Vector Spaces and Dual Spaces

A vector space is a set whose elements can be added and multiplied by scalars while obeying the rules of linear algebra. Its dual space consists of scalar-valued linear functions on those vectors. Vectors describe directions and superpositions; dual vectors extract linear information from them.

Why Quantum Mechanics Needs This

The superposition principle begins with vector-space structure. If two kets are allowed state vectors, then their linear combinations are vectors in the same space. Quantum operators are linear maps, and probability amplitudes pair bras with kets to produce complex numbers.

This algebraic structure is necessary but not sufficient for quantum theory. An inner product adds lengths and orthogonality, completeness turns an inner-product space into a Hilbert space, and the physical pure states are rays rather than individual nonzero vectors. Keeping these layers separate prevents the word “vector” from silently carrying assumptions that have not yet been introduced.

Definition of a Vector Space

Let $\mathbb F$ be a field, usually $\mathbb R$ or $\mathbb C$. A vector space $V$ over $\mathbb F$ has two operations:

$$\begin{aligned} V\times V&\longrightarrow V, & (u,v)&\longmapsto u+v, \\ \mathbb F\times V&\longrightarrow V, & (a,v)&\longmapsto av. \end{aligned}$$

For all $u,v,w\in V$ and $a,b\in\mathbb F$, these operations obey

$$\begin{aligned} u+v&=v+u, \\ (u+v)+w&=u+(v+w), \\ u+0&=u, \\ u+(-u)&=0, \\ a(u+v)&=au+av, \\ (a+b)u&=au+bu, \\ (ab)u&=a(bu), \\ 1u&=u. \end{aligned}$$

The symbols $0$ and $1$ refer to different objects in different positions: $0\in V$ is the zero vector, while $0,1\in\mathbb F$ are scalars. Context usually makes the distinction harmless, but it matters when checking types in an abstract argument.

The same underlying set can carry different vector-space structures. For example, $\mathbb C$ has dimension one over $\mathbb C$ and dimension two over $\mathbb R$. The scalar field must therefore be part of the declaration, not an afterthought. Complex Vector Spaces develops the specifically complex structure used in quantum mechanics.

Linear Combinations and Span

Given vectors $v_1,\ldots,v_n\in V$ and scalars $a_1,\ldots,a_n\in\mathbb F$, a finite sum

$$\sum_{j=1}^{n}a_jv_j$$

is a linear combination. The span of a subset $S\subseteq V$ is the set of all finite linear combinations of elements of $S$:

$$\begin{aligned} \operatorname{span}S &= \left\lbrace \sum_{j=1}^{n}a_js_j :\right. \\ &\qquad\left. n\in\mathbb N,\ a_j\in\mathbb F,\ s_j\in S \right\rbrace. \end{aligned}$$

The word finite is important. Infinite series require a notion of convergence, which belongs to normed and Hilbert spaces rather than bare vector spaces. In finite-dimensional calculations this distinction is invisible; for wavefunctions it is essential.

The span is the smallest vector subspace containing $S$. It answers a constructive question: which vectors can be assembled from the available directions using superposition?

Subspaces and Homogeneous Constraints

A subset $W\subseteq V$ is a vector subspace when it contains the zero vector and is closed under linear combinations. Equivalently, $W$ is nonempty and

$$\begin{gathered} u,v\in W,\qquad a,b\in\mathbb F, \\ \Longrightarrow\quad au+bv\in W. \end{gathered}$$

Solution sets of homogeneous linear equations are subspaces. If $T:V\to W'$ is linear, then

$$\ker T = \lbrace v\in V:T(v)=0\rbrace$$

is a subspace of $V$. By contrast, the solution set of $T(v)=w_0$ with $w_0\ne0$ is generally an affine translate of a subspace, not a vector subspace: it does not contain the zero vector.

This distinction appears in quantum mechanics in several forms:

• solutions of a homogeneous linear wave equation form a vector space;
• normalized solutions do not form a vector subspace, because sums need not remain normalized;
• density operators of unit trace form a convex set, not a vector subspace;
• physical pure states form a projective space of rays, not the original Hilbert space.

The ambient vector space remains the algebraic setting in which superpositions and operators are defined.

Linear Independence

A finite list $v_1,\ldots,v_n$ is linearly independent when

$$\sum_{j=1}^{n}a_jv_j=0 \quad\Longrightarrow\quad a_1=\cdots=a_n=0.$$

If a nonzero coefficient relation exists, the list is linearly dependent. At least one vector can then be expressed as a linear combination of the others, so it contributes no new direction to the span.

Independence is a property of vectors, not of how far apart they look in a particular coordinate drawing. It also does not require orthogonality. Two nonparallel vectors in $\mathbb R^2$ are independent even if their angle is small. Orthogonality requires an inner product, while independence is defined in every vector space.

For a finite list, a practical test is to solve the homogeneous coefficient equation. After coordinates are chosen, this becomes a null-space or rank calculation. The coordinate machinery is developed in Bases and Coordinates and Matrices as Linear Maps.

Bases and Dimension

A basis is a linearly independent spanning set. These two properties imply that every vector has a unique finite expansion in the basis. For a finite-dimensional space,

$$v=\sum_{j=1}^{n}v^je_j,$$

and the number $n$ of basis vectors is the dimension of $V$.

The basis vectors $e_j$ and the coordinates $v^j$ play different roles. The vector $v$ is basis-independent; its coordinate list changes when the basis changes. This page uses bases only to construct dual vectors. The canonical treatment of coordinate maps, basis changes, and component transformations is Bases and Coordinates.

Infinite-dimensional spaces require more care. An algebraic basis, also called a Hamel basis, still uses finite sums, but it is usually not the basis meant in wave mechanics. Orthonormal expansions use convergent infinite series and the topology of a Hilbert space. See Hilbert Spaces for that distinction.

Examples and Nonexamples

Coordinate vectors. The set $\mathbb F^n$ with componentwise addition and scalar multiplication is a vector space.

Matrices. The set of all $m\times n$ matrices over $\mathbb F$ is a vector space. The Hermitian $n\times n$ matrices form a real vector space, but not a complex vector subspace: multiplying a nonzero Hermitian matrix by $i$ makes it anti-Hermitian.

Functions. Complex-valued functions on a fixed domain form a vector space under pointwise operations:

$$(af+bg)(x)=af(x)+bg(x).$$

Continuity, differentiability, square-integrability, or boundary conditions can select subspaces when those conditions are preserved by linear combinations.

Normalized vectors. The set

$$\lbrace v\in V:\lVert v\rVert=1\rbrace$$

is not a vector subspace. It excludes the zero vector and is not closed under addition or arbitrary scalar multiplication.

Probability distributions. Normalized probability vectors form a convex set. Convex combinations preserve normalization and positivity, but arbitrary linear combinations do not.

The Algebraic Dual

The algebraic dual of $V$ is

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

the set of all linear functionals $\ell:V\to\mathbb F$. Linearity means

$$\ell(au+bv)=a\ell(u)+b\ell(v).$$

The dual is itself a vector space. For $\ell,m\in V^*$ and $a,b\in\mathbb F$, define

$$(a\ell+bm)(v)=a\ell(v)+bm(v).$$

A functional consumes a vector and returns a scalar. The action is often written as a pairing,

$$\langle\ell,v\rangle_{\mathrm{dual}} = \ell(v).$$

This dual pairing is bilinear. It should not be confused with an inner product, which takes two vectors from the same inner-product space and is sesquilinear over $\mathbb C$.

Dual Bases

Let $(e_1,\ldots,e_n)$ be a basis of a finite-dimensional vector space $V$. The dual basis $(e^1,\ldots,e^n)$ in $V^*$ is defined by

$$e^i(e_j)=\delta^i{}_j.$$

If

$$v=\sum_jv^je_j,$$

then the dual basis extracts the coordinates:

$$e^i(v)=v^i.$$

Every functional has a unique expansion

$$\ell=\sum_i\ell_i e^i,$$

and its action is

$$\ell(v)=\sum_i\ell_i v^i.$$

The individual components $\ell_i$ and $v^i$ depend on the basis, but the scalar $\ell(v)$ does not. When the vector basis changes, the dual basis changes in the compensating way required to preserve $e^i(e_j)=\delta^i{}_j$. Index Notation and Summation Conventions develops this covector–vector pairing in component language.

In finite dimension,

$$\dim V^*=\dim V.$$

Equal dimensions do not provide a preferred identification between $V$ and $V^*$. A basis can create one, but changing the basis changes that identification. An inner product provides a geometrically meaningful identification of a different kind.

Functionals on Function Spaces

Dual vectors need not look like rows of numbers. Let $V$ be a vector space of complex-valued functions. Evaluation at a fixed point,

$$E_{x_0}(f)=f(x_0),$$

is linear whenever point evaluation is defined on the chosen function space. Integration against a fixed weight is another functional:

$$L_w(f) = \int_a^b w(x)f(x)\,dx.$$

By contrast,

$$Q_{x_0}(f)=\lvert f(x_0)\rvert^2$$

is not linear. It is quadratic in the function. This is the same structural distinction that separates a probability amplitude from its squared magnitude.

Topology matters in infinite-dimensional function spaces. Point evaluation is continuous on some spaces of functions and not even well-defined on $L^2$ equivalence classes, whose elements are equal when they differ only on a set of measure zero. The relevant Hilbert-space dual is therefore the continuous dual, not the full algebraic dual. See $L^2$ Spaces for the function-space setting.

Bras and the Riesz Map

In a complex inner-product space using the physics convention, $\langle u\vert v\rangle$ is conjugate-linear in $u$ and linear in $v$. Holding $u$ fixed therefore defines a linear functional of $v$:

$$R(u):v\longmapsto\langle u\vert v\rangle.$$

The map

$$R:V\longrightarrow V^*$$

is conjugate-linear:

$$R(au+bw) = a^*R(u)+b^*R(w).$$

In finite-dimensional inner-product spaces, every linear functional is $R(u)$ for a unique $u$. In a Hilbert space, the Riesz representation theorem makes the corresponding statement for continuous linear functionals. Dirac notation writes

$$u\longleftrightarrow\lvert u\rangle, \qquad R(u)\longleftrightarrow\langle u\vert.$$

Thus a bra is not obtained from a ket by merely rotating a printed column into a row. The inner product supplies the ket-to-bra map, complex coefficients are conjugated, and an orthonormal coordinate basis turns that functional into a conjugate-transposed row.

The canonical treatments of the inner product and notation are Inner Products, Dirac Notation as Linear Algebra, and Bra-Ket Notation.

Worked Example: A Plane and Its Constraint

In $\mathbb C^3$, let

$$u= \begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix}, \qquad w= \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}.$$

The vectors are linearly independent. If $au+bw=0$, the first two components give $a=b=0$. Their span is

$$S = \left\lbrace \begin{pmatrix} a\\ b\\ a+b \end{pmatrix} : a,b\in\mathbb C \right\rbrace.$$

The vector

$$z= \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}$$

belongs to $S$ because $z=u+w$. The list $(u,w,z)$ is therefore linearly dependent even though no vector is zero and no two are scalar multiples.

Now define the dual vector

$$\ell(x)=x_3-x_1-x_2.$$

For every $x=au+bw\in S$,

$$\ell(x) =(a+b)-a-b =0.$$

The functional $\ell$ annihilates the whole subspace:

$$S=\ker\ell.$$

This example shows two complementary descriptions of the same plane. The vectors $u$ and $w$ generate it, while the dual vector $\ell$ imposes the homogeneous linear constraint that defines it.

Physical Interpretation

Vector-space structure explains why amplitudes, rather than probabilities, are combined linearly. If $\lvert\psi_1\rangle$ and $\lvert\psi_2\rangle$ are vectors, then

$$\alpha\lvert\psi_1\rangle +\beta\lvert\psi_2\rangle$$

is another vector. Whether it is normalized is a separate question, and whether two nonzero vectors represent the same physical pure state depends on their ray. See State Vectors and Rays and Global Phase for the physical postulates and interpretation.

Dual vectors explain the algebraic shape of amplitudes. A bra acts linearly on the ket in its slot and produces a scalar. The Born rule then assigns physical probabilities to squared amplitudes; that probability assignment is an additional quantum postulate, not a consequence of the vector-space axioms.

Finite and Infinite Dimensions

Several familiar finite-dimensional facts need qualification in infinite dimensions:

• the algebraic dual is generally much larger than the continuous dual;
• an algebraic basis uses finite sums, while Hilbert bases use norm-convergent expansions;
• equal dimension no longer gives a useful elementary route from vectors to all linear functionals;
• generalized position and momentum bras are distributions, not continuous functionals on the Hilbert space itself;
• convergence, boundedness, and operator domains require topological structure absent from a bare vector space.

Begin with Finite-Dimensional Hilbert Spaces for matrix quantum mechanics and Hilbert Spaces when limits, wavefunctions, or unbounded operators enter.

Common Mistakes

• Omitting the scalar field. Independence and dimension can change when the field changes.
• Treating coordinates as vectors. A coordinate column represents a vector only after a basis has been chosen.
• Equating independence with orthogonality. Independence is algebraic; orthogonality requires an inner product.
• Calling every constrained set a subspace. Normalization, positivity, or an inhomogeneous equation usually defines a nonlinear, convex, or affine set.
• Identifying $V$ with $V^*$ canonically. Equal finite dimensions do not supply a preferred identification.
• Calling every dual functional a bra without qualification. The ket-to-bra relation uses an inner product; in infinite dimensions the Hilbert dual contains continuous functionals.
• Using infinite sums in a bare vector space. Convergence requires a topology or norm.
• Deriving the Born rule from linear algebra. Vector and dual spaces supply amplitudes, not their physical probability interpretation.

Exercises

1. Let

$$W_0 = \lbrace(x,y,z)\in\mathbb C^3:x+y+z=0\rbrace$$

and

$$W_1 = \lbrace(x,y,z)\in\mathbb C^3:x+y+z=1\rbrace.$$

Which set is a vector subspace?

Solution

$W_0$ is the kernel of the linear functional $\ell(x,y,z)=x+y+z$, so it is a subspace. Explicitly, the constraint is preserved by arbitrary linear combinations.

$W_1$ is not a subspace because it does not contain the zero vector. It is an affine translate of $W_0$.

2. In $\mathbb C^3$, determine whether

$$\begin{gathered} v_1= \begin{pmatrix} 1\\ i\\ 0 \end{pmatrix}, \qquad v_2= \begin{pmatrix} 0\\ 1\\ i \end{pmatrix}, \\ v_3= \begin{pmatrix} 1\\ 1+i\\ i \end{pmatrix}. \end{gathered}$$

are linearly independent.

Solution

They are linearly dependent because

$$v_3=v_1+v_2.$$

Equivalently,

$$v_1+v_2-v_3=0$$

is a nontrivial coefficient relation. The three vectors span the same two-dimensional subspace as $v_1$ and $v_2$.

3. Let

$$e_1= \begin{pmatrix} 1\\ 1 \end{pmatrix}, \qquad e_2= \begin{pmatrix} 1\\ -1 \end{pmatrix}$$

be a basis of $\mathbb R^2$. Find the dual basis functionals $e^1$ and $e^2$.

Solution

For $v=(x,y)$, solve

$$v=a e_1+b e_2 = \begin{pmatrix} a+b\\ a-b \end{pmatrix}.$$

Thus

$$a=\frac{x+y}{2}, \qquad b=\frac{x-y}{2}.$$

The dual basis extracts these coefficients:

$$\begin{aligned} e^1(x,y)&=\frac{x+y}{2}, \\ e^2(x,y)&=\frac{x-y}{2}. \end{aligned}$$

Direct substitution gives $e^i(e_j)=\delta^i{}_j$.

4. Let $V$ be a complex inner-product space using the physics convention. If

$$\lvert\chi\rangle = a\lvert u\rangle+b\lvert w\rangle,$$

derive the corresponding bra and explain why the map from kets to bras is not complex-linear.

Solution

For every $\lvert v\rangle$,

$$\begin{aligned} \langle\chi\vert v\rangle &= \langle a u+b w \vert v\rangle \\ &= a^*\langle u\vert v\rangle +b^*\langle w\vert v\rangle. \end{aligned}$$

Therefore

$$\langle\chi\vert = a^*\langle u\vert+b^*\langle w\vert.$$

The coefficients are conjugated, so $R(a u+b w)=a^*R(u)+b^*R(w)$. The Riesz map from vectors to their associated dual functionals is conjugate-linear, not complex-linear.

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• P. R. Halmos, Finite-Dimensional Vector Spaces, Springer, 1974.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.