Vector Spaces, Duals, Linear Maps, and Bases

The basis-independent statements are the ones formulated in terms of spaces, vectors, covectors, and maps: membership in a subspace, evaluation of a covector, composition of maps, kernels, images, rank, and linear dependence. A basis turns these objects into arrays, but a change of basis changes the arrays without changing the objects. The decisive check is therefore not whether a formula has indices; it is whether every index formula represents a well-typed map and transforms back to the same geometric statement.

This page works over a finite-dimensional vector space over $\mathbb F=\mathbb R$ or $\mathbb C$. It uses no norm, topology, or inner product unless one is introduced explicitly. Topological duals and unbounded operators require the separate framework of Functional and Spectral Analysis.

Vector spaces before coordinates

A vector space $V$ over $\mathbb F$ is a set with addition and scalar multiplication satisfying the usual linearity axioms. A subspace $U\subseteq V$ is closed under those operations. Given $v_1,\ldots,v_n\in V$, their span is

$$\operatorname{span}\{v_1,\ldots,v_n\} = \left\{ \sum_{a=1}^n c^a v_a:c^a\in\mathbb F \right\}.$$

The vectors are linearly independent when $\sum_a c^a v_a=0$ forces every $c^a=0$. A basis is an independent spanning list. It gives a unique component expansion

$$v=v^a e_a,$$

but $v$ is the vector and $(v^1,\ldots,v^n)$ is only its component list in the chosen basis $\{e_a\}$. The dimension $\dim V$ is the number of elements in any basis of finite-dimensional $V$.

The algebraic dual

$$V^*=\operatorname{Hom}_{\mathbb F}(V,\mathbb F)$$

is the vector space of linear functionals on $V$. Its elements are covectors. For a basis $\{e_a\}$, the dual basis $\{e^a\}$ is defined by

$$e^a(e_b)=\delta^a{}_b.$$

Thus a covector $\alpha=\alpha_a e^a$ evaluates a vector by

$$\alpha(v)=\alpha_a v^a.$$

This pairing $V^*\times V\to\mathbb F$ is part of the definition of the dual. By contrast, a map $V\to V^*$ is extra structure. A nondegenerate bilinear or Hermitian form can supply such a map, but there is no preferred identification of $V$ with $V^*$ for a bare vector space. This distinction is the reason that “raising” or “lowering” an index is never a typographical operation.

A linear map $T:V\to W$ satisfies

$$T(av+bw)=aT(v)+bT(w).$$

Its kernel and image are the intrinsic subspaces

$$\ker T=\{v\in V:T(v)=0\}, \qquad \operatorname{im}T=\{T(v):v\in V\}.$$

The rank is $\dim(\operatorname{im}T)$, not the number of nonzero entries in some matrix. These definitions make sense before either space has a basis. The definitions and finite-dimensional scope here follow Axler 2024, Chapters 1–3, with the vector–covector distinction cross-checked against Frankel 2012, §§2.1a and 2.4.

What a basis change actually does

Let $\{e_a\}$ be a basis of $V$ and choose a new basis

$$e'_a=e_b M^b{}_a,$$

where $M$ is invertible. Since the vector itself is unchanged,

$$v=v^a e_a=v'^a e'_a, \qquad v'^a=(M^{-1})^a{}_b v^b.$$

The dual basis must satisfy $e'^a(e'_b)=\delta^a{}_b$, so

$$e'^a=(M^{-1})^a{}_b e^b, \qquad \alpha'_a=\alpha_b M^b{}_a.$$

The opposite transformations make evaluation invariant:

$$\alpha'_a v'^a = \alpha_b M^b{}_a (M^{-1})^a{}_c v^c = \alpha_b v^b.$$

This cancellation is the invariant content of an upper–lower contraction. It does not depend on treating upper and lower indices as decorations.

For a map $T:V\to W$, choose a basis $\{f_i\}$ of $W$ and write

$$T(e_a)=f_i T^i{}_a.$$

If $f'_i=f_jN^j{}_i$, then the new component matrix is

$$T'=N^{-1}TM.$$

Indeed, $v'=M^{-1}v$ and $w'=N^{-1}w$ give

$$w'=T'v' =N^{-1}TM M^{-1}v =N^{-1}Tv.$$

Using one change-of-basis matrix on both sides is justified only when the domain and codomain have actually been identified and their bases are being changed together. For an endomorphism $T:V\to V$ with the same old and new basis in domain and codomain, the formula reduces to the similarity transformation $T'=M^{-1}TM$.

Kernels, images, and the rank–nullity check

The kernel and image do not move when a matrix is rewritten in another basis; only their component descriptions do. Their dimensions are connected by the rank–nullity theorem:

$$\dim V=\dim\ker T+\dim\operatorname{im}T.$$

For a short proof, take a basis $u_1,\ldots,u_k$ of $\ker T$ and extend it to a basis $u_1,\ldots,u_n$ of $V$. Then

$$T(u_{k+1}),\ldots,T(u_n)$$

spans $\operatorname{im}T$. It is also independent: if $\sum_{a=k+1}^n c^aT(u_a)=0$, then $\sum_{a=k+1}^n c^a u_a\in\ker T$, which is impossible unless every $c^a$ vanishes because the full list of $u_a$ is independent. Hence $\dim\operatorname{im}T=n-k$.

This theorem is an efficient consistency test. For example, a map from a five-dimensional space with a two-dimensional kernel must have rank three, regardless of the bases or row-reduction strategy used. If $V$ and $W$ have the same finite dimension, then injectivity, surjectivity, and invertibility of $T:V\to W$ are equivalent. Without equality of dimensions, those implications must be checked separately.

Dual maps and the absence of a preferred transpose

Every linear map $T:V\to W$ induces a dual, or pullback, map

$$T^*:W^*\longrightarrow V^*, \qquad (T^*\beta)(v)=\beta(Tv).$$

The direction reverses because a functional on $W$ can be composed with a map into $W$. Functoriality follows directly:

$$(S\circ T)^*=T^*\circ S^*.$$

In dual bases, $T^*$ is represented by the transpose of the component matrix of $T$. This statement uses no inner product and no complex conjugation. An adjoint is a different construction: it requires specified forms on the domain and codomain. In a complex inner-product space its matrix generally involves conjugate transpose, not transpose. The chapter’s later page on forms, adjoints, and isometries develops that extra structure.

There is, however, a canonical evaluation map into the double dual,

$$\iota_V:V\longrightarrow V^{**}, \qquad \iota_V(v)(\alpha)=\alpha(v).$$

For finite-dimensional $V$, $\iota_V$ is an isomorphism. It is natural because it was defined without a basis or form. This does not produce a canonical isomorphism $V\simeq V^*$.

QFT-facing example: a field multiplet and its source

Consider a finite multiplet of regulated or classical field variables taking values in an internal vector space $R$:

$$\phi(x)=\phi^a(x)e_a.$$

A linear source belongs to the dual space, $J(x)=J_a(x)e^a\in R^*$, so the source coupling is the scalar

$$S_{\mathrm{source}} = \int\mathrm d^d x\,J(x)(\phi(x)) = \int\mathrm d^d x\,J_a(x)\phi^a(x).$$

Under $e'_a=e_bM^b{}_a$,

$$\phi'^a=(M^{-1})^a{}_b\phi^b, \qquad J'_a=J_bM^b{}_a,$$

and therefore $J'_a\phi'^a=J_a\phi^a$. The invariant pairing, rather than the appearance of a row and a column, is what makes the coupling independent of the chosen field coordinates.

For complex fields, one must say whether the source is in the algebraic dual, the conjugate dual, or a space identified using a Hermitian form. Those choices lead to different component conjugations even though the type check is the same. For quantum fields, both the field and source pairing are distributional; the finite-dimensional internal-space argument here does not turn $\phi(x)$ into an ordinary pointwise operator. For the physics-facing linear-algebra conventions used in this distinction, see Nakahara 2003, § 2.2.

The developed physical treatment—field configuration space, locality, engineering dimensions, and dynamics—belongs to Fields, Configurations, Dimensions, and Local Dynamics. The source-coupling application is the invariant version of the component couplings discussed in Schwartz 2014, §§ 3.1–3.4.

Failure modes and invariant checks

A row is not intrinsically a covector. A row of numbers represents a covector only after the basis and transformation law are specified. The same array can represent a different type of object in another convention.

A metric is not hidden in index position. Writing $v_a=G_{ab}v^b$ uses the map $V\to V^*$ defined by $G$. If $G$ changes, the identified covector changes even when $v$ does not.

A transpose is not automatically an adjoint. The dual map is defined by composition. An adjoint depends on chosen bilinear or Hermitian forms and their linearity convention.

Finite-dimensional algebra does not settle domain questions. In infinite dimensions, algebraic and continuous duals can differ, ranges need not be closed, and an unbounded operator is inseparable from its domain. Those are not small corrections to the statements above; they require new hypotheses.

A useful round trip is to choose an invertible, non-orthogonal $M$, transform $v$, $\alpha$, and $T$, and then verify both $\alpha(v)$ and $T(v)$ in the original basis. If either reconstructed object changes, a domain, codomain, inverse, or dual transformation has been misidentified.

Exercises

1. Let $T:\mathbb R^3\to\mathbb R^2$ be $T(x,y,z)=(x+y,y+z)$. Find bases of $\ker T$ and $\operatorname{im}T$, then verify rank–nullity.

   Check

   The equations $x+y=0$ and $y+z=0$ give $(x,y,z)=t(-1,1,-1)$, so $\ker T=\operatorname{span}\{(-1,1,-1)\}$. The image is all of $\mathbb R^2$, for example because $T(1,0,0)=(1,0)$ and $T(0,0,1)=(0,1)$. Thus nullity $1$ plus rank $2$ equals $\dim\mathbb R^3=3$.

2. Explain why the expression $\alpha_a v^a$ is invariant but $\delta_{ab}v^av^b$ is not invariant under a general change of basis unless $\delta_{ab}$ is also transformed as a bilinear form.

   Check

   The covector and vector components transform contragrediently, so their matrices cancel. Keeping the numerical array $\delta$ fixed under a non-orthogonal basis change silently replaces the original bilinear form. The invariant expression is $G_{ab}v^av^b$ with $G'_{ab}=G_{cd}M^c{}_aM^d{}_b$.

3. For $T:V\to W$ and $\beta\in W^*$, verify directly that $T^*\beta\in V^*$ and that $(S\circ T)^*=T^*\circ S^*$.

   Check

   Linearity of $T^*\beta=\beta\circ T$ follows from linearity of both maps. For $\gamma$ in the dual of the codomain of $S$, $((S\circ T)^*\gamma)(v)=\gamma(S(Tv))$ and $(T^*(S^*\gamma))(v)=(S^*\gamma)(Tv)=\gamma(S(Tv))$.

References

• Sheldon Axler, Linear Algebra Done Right, 4th ed., Chapters 1–3 (especially the sections on linear maps and duality), Springer, 2024. Open-access book and chapter records. This is the structural source for finite-dimensional spaces, linear maps, rank–nullity, and duality.
• Theodore Frankel, The Geometry of Physics, 3rd ed., §§2.1a and 2.4, Cambridge University Press, 2012. Book record. This source emphasizes the distinction between vectors, covectors, and their coordinate components.
• Mikio Nakahara, Geometry, Topology and Physics, 2nd ed., § 2.2, Institute of Physics Publishing, 2003. This is the QFT-facing source for vectors, maps, kernels, images, duals, and tensors in a physics convention.
• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, §§3.1–3.4, Cambridge University Press, 2014. Book record. The source-current coupling supplies the physical application used above.