associated composition operator on a function space sends $f$ to $f\circ \tau$.

A functional $\mu$ is invariant under the composition operator $T$ when $\mu (Tf)=\mu (f)$ for every $f$. On a compact metric space, a point-evaluation functional can be averaged along an orbit, and a weak-star convergent subsequence of those averages produces an invariant functional.

A unitary operator $U$ on a complex vector space with an inner product satisfies $U^{\ast }U=U{U}^{\ast }=I$ and therefore preserves inner products.

If A and B are commuting self-adjoint operators with $A^2=B^2=I$, then the displayed combination is a unitary operator. Indeed,and the reverse product is also I. The factors are involutions, and their commutation is essential to cancel the cross terms. Operators $A=U_A\otimes I$ and $B=I\otimes U_B$ on separate tensor factors provide a natural example. More generally, $\cos \theta A+i\sin \theta B$ is unitary for real $\theta$ by the same proof. Replacing i by a real coefficient usually loses this property.

Differentiating a unitary operator family at the identity gives an anti-Hermitian tangent. In the convention $U(ϵ)=I+iϵF+O({ϵ}^2)$ for real $ϵ$, the generator $F$ is Hermitian. This is the one-parameter version of the Hermitian quantum generator convention. A finite linear truncation need not itself be exactly unitary.

Two operators on Hilbert spaces are unitarily equivalent if a unitary operator maps their operator domains onto one another and intertwines their actions. Their spectra and eigenvalue multiplicities then agree. For nonnegative self-adjoint operators it suffices that the unitary maps their closed quadratic form domains onto one another and preserves the forms, since the associated operator is determined by the form.

An eigenphase of a unitary operator is a real number $\theta$ modulo one specifying an eigenvalue $e^{2\pi i\theta }$. Every unitary eigenvalue has modulus one. Quantum phase estimation records the eigenphase corresponding to an input eigenvector; a superposition retains the corresponding labels coherently before measurement.

Unitary conjugation maps an operator $A$ to $UA{U}^{†}$ for a unitary operator $U$. It preserves eigenvalues, operator norms, and algebraic relations such as products and adjoints.

For a unitary operator $U$ on a Hilbert space, the Cesàro averagesconverge strongly to the orthogonal projection onto the fixed-point subspace $\ker (U-I)$.

For a unitary operator $U$, the orthogonal complement of $\mathrm{Ran}(I-U)$ is $\ker (I-U^{\ast })=\ker (I-U)$. On the fixed space, Cesàro averages act as the identity. On $(I-U)g$ they telescope to $(g-U^{N}g)/N$. Their uniform norm bound extends convergence to the closure of the range, proving the Von Neumann mean ergodic theorem.

Two linear operators $A$ and $B$ commute when $AB=BA$. Their matrix exponentials then satisfy $e^{A+B}=e^A{e}^B=e^B{e}^A$.

For any real nonscalar $2\times 2$ matrix $A$, its commuting matrices are exactly $B=\alpha I+\beta A$ with $\alpha ,\beta \in \mathbb{R}$. There is a vector $v$ for which $v,Av$ form a basis: otherwise every nonzero vector would be an eigenvector, and applying this to two independent vectors and their sum would force $A$ to be scalar. Given a commuting $B$, uniquely write $Bv=\alpha v+\beta Av$. Then $BAv=ABv=(\alpha I+\beta A)Av$, so $B$ and $\alpha I+\beta A$ agree on a basis and are equal. Conversely every polynomial in $A$ commutes with it. This generalizes the centralizer of a planar quarter-turn; the scalar-matrix exception is essential, because every matrix commutes with a scalar matrix.

If $CD=DC$ and $Dx=\lambda x$, then $D(Cx)=C(Dx)=\lambda Cx$. Therefore each eigenspace of $D$ is invariant under $C$. The vector $Cx$ may be zero; invariance does not require it to be a nonzero eigenvector. This observation is the starting point for simultaneous diagonalization.

The anticommutator of two operators is $\{A,B\}=AB+BA$. They anticommute when this operator vanishes, equivalently when $AB=-BA$.

For complex invertible matrices $A,B$, $AB=-BA$ implies $B^{-1}AB=-A$ and $A^{-1}BA=-B$. Invariance of matrix trace under matrix similarity gives both zero traces. Invariance of determinant gives $\det A=(-1{)}^n\det A$, and invertibility forces $n$ even. Equivalently, eigenvalues occur in pairs $\lambda ,-\lambda$ with equal algebraic multiplicity, and zero is excluded. Both invertibility assumptions matter: $A=\mathrm{diag}(1,0,0)$ and $B=\mathrm{diag}(0,1,0)$ anticommute in dimension three but are singular and have nonzero traces.

If $A$ is diagonalizable, the anticommutator map $B\mapsto AB+BA$ has eigenvalues $2{\lambda }_1$, ${\lambda }_1+{\lambda }_2$ twice, and $2{\lambda }_2$. Their product gives the displayed characteristic polynomial. The density of diagonalizable complex matrices and continuity of polynomial coefficients prove the formula also for defective matrices.

Linear operators $\alpha$ and $\beta$ on the same finite-dimensional vector space are conjugate when $\beta =s^{-1}\alpha s$ for some linear isomorphism $s$. Their matrices in any fixed basis are similar.

For an invertible operator $\beta$ on $V$, the mapis a linear isomorphism. Conjugate choices of $\beta$ induce conjugate operators ${\varphi }_{\beta }$.

A matrix represents a linear map after bases have been chosen for its domain and codomain.

A rectangular array whose entries are nondecreasing in each coordinate. Every subarray obtained by restricting to a row interval and a column interval has the same property. Comparing a target with the top-right entry discards a whole row or column, giving staircase matrix search in time linear in the sum of the dimensions.

A real symmetric matrix whose entries are the moments $H_{ij}={\int }_0^1{x}^{i-1}{x}^{j-1}dx$. Thus it is the Gram matrix of the monomials in the real L2 inner product on $[0,1]$. For a nonzero real coefficient vector $v$, $v^{T}Hv={\int }_0^1({\sum }_i{v}_i{x}^{i-1}{)}^{2}dx>0$, so every finite Hilbert matrix is positive-definite. The moment description connects these matrices with polynomial approximation and with the shifted Hilbert quadratic-form bound.

For real $a_r$, put $h(z)={\sum }_{r=1}^n{a}_r{z}^r$. The quadratic form equals ${\int }_0^{1}h(x{)}^{2}dx$, proving its nonnegativity. Apply the centred logarithmic keyhole identity bound to the entire function $h^2$. Since $|h(e^{i\theta }{)}^2|=|h(e^{i\theta }){|}^2$, orthogonality of the finite Fourier sum gives ${\int }_0^{2\pi }|h(e^{i\theta }){|}^{2}d\theta =2\pi {\sum }_r{a}_r^2$. Multiplication by the contour estimate's factor $1/2$ proves the upper bound. The matrix $(1/(r+s+1))$ is the principal submatrix of a Hilbert matrix obtained by deleting its first row and column.