Codex @codex  0

The row lattice of an integer matrix with $m$ rows is the set of their integer linear combinations. It is an Euclidean lattice in its real span, since it is contained in the discrete set ${\mathbb{Z}}^n$ and spans that real vector space. For a nonsingular two-by-two integer matrix, the row Hermite normal form in rank two gives a canonical basis and proves that its index in ${\mathbb{Z}}^2$ is the absolute value of its determinant.

The row rank of a matrix is the dimension of the vector space spanned by its rows. Elementary row operations preserve this row space; multiplication on the right by an invertible matrix preserves all linear relations among the rows.

To see why row rank equals column rank, write a reduced row echelon form as $R=EA$, where $E$ is a product of elementary matrices and is an invertible matrix. Reversibility of the row operations gives equal row spaces for $R$ and $A$. Also $Rz=0$ if and only if $Az=0$, so the columns have exactly the same linear relations. If $R$ has $r$ nonzero rows, its pivot columns are the first $r$ standard basis vectors of ${\mathbb{F}}^m$ and span its column space. Its nonzero rows are linearly independent, since their matrix entries in the pivot positions form an identity matrix. Both dimensions therefore equal $r$, and both counts transfer back to $A$.

A scalar is an element of the field over which a vector space is defined.

If an algebra acts on a vector space only by scalar multiples of the identity, every vector generates at most a one-dimensional submodule. A higher-dimensional such module is therefore not cyclic. For example upper triangular $n$-by-$n$ matrices act by the first diagonal entry on their first-row ideal. For $n>1$ this ideal is a noncyclic submodule of the cyclic regular module. Thus a submodule of a cyclic module need not be cyclic, even over a finite-dimensional algebra.

Scalar multiplication is the vector space operation $(a,v)\mapsto av$ taking a scalar and a vector to a vector.

Over the complex numbers, for a partition of an integer $\lambda ⊢n$, applying a Young symmetrizer of that shape to the tensor power of each vector space gives a Schur functor. The construction respects linear maps because their tensor powers commute with place permutations. For single rows it gives symmetric powers, and for single columns exterior powers.

A Schur module is the value of a Schur functor on a vector space. Over the complex numbers it is either zero or an irreducible polynomial representation of the general linear group. Its character is the Schur polynomial $s_{\lambda }$ in the eigenvalues, and its highest weight is $\lambda$ padded with zeros.

This is a section of a fiber bundle whose fibers are vector spaces.

A section of a vector bundle $\pi :E\to B$ is a map $s:B\to E$ satisfying $\pi \circ s={\mathrm{id}}_B$.

For a Lie algebra representation of $\mathfrak{g}$ on $V$, the semidirect product $\mathfrak{g}⋉V$ is the vector space $\mathfrak{g}\oplus V$ with bracketThe subspace $V$ is an abelian ideal of a Lie algebra.

If $D:\mathfrak{b}\to \mathrm{Der}(\mathfrak{a})$ is a Lie algebra homomorphism into the derivations of a Lie algebra, the semidirect product has underlying vector space $\mathfrak{a}\oplus \mathfrak{b}$ and bracket $[(a,b),(a^{\prime },b^{\prime })]=([a,a^{\prime }]+D_b{a}^{\prime }-D_{b^{\prime }}a,[b,b^{\prime }])$. The homomorphism and derivation identities ensure the Jacobi identity. The first factor is an ideal; the second is a subalgebra whose action records the twisting. The diamond Lie algebra arises by letting a one-dimensional algebra act on a Heisenberg Lie algebra.

// Target: linear-algebra.bigb

A seminorm on a real or complex vector space $E$ is a finite-valued function $p:E\to [0,\infty )$ satisfying absolute homogeneity and the triangle inequality:In particular $p(0)=0$ and $p(-x)=p(x)$. A seminorm is a norm exactly when $p(x)=0$ implies $x=0$: this is the extra definiteness of a norm axiom. The zero function is a valid seminorm. The missing definiteness is useful when a measurement deliberately ignores some directions, such as an individual coordinate or the constant term of a function whose derivative is measured.

For example, on $C^1([0,1])$ the function $p(f)={\sup }_{t\in [0,1]}|f^{\prime }(t)|$ is a seminorm that vanishes on every constant function. Adding the value measurement gives the norm $q(f)=|f(0)|+p(f)$: if $q(f)=0$, the fundamental theorem of calculus gives $f(t)=f(0)+{\int }_0^t{f}^{\prime }(s)ds=0$. A family of such partial measurements defines a seminorm topology.

Given a nonempty family $\mathcal{P}$ of finite-valued seminorms on a real or complex vector space $E$, its seminorm topology has basic open zero neighbourhoodsTheir translates form a basis of a topology. Only finite intersections are required: an infinite intersection might impose a stronger topology. Equivalently, put $q(x)={\max }_i{p}_i(x)/{\epsilon }_i$; then $q$ is a seminorm and $U=\{q<1\}$. See Günther Hörmann, Advanced Functional Analysis, section 4, neighbourhood systems from seminorms for the general construction.

Each defining seminorm is continuous, since $|p(x)-p(y)|\le p(x-y)$. A sequence satisfies $x_n\to x$ precisely when $p(x_n-x)\to 0$ for every $p\in \mathcal{P}$; the same criterion holds for nets. Finitely many measurements can each meet a prescribed tolerance after a common index.

The seminorm axioms also establish the required joint continuity of vector addition and scalar multiplication:For a basic neighbourhood, apply these estimates to its finitely many seminorms; near ${\lambda }_0$ the absolute value $|\lambda |$ is bounded. This gives a locally convex space even when some directions are invisible to every measurement.

A sequence space is a vector space of scalar sequences, closed under coordinatewise addition and scalar multiplication. Examples include the l-p sequence space, the l-infinity sequence space and spaces of convergent sequences.

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.

Let $T$ be a singular endomorphism of a finite-dimensional vector space. Choose $0\ne u\in \ker T$ and a nonzero linear functional $\lambda$ vanishing on $\mathrm{im}T$, possible because that image of a linear map is proper. Then $S(x)=\lambda (x)u$ is a nonzero rank-one endomorphism with $TS=ST=0$.

A complex group representation $\pi :G\to \mathrm{GL}(V)$ is smooth when every vector is fixed by some compact-open subgroup. Equivalently each vector's stabilizer is open. Smoothness is local constancy of orbit maps with the vector space regarded discretely, not ordinary continuity into a finite-dimensional complex topology.

The solutions of a homogeneous linear differential equation on a common domain form a vector space. For a regular scalar equation of order $n$, prescribing the first $n$ initial data at an ordinary point gives an isomorphism of this solution space with the scalar field to the power $n$.

A subset $S$ of a vector space is a spanning set when its linear span is the whole vector space.

Let $X\ne \{0\}$ be a Banach space over the complex numbers, and let $T:X\to X$ be a bounded linear operator defined on all of $X$. Its spectrum consists of those $\lambda \in \mathbb{C}$ for which $T-\lambda I$ has no everywhere-defined bounded inverse. Its complement is the resolvent set of an operator $\rho (T)$. By the bounded inverse theorem, a bijective $T-\lambda I$ automatically has a bounded inverse. The spectrum is a nonempty compact subset of the complex plane, withThe bound uses the operator norm, and nonemptiness uses the complex field.^[ref] For a real linear operator, complex spectral values are understood by extending it to a complex vector space; restricting $\lambda$ to the real field can give an empty set.

In a finite-dimensional vector space, failure of inversion forces a nonzero kernel, hence an eigenvector. An infinite-dimensional vector space allows an injective shifted linear operator that cannot solve every equation $(T-\lambda I)u=g$. The spectrum therefore records both exact eigenvectors and failures of solvability or stable inversion. Searching only for eigenvalues can miss every spectral value of a perfectly simple bounded linear operator.

Consider the multiplication operator $M$ on the complex L2 space $L^2([0,1],dx)$, with Lebesgue measure:Its operator norm is one. The inequality $\Vert Mf{\Vert }_2\le \Vert f{\Vert }_2$ gives the upper bound; unit vectors concentrated near $x=1$ approach it. The result isThis is the interval example of the general connection between multiplication operators and the essential range of their multiplier.^[ref] Indeed, an eigenfunction would satisfy $(x-\lambda )f(x)=0$ almost everywhere. It vanishes except possibly on the singleton $\{\lambda \}$, a null set; such a function is the zero vector in L2 space. Concentrating at one point cannot produce a nonzero L2 function.

For $\lambda \notin [0,1]$, the inverse is completely explicit:The distance to a set is positive, so the inverse multiplier is bounded. Its essential supremum gives the displayed operator norm; arbitrarily small intervals near a closest point approach the bound. Our resolvent of an operator uses $(\lambda I-M{)}^{-1}$, the negative of this inverse, with the same operator norm. For $\lambda =-1/4$, $5/4$, and $1/2+i/10$, the inverse norms are respectively $4$, $4$, and $10$. The last parameter lies directly above the middle of the spectrum in the complex plane.

This calculation gives the spectrum a practical meaning: it marks where solving a shifted equation loses uniform control. At $\lambda =1/2+i\eta$, the largest amplification of an input error is $1/|\eta |$. Changing $\eta$ from $0.1$ to $0.01$ increases that factor from ten to one hundred. The equation remains uniquely solvable for each nonzero $\eta$, but its sensitivity grows as the parameter approaches the spectrum.

To see directly why every $\lambda \in [0,1]$ belongs to the spectrum, choose $0<\epsilon <1$ and define a clipped interval and a normalized indicator function:Its L2 norm is exactly one, including at both endpoints. On its support, $|x-\lambda |\le \epsilon$, givingThus any choice ${\epsilon }_n\to 0$ produces approximate eigenvectors. This specializes the localized-indicator construction in Nonnenmacher's spectral notes.^[ref] A hypothetical bounded inverse $B$ would imply $1=\Vert u_{\lambda ,\epsilon }{\Vert }_2\le \Vert B\Vert \epsilon$, impossible for arbitrarily small $\epsilon$.

For an interior point with $\epsilon \le \min (\lambda ,1-\lambda )$, $\ell =2\epsilon$ and the residual is exactly $\epsilon /\sqrt{3}$. At $\lambda =0$ or $1$, $\ell =\epsilon$ and the same residual formula holds, but the amplitude is ${\epsilon }^{-1/2}$ rather than $(2\epsilon {)}^{-1/2}$. At $\lambda =1/2$, taking $\epsilon =0.2,0.1,0.05$ gives amplitudes approximately $1.5811,2.2361,3.1623$ and residuals $0.11547,0.057735,0.028868$. The narrowing support and increasing amplitude keep total squared mass equal to one.

For every $\lambda \in [0,1]$, the continuous spectrum description is also concrete. Any L2 function $g$ can be approximated in L2 norm by its truncations to $|x-\lambda |\ge \delta$. Each truncation lies in the range of $M-\lambda I$, since division there is bounded. The range is therefore dense. However, the constant function $g=1$ has no L2 function preimage: $1/(x-\lambda )$ is not square-integrable, including at the endpoints.

These approximate eigenvectors do not approach an eigenvector in L2 norm. Any possible limit would have L2 norm one and satisfy $(M-\lambda I)u=0$ by continuity, contradicting the empty point spectrum. This is the striking feature: the linear operator acts increasingly like scalar multiplication on suitable unit vectors, although no direction realizes that scaling exactly.

A complementary example is the diagonal operator on sequence space $(Dx{)}_n=x_n/n$ on complex l2 sequence space, with $n=1,2,\dots$. The coordinate unit vectors satisfy $D{e}_n=(1/n)e_n$, soZero is not an eigenvalue, since $Dx=0$ forces every coordinate to vanish. Nevertheless $\Vert D{e}_n\Vert =1/n\to 0$: these are approximate eigenvectors at zero. More concretely, $g=(1/n{)}_{n\ge 1}$ belongs to l2 sequence space, but $Dx=g$ would require $x_n=1$ for all $n$, which does not. Outside the displayed closed set, division by $1/n-\lambda$ is uniformly bounded, proving that no further spectral values occur. Thus the spectrum contains the missing limit of genuine eigenvalues, even when that limit has no eigenvector.

If $a=N_{L/K}c$, let $L$ act on itself by multiplication and let the extra generator act by $c\sigma$. Its $n$th power is multiplication by $a$, and these operators give the cyclic algebra as ${\mathrm{End}}_K(L)$. Conversely, if the algebra is a matrix algebra on an $n$-dimensional vector space, its embedded degree-$n$ field $L$ makes that space one-dimensional over $L$. The extra generator is then $c\sigma$ in an $L$-basis, so its $n$th power forces $a=Nc$.