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The zero-composition subspaces of a linear map are subspaces because the defining equations are homogeneous linear equations in $\beta$: they hold for the zero map and are preserved by addition and scalar multiplication. A map with $\beta \alpha =0$ vanishes on $\mathrm{im}\alpha$, so it factors uniquely through the quotient vector space $V/\mathrm{im}\alpha$. Conversely every map from that quotient to $U$ gives such a $\beta$. ThereforeThe condition $\alpha \beta =0$ says exactly that the image of $\beta$ lies in $\ker \alpha$. ConsequentlyBoth dimension formulas use the fact that the space of maps between spaces of dimensions $p,q$ has dimension $pq$: matrix units in any chosen bases supply a basis.

Use the standard coroot normalization $\alpha (H_{\alpha })=2$. The symbols $H_{\alpha }$ are vectors in the Cartan subalgebra, not the root-space vectors $X_{\alpha }$. The root-string theorem gives $\beta (H_{\alpha })\in \mathbb{Z}$, so every root takes real values onThe simple coroots form a basis of this real space.

In the root-space decomposition, $\mathrm{ad}H$ is zero on $\mathfrak{h}$ and scalar multiplication by $\beta (H)$ on the one-dimensional space ${\mathfrak{g}}_{\beta }$. Therefore, for $H,K\in \mathfrak{h}$,For $H,K\in {\mathfrak{h}}_{\mathbb{R}}$ every term is real, so the restricted form is real. Moreover,If it is zero, all $\beta (H)$ vanish. Then $H$ commutes with every root space and with $\mathfrak{h}$, hence is central in $\mathfrak{g}$. The center of a Lie algebra vanishes for a semisimple Lie algebra, so $H=0$. Thus the Euclidean subspace of a Cartan subalgebra hasEquivalently, positivity follows because the roots span ${\mathfrak{h}}^{\ast }$.

Because $Y$ is closed, it is a nowhere dense set exactly when its interior is empty. If a ball about $y\in Y$ lay inside $Y$, translating it by $-y$ would put a ball about $0$ inside $Y$. Every vector of $X$ has a small scalar multiple in that ball, and closure under scalar multiplication would force $Y=X$. This contradiction proves a proper closed linear subspace is nowhere dense.

Put $U=⟨u_1,\dots ,u_k⟩$. Taking all coefficients zero puts the zero vector in $U$. If $s={\sum }_i{\alpha }_i{u}_i$, $t={\sum }_i{\beta }_i{u}_i$ lie in $U$ and $a,b\in \mathbb{R}$, thenThus $U$ is closed under vector addition and scalar multiplication, contains the zero vector, and is therefore a linear subspace of $V$. This is precisely why the linear span is a linear subspace.

A map $T:V\to W$ between vector spaces over the same field $\mathbb{F}$ is a linear map precisely when it preserves vector addition and scalar multiplication:Equivalently, $T(au+bv)=aT(u)+bT(v)$ for all $u,v\in V$ and $a,b\in \mathbb{F}$.

A vector space over a field $\mathbb{F}$ is a set $V$ with vector addition $V\times V\to V$ and scalar multiplication $\mathbb{F}\times V\to V$ satisfying the following vector space axioms. Under vector addition, $V$ is an abelian group: addition is associative and commutative, there is a zero vector $0_V$, and each $v$ has an additive inverse $-v$. For all $a,b\in \mathbb{F}$ and $u,v\in V$,The operations are defined for every indicated pair and take their values in $V$.

Take the vector subspace of skew-symmetric matrices,It is closed under vector addition and scalar multiplication, and a basis isIndeed, its diagonal entries vanish, and its entries below the diagonal are forced by those above it. Hence $\boxed{\dim K=\left(\genfrac{}{}{0ex}{}{n}{2}\right)}$. Part (a)(ii) proves that the only diagonalisable matrix in $K$ is $0_n$.

Apply scalar multiplication and vector addition componentwise:

Let $e_1,\dots ,e_n$ denote the standard basis of a module of the free module $R^n$. Since the module homomorphism $\varphi :R^m\to M$ is surjective, choose $v_j\in R^m$ such that $\varphi (v_j)=\psi (e_j)$ for each $j$. DefineUniqueness of coordinates in a basis of a module makes this well-defined, and the formula preserves addition and scalar multiplication, so $q$ is a module homomorphism. For every $x={\sum }_j{r}_j{e}_j$,Thus $\boxed{\varphi \circ q=\psi }$. This is the lifting property of a free module; only the surjectivity of $\varphi$ is needed for this part.

Put $K=\ker \varphi$, $H=\ker \psi$, and $Q=\mathrm{Im}q$. The kernel of a module homomorphism $H$ is a submodule, so its image $L=q(H)$ is a submodule. For $x\in H$, $\varphi (q(x))=\psi (x)=0$, showing $L\subseteq K$. Thus the quotient module $K/L$ exists. Also $L\subseteq Q$.

Consider the restriction to $K$ of the natural module homomorphism $R^m\to R^m/Q$:Its kernel of a module homomorphism is $K\cap Q$. We have $L\subseteq K\cap Q$. Conversely, if $y\in K\cap Q$, choose $x\in R^n$ with $y=q(x)$. Thenso $x\in H$ and $y\in L$. Hence $\boxed{K\cap Q=L}$.

The map $\alpha$ is surjective. Given $v+Q\in R^m/Q$, use the surjectivity of $\psi$ to choose $x\in R^n$ with $\psi (x)=\varphi (v)$. Set $y=v-q(x)$. ThenThus $y\in K$ is a preimage of the arbitrary class $v+Q$.

By the first isomorphism theorem for modules, $\alpha$ induces the module isomorphismExplicitly, replacing $y$ by $y+\ell$ with $\ell \in L$ leaves its class modulo $Q$ unchanged, proving well-definedness. Its formula preserves addition and scalar multiplication. If its value is zero, $y\in K\cap Q=L$, so $y+L=0$, proving injectivity; surjectivity was proved above. These checks give all the requested properties of the induced map.

The zero matrix has the required pattern, and matrix addition and scalar multiplication preserve every prescribed zero entry. This proves the vector subspace condition. A basis consists of the matrix unitsThey independently select every permitted entry of a tridiagonal matrix, so the dimension is $n+2(n-1)=3n-2$ for $n\ge 1$.

For every additive function $f:\mathbb{R}\to \mathbb{R}$, every real number $x$, and every rational number $q$, one has $f(qx)=qf(x)$. This is the scalar multiplication part of being a linear map over the rational numbers.

First $f(0)=f(0)+f(0)$ gives $f(0)=0$, and $f(x)+f(-x)=f(0)$ gives $f(-x)=-f(x)$. Repeated addition gives $f(nx)=nf(x)$ for each positive integer $n$, and the negative case follows from the sign identity. If $q=m/n$ with $m\in \mathbb{Z}$ and $n>0$, thenNo continuity was used. In particular $f(1)=3$ fixes $f(-2/3)=-2$, $f(5/6)=5/2$, and every other rational value, even for a discontinuous additive function.

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.

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.

At a fixed point $p$ of an $n$-dimensional smooth manifold $M$ without boundary, the tangent space $T_{p}M$ is the real vector space of all tangent vectors at $p$. In manifold charts $x=(x^1,\dots ,x^n)$ and $y=(y^1,\dots ,y^n)$, a tangent vector has componentsThe coordinate basis transforms inversely, so the tangent vector is unchanged. This invertible Jacobian matrix also preserves vector addition and scalar multiplication, giving a vector space of dimension $n$ independently of coordinates. ^[ref]

For example, at $(x,y)=(1,0)$ in ${\mathbb{R}}^2$, use the alternative manifold chart $u=x$, $w=y+x^2$. The Jacobian matrix sends $(v_x,v_y)=(2,3)$ to $(v_u,v_w)=(2,7)$. Indeed, the smooth curve $(x(t),y(t))=(1+2t,3t)$ gives $w(t)=1+7t+4t^2$. Conversely, ${\partial }_u{|}_p={\partial }_x{|}_p-2{\partial }_y{|}_p$ and ${\partial }_w{|}_p={\partial }_y{|}_p$, so $2{\partial }_u{|}_p+7{\partial }_w{|}_p=2{\partial }_x{|}_p+3{\partial }_y{|}_p$. The numbers change; their action on every smooth function does not.

For an embedded submanifold $M\subseteq {\mathbb{R}}^N$, the tangent space can be identified with the vector subspaceA proof outline explains why every coordinate vector is possible: in a manifold chart $x$, lift the short straight path $x(p)+ta$ using $x^{-1}$. Its tangent vector has components $a$. For a regular local parametrization $X:U\subseteq {\mathbb{R}}^n\to M$, the same chain rule gives $T_{p}M=\mathrm{im}D{X}_q$, where $X(q)=p$. This is the tangent space from a local parametrization.

For the sphere $S_R^2=\{p\in {\mathbb{R}}^3:p\cdot p=R^2\}$, differentiating that equation gives $p\cdot v=0$. Conversely, for any nonzero $v$ with $p\cdot v=0$, the smooth curvelies on the sphere and has ${\gamma }_v^{\prime }(0)=v$; use a constant smooth curve for $v=0$. Thus $T_p{S}_R^2=\{v:p\cdot v=0\}$, not merely a subset of that plane. At $p=(0,0,R)$ this vector subspace is $v_z=0$, while the affine tangent plane drawn through $p$ is $z=R$.

This also explains a mechanical constraint without requiring a force calculation. In the ideal mathematical model of a particle confined to a stationary sphere, its allowed velocities at $p$ are exactly $T_p{S}_R^2$: the radial component must vanish, but either independent tangential direction is available. For a general stationary holonomic constraint $F(q)=0$ with independent equations, the regular level set theorem gives $T_{q}M=\ker D{F}_q$. These are instantaneous kinematic possibilities; they do not determine the actual subsequent motion. On a general smooth manifold, tangent spaces at different base points cannot be identified canonically; an affine connection supplies parallel transport when such comparisons are needed.

A tensor product turns a bilinear map into a linear map. For vector spaces $V,W$ over the same field $F$, its elements are finite sums of pure tensors $v\otimes w$, subject to bilinearity:The scalar $a$ belongs to $F$; writing $V{\otimes }_{F}W$ makes this choice explicit. These rules preserve separate linearity while allowing arbitrary linear combinations of products.^[ref]

For a concrete example, take $V=W={\mathbb{R}}^2$, with basis vectors $e_1,e_2$ in the first copy and $f_1,f_2$ in the second. The four tensor-product basis elements areThe pure tensor made from $v=2e_1+e_2$ and $w=-f_1+3f_2$ expands toEach basis vector in one factor can interact with each basis vector in the other. This is why dimensions multiply; the tensor-product coefficient matrix displays these four interactions as a two-by-two grid.

The tensor product is different from the Cartesian product $V\times W$. With componentwise vector addition and scalar multiplication, the latter contains pairs and has dimension $\dim V+\dim W$. The map $\tau (v,w)=v\otimes w$ is bilinear, generally neither linear on that Cartesian product nor injective nor surjective. In our example, $\tau (2v,2w)=4\tau (v,w)$, and $\tau (2v,w/2)=\tau (v,w)$; the separability of a tensor explains why some outputs cannot be reached at all. Both spaces happen to have dimension four when $V=W={\mathbb{R}}^2$, but this coincidence does not identify their structures.

A topological vector space is a vector space $E$ over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, equipped with a topology such that vector addition and scalar multiplication are jointly continuous:The domains carry the product topology, and the scalar field carries its usual topology. Joint continuity permits both inputs to vary at once. Here the definition does not impose the Hausdorff condition; we state it separately when needed. ^[ref].

Every normed vector space is a topological vector space: the triangle inequality gives $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$, whilecontrols scalar multiplication near any $({\lambda }_0,x_0)$. More generally, a seminorm topology supplies many measurements of smallness, rather than one norm. Translation by any vector is a homeomorphism, so a neighbourhood basis at zero determines all neighbourhoods by translation.

The compatibility matters: on a nonzero real vector space, the discrete topology makes vector addition continuous, but not scalar multiplication. For $x\ne 0$, the sequence of scalars $1/n$ tends to zero, whereas the sequence $x/n$ never enters the open set $\{0\}$. Thus choosing a topology arbitrarily need not respect the linear algebra.

This framework lets functional analysis choose the appropriate notion of limit of a sequence. Norm topology measures one global error, weak topology tests linear functionals, and the Fréchet space of smooth functions controls every order of derivative. In each case the operations forming linear combinations are continuous, making approximation compatible with the underlying equations.

A vector is an element of a vector space. Its defining operations are vector addition and scalar multiplication. An arrow in a Euclidean space is one example; a polynomial or a function can also be a vector when the chosen vector space uses the corresponding addition and scalar multiplication.

For instance, the polynomials $p(x)=1+x$ and $q(x)=x-x^2$ belong to the real vector space of polynomials of degree at most two. Their vector addition gives $(p+q)(x)=1+2x-x^2$, and scalar multiplication gives $(2p)(x)=2+2x$. In the monomial basis $(1,x,x^2)$ these are the coordinate vectors $(1,1,0)$, $(0,1,-1)$, $(1,2,-1)$ and $(2,2,0)$ respectively. The vectors are the polynomials; these tuples are their representations in a chosen basis.

A general vector space does not itself prescribe lengths or angles. A norm supplies a size, while an inner product also supplies angle information and orthogonality. The familiar arrow geometry below belongs to Euclidean vectors, where this extra structure is available.

Vector addition is the vector space operation combining two vectors into their sum. It is commutative and associative, has the zero vector as identity, and distributes with scalar multiplication as specified by the vector space axioms. Coordinatewise addition describes it after choosing a basis.

A vector space over a field $\mathbb{F}$ is a set $V$ equipped with vector addition and scalar multiplication satisfying the vector space axioms. Its vectors can represent directed displacements, lists of measurements, polynomials, or entire functions. The common structure is that adding two elements and multiplying an element by a scalar remain meaningful operations inside the same space.

For example, ${\mathbb{R}}^2$ uses coordinatewise addition and scaling. Withwe have $v=2b_1+b_2$. Moving twice along $b_1$ and once along $b_2$ reaches the same vector as five horizontal and four vertical steps. The two descriptions refer to different bases of the same space. A basis makes such descriptions unique; a spanning set makes them possible.

The space $\mathbb{R}[x{]}_{\le 2}$ of real polynomials of degree at most two has basis $1,x,x^2$: the polynomial $3-2x+x^2$ is the linear combination with coordinates $(3,-2,1)$. The zero vector here is the zero polynomial, and addition combines coefficients. A derivative is a linear map on this space; its matrix records how it acts on the three basis vectors. The same coordinate machinery therefore handles both geometric motion and operations on functions.

The choice of field matters. The complex numbers form a one-dimensional vector space over $\mathbb{C}$ with basis $1$, but a two-dimensional vector space over $\mathbb{R}$ with basis $1,i$. The question "how many independent directions?" has an answer only after the allowed scalars are specified. Likewise, polynomials of degree exactly two do not form a vector space: subtraction can remove the leading term, and the zero polynomial is absent.

The usefulness of the abstraction is that a linear map is determined by its values on a basis. Once those images are known, linear combinations give its action on every vector. The rank-nullity theorem relates the number of attainable outputs to the number of invisible input directions, while change of basis lets us choose coordinates suited to the problem. Eigenvectors, for example, choose directions on which repeated application of a linear operator becomes scalar multiplication.

A historical source is Hermann Grassmann's Die lineale Ausdehnungslehre (1844). Its full title advertises applications to other branches of mathematics, statics, mechanics, magnetism and crystallography; the original edition is digitized by ETH-Bibliothek. This gives an early setting for treating geometric quantities algebraically across different applications, rather than attributing the modern axiom list to that title alone.^[ref]