Vector Spaces

Concrete models

1.The plane $\mathbb{R}^2$ is a vector space whose elements are ordered pairs. Its addition and scalar multiplication are:

$$(a_1, a_2) + (b_1, b_2) = (a_1 + b_1, a_2 + b_2)$$

$$\alpha(a_1, a_2) = (\alpha a_1, \alpha a_2)$$

2.Addition combines two vectors, while the real number $\alpha$ scales a vector. With componentwise addition and scalar multiplication, $F^n$ is a vector space over any field $F.$ Examples are $\mathbb{Q}^n$ over $\mathbb{Q},$ $\mathbb{R}^n$ over $\mathbb{R},$ and $\mathbb{F}_p^n$ over the finite field $\mathbb{F}_p.$

3.The elements need not be lists of numbers. The set $M_{m \times n}(F)$ of all $m \times n$ matrices over $F$ has entrywise addition and scalar multiplication. The set $F[x]$ of polynomials has polynomial addition and multiplication by constants from $F.$ If $X$ is a nonempty set, the set $F^X$ of all functions $f : X \to F$ has pointwise operations:

$$(f + g)(x) = f(x) + g(x)$$

$$(\alpha f)(x) = \alpha f(x)$$

4.A sequence of real numbers is an element of $\mathbb{R}^{\mathbb{N}}.$ For an interval $I,$ the continuous functions and the differentiable functions from $I$ to $\mathbb{R}$ are subsets of $\mathbb{R}^I$ closed under both operations. In every case, addition and scalar multiplication satisfy the same identities. These operations do not define lengths or angles, which require the additional structure of an inner product. The abstract definition lists the vector space axioms.

Abstract definition

5.A vector space over a field $F$ is a set $V$ equipped with vector addition $+ : V \times V \to V$ and scalar multiplication $\cdot : F \times V \to V.$ These operations satisfy the following axioms:

7.The identity of the abelian group is the zero vector $\mathbf{0},$ and the elements of $F$ are the scalars. The choice of $F$ is part of the structure. For example, $\mathbb{C}^n$ is a vector space over the complex numbers and over $\mathbb{R},$ but the two structures have different dimensions.

8.The axioms imply the rules for zero and additive inverses. Distributivity over scalar addition gives:

$$0 \cdot \mathbf{v} = (0 + 0) \cdot \mathbf{v} = 0 \cdot \mathbf{v} + 0 \cdot \mathbf{v}$$

9.Cancellation in the abelian group gives $0 \cdot \mathbf{v} = \mathbf{0}.$ Distributivity applied to $\alpha \cdot (\mathbf{0} + \mathbf{0})$ gives $\alpha \cdot \mathbf{0} = \mathbf{0}.$ The distributive laws also give $(-1) \cdot \mathbf{v} = -\mathbf{v}$ and $(-\alpha) \cdot \mathbf{v} = -(\alpha \cdot \mathbf{v}).$

10.Suppose that $\alpha \cdot \mathbf{v} = \mathbf{0}$ and $\alpha \neq 0.$ Multiplication by $\alpha^{-1}$ gives:

$$\mathbf{v} = (\alpha^{-1}\alpha) \cdot \mathbf{v} = \alpha^{-1} \cdot (\alpha \cdot \mathbf{v}) = \mathbf{0}$$

11.Hence $\alpha \cdot \mathbf{v} = \mathbf{0}$ implies $\alpha = 0$ or $\mathbf{v} = \mathbf{0}.$ Distinct scalars give distinct multiples of a nonzero vector, so every nonzero vector space over an infinite field is infinite.

12.

A vector space has an underlying abelian group and a separate scalar field. A module has a scalar ring instead of a field, so several results below fail for modules.

Subspaces and span

13.A subspace of $V$ is a nonempty subset $W \subseteq V$ that is closed under addition and scalar multiplication. With the inherited operations, $W$ is a vector space over the same field. The zero vector belongs to $W,$ since $0 \cdot \mathbf{w} = \mathbf{0}$ for any $\mathbf{w} \in W.$

14.Given a subset $S \subseteq V,$ its span consists of all finite linear combinations of elements of $S$:

$$\mathrm{span}(S) = \{\ \alpha_1\mathbf{v}_1 + \cdots + \alpha_n\mathbf{v}_n \mid n \geq 1,\ \alpha_i \in F,\ \mathbf{v}_i \in S\ \}$$

15.The convention $\mathrm{span}(\varnothing) = \{\ \mathbf{0}\ \}$ covers the empty set. The span of $S$ is the smallest subspace that contains $S,$ since every such subspace contains all finite linear combinations of elements of $S.$

16.For example, the vector $(1, 2)$ generates the subspace:

$$W = \mathrm{span}\{\ (1, 2)\ \} = \{\ (t, 2t) \mid t \in \mathbb{R}\ \}$$

17.This is the line through the origin with slope $2.$ If $s, t, \alpha \in \mathbb{R},$ then $(s, 2s) + (t, 2t) = (s + t, 2(s + t))$ and $\alpha(t, 2t) = (\alpha t, 2\alpha t),$ so the set is closed under both operations.

18.The page on subspaces contains the closure criterion, sums and intersections, Grassmann's formula, direct sums and complements.

Linear maps

19.A function $T : V \to W$ between vector spaces over the same field is a linear map when it preserves addition and scalar multiplication:

$$T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$$

$$T(\alpha \mathbf{v}) = \alpha T(\mathbf{v})$$

20.Equivalently, $T(\alpha\mathbf{u} + \beta\mathbf{v}) = \alpha T(\mathbf{u}) + \beta T(\mathbf{v})$ for all vectors $\mathbf{u}, \mathbf{v}$ and scalars $\alpha, \beta.$ Compositions of linear maps are linear, and a linear map from a space to itself is an endomorphism. A bijective linear map is a linear isomorphism, and its inverse is linear.

21.The coordinate projection $P : F^3 \to F^2,$ defined by $P(x, y, z) = (x, y),$ is linear. The trace map $\mathrm{tr} : M_n(F) \to F$ is linear because $\mathrm{tr}(A + B) = \mathrm{tr}(A) + \mathrm{tr}(B)$ and $\mathrm{tr}(\alpha A) = \alpha\mathrm{tr}(A).$ The formal derivative $D : F[x] \to F[x]$ is linear and satisfies:

$$D\left(\sum_{k=0}^n a_kx^k\right) = \sum_{k=1}^n ka_kx^{k-1}$$

22.The kernel and image of $T$ are:

$$\ker(T) = \{\ \mathbf{v} \in V \mid T(\mathbf{v}) = \mathbf{0}\ \}$$

$$\mathrm{im}(T) = \{\ T(\mathbf{v}) \mid \mathbf{v} \in V\ \}$$

23.Both sets are subspaces. If $T(\mathbf{u}) = T(\mathbf{v}) = \mathbf{0},$ then $T(\alpha\mathbf{u} + \beta\mathbf{v}) = \mathbf{0}.$ For the image, $\alpha T(\mathbf{u}) + \beta T(\mathbf{v}) = T(\alpha\mathbf{u} + \beta\mathbf{v}),$ which belongs to $\mathrm{im}(T).$

Bases, coordinates, and dimension

24.A subset $B \subseteq V$ is linearly independent when every vanishing finite linear combination of distinct elements of $B$ has all coefficients equal to zero. The empty set is linearly independent, while a set containing $\mathbf{0}$ is dependent. A basis of $V$ is a linearly independent subset whose span is $V.$

25.If $B = \{\ \mathbf{v}_1, \ldots, \mathbf{v}_n\ \}$ is a basis, every vector has a unique expansion:

$$\mathbf{v} = \alpha_1\mathbf{v}_1 + \cdots + \alpha_n\mathbf{v}_n$$

26.Existence follows from the spanning property. If a second expansion has coefficients $\beta_1, \ldots, \beta_n,$ subtraction gives:

$$(\alpha_1 - \beta_1)\mathbf{v}_1 + \cdots + (\alpha_n - \beta_n)\mathbf{v}_n = \mathbf{0}$$

27.Linear independence then gives $\alpha_i = \beta_i$ for every $i.$

28.The standard basis of $F^n$ consists of the vectors $\mathbf{e}_1, \ldots, \mathbf{e}_n,$ where $\mathbf{e}_i$ has entry $1$ in position $i$ and zero elsewhere. The matrices with one entry equal to $1$ and all other entries equal to zero form a basis of $M_{m \times n}(F).$ The monomials $1, x, \ldots, x^n$ form a basis of the polynomials of degree at most $n.$ These bases have $n,$ $mn$ and $n + 1$ elements respectively.

29.The polynomial space $F[x]$ has the infinite basis $\{\ 1, x, x^2, \ldots\ \}.$ The space of continuous real-valued functions on an interval with more than one point is infinite-dimensional, since it contains the linearly independent monomials of every degree. The scalar field affects dimension. The vectors $\mathbf{e}_1, \ldots, \mathbf{e}_n$ form a basis of $\mathbb{C}^n$ over $\mathbb{C},$ while $\mathbf{e}_1, \ldots, \mathbf{e}_n, i\mathbf{e}_1, \ldots, i\mathbf{e}_n$ form a basis over $\mathbb{R}.$ Hence $\dim_{\mathbb{C}}\mathbb{C}^n = n$ and $\dim_{\mathbb{R}}\mathbb{C}^n = 2n.$

30.A vector space is finite-dimensional when it has a finite spanning set. Removing dependent vectors from such a set produces a basis. If a set of $m$ vectors spans $V$ and a set of $n$ vectors is linearly independent, the exchange argument gives $n \leq m.$ Applying this inequality in both directions to two bases shows that they have the same number of elements. This number is the dimension $\dim V.$ The zero space $\{\ \mathbf{0}\ \}$ has the empty basis and dimension $0.$

31.A basis has two equivalent characterisations:

33.Removing a vector from a basis leaves a set that does not span $V.$ Adjoining a vector to a basis gives a dependent set. Conversely, a maximal independent set spans $V,$ since a vector outside its span could be adjoined without creating a relation.

34.If $W$ is a subspace of a finite-dimensional space $V,$ a linearly independent subset of $W$ extends to a basis of $W,$ and that basis extends to a basis of $V.$ Hence $\dim W \leq \dim V,$ with equality only when $W = V.$

35.

Assuming the axiom of choice, Zorn's lemma extends every linearly independent subset of an arbitrary vector space to a basis. Finite-dimensional spaces use the finite extension procedure above.

36.Ordering a basis $B = (\mathbf{v}_1, \ldots, \mathbf{v}_n)$ turns the coefficients of a vector into its coordinate vector:

$$[\mathbf{v}]_B = (\alpha_1, \ldots, \alpha_n)$$

37.The coordinate map $C_B : V \to F^n,$ defined by $C_B(\mathbf{v}) = [\mathbf{v}]_B,$ is a linear isomorphism. Every $n$-dimensional vector space over $F$ is therefore isomorphic to $F^n,$ although the isomorphism depends on the ordered basis.

38.A function on a basis determines one linear map on the whole space. Given $f : B \to U,$ the formula:

$$T\left(\sum_i \alpha_i\mathbf{v}_i\right) = \sum_i \alpha_i f(\mathbf{v}_i)$$

39.defines the unique linear map $T : V \to U$ whose restriction to $B$ is $f.$ Uniqueness of coordinates makes the formula well defined.

40.Suppose that $T : V \to U$ has finite-dimensional domain. If $\mathbf{k}_1, \ldots, \mathbf{k}_r$ is a basis of $\ker(T),$ extend it to a basis $\mathbf{k}_1, \ldots, \mathbf{k}_r, \mathbf{v}_{r+1}, \ldots, \mathbf{v}_n$ of $V.$ The vectors $T(\mathbf{v}_{r+1}), \ldots, T(\mathbf{v}_n)$ span $\mathrm{im}(T)$ because $T$ sends the kernel components to zero. If a linear combination of these images is zero, the corresponding combination of $\mathbf{v}_{r+1}, \ldots, \mathbf{v}_n$ lies in the kernel and is therefore a linear combination of $\mathbf{k}_1, \ldots, \mathbf{k}_r.$ Independence of the extended basis forces every coefficient to be zero. The images form a basis of $\mathrm{im}(T),$ so:

$$\dim V = \dim \ker(T) + \dim \mathrm{im}(T)$$

41.This is the rank-nullity theorem. The two terms on the right are the nullity and rank of $T,$ respectively.

42.For a matrix map $A : F^n \to F^m,$ $\dim \mathrm{im}(A)$ is the rank of the matrix, while $\ker(A)$ is the solution space of the homogeneous system $A\mathbf{x} = \mathbf{0}.$

Quotient spaces

43.A quotient space treats two vectors as equivalent when their difference lies in a subspace $N.$ The equivalence class of $\mathbf{v}$ is the coset $\mathbf{v} + N,$ and the set of all classes is:

$$V/N = \{\ \mathbf{v} + N \mid \mathbf{v} \in V\ \}$$

44.The vector space operations on classes are defined by:

$$(\mathbf{u} + N) + (\mathbf{v} + N) = (\mathbf{u} + \mathbf{v}) + N$$

$$\alpha(\mathbf{v} + N) = \alpha\mathbf{v} + N$$

45.If $\mathbf{v} + N = \mathbf{v}' + N,$ then $\mathbf{v} - \mathbf{v}' \in N.$ Closure of $N$ under addition and scalar multiplication shows that replacing representatives does not change either result. The quotient map $\pi_N : V \to V/N,$ defined by $\pi_N(\mathbf{v}) = \mathbf{v} + N,$ is linear and surjective, with kernel $N.$

46.The quotient map has a factorisation property. If $T : V \to U$ is linear and $N \subseteq \ker(T),$ then vectors in the same coset of $N$ have the same image under $T.$ The formula:

$$\widetilde{T}(\mathbf{v} + N) = T(\mathbf{v})$$

47.therefore defines the unique linear map $\widetilde{T} : V/N \to U$ such that $T = \widetilde{T} \circ \pi_N.$ Taking $N = \ker(T)$ and restricting the codomain to $\mathrm{im}(T)$ gives the homomorphism theorem:

$$V/\ker(T) \cong \mathrm{im}(T)$$

48.The isomorphism sends $\mathbf{v} + \ker(T)$ to $T(\mathbf{v}).$ Its kernel is zero, and it is surjective by the definition of $\mathrm{im}(T).$

49.When $T : V \to U$ is surjective, inverse image gives a bijection between the subspaces of $U$ and the subspaces of $V$ that contain $\ker(T).$ The inverse assignments send $L \subseteq U$ to $T^{-1}(L)$ and $M \subseteq V$ to $T(M).$ This is the correspondence theorem for vector spaces.

50.The homomorphism theorem gives two quotient identities. If $N \subseteq M \subseteq V,$ then:

$$(V/N)/(M/N) \cong V/M$$

51.If $A$ and $N$ are subspaces of $V,$ restricting the quotient map to $A$ gives:

$$(A + N)/N \cong A/(A \cap N)$$

52.The kernel of the restricted map is $A \cap N,$ while every coset in $(A + N)/N$ has a representative in $A.$

Direct sums and complements

53.For subspaces $A$ and $B,$ the sum $A + B$ is direct when $A \cap B = \{\ \mathbf{0}\ \}.$ In this case every vector of $A + B$ has a unique expression $\mathbf{a} + \mathbf{b},$ with $\mathbf{a} \in A$ and $\mathbf{b} \in B.$ Grassmann's formula for finite-dimensional subspaces is:

$$\dim(A + B) = \dim A + \dim B - \dim(A \cap B)$$

54.When the intersection is trivial, the formula reduces to $\dim(A \oplus B) = \dim A + \dim B.$

55.Suppose that $N$ is a subspace of a finite-dimensional vector space $V.$ Choose a basis $\mathbf{n}_1, \ldots, \mathbf{n}_r$ of $N$ and extend it to a basis:

$$\mathbf{n}_1, \ldots, \mathbf{n}_r, \mathbf{m}_{r+1}, \ldots, \mathbf{m}_n$$

56.of $V.$ If $M = \mathrm{span}\{\ \mathbf{m}_{r+1}, \ldots, \mathbf{m}_n\ \},$ then each $\mathbf{v} \in V$ has a unique expression $\mathbf{v} = \mathbf{n} + \mathbf{m},$ with $\mathbf{n} \in N$ and $\mathbf{m} \in M.$ Thus $V = N \oplus M,$ and $M$ is a complement of $N.$ The dimensions satisfy:

$$\dim V = \dim N + \dim M$$

57.For a linear map $T : V \to U,$ take $N = \ker(T).$ The restriction $T|_M : M \to \mathrm{im}(T)$ is an isomorphism. It is injective because $M \cap \ker(T) = \{\ \mathbf{0}\ \},$ and it is surjective because every vector of $V$ has a component in $M$ with the same image. Hence:

$$V \cong \ker(T) \oplus \mathrm{im}(T)$$

58.If $T$ is surjective, the inverse of $T|_M$ followed by the inclusion $M \subseteq V$ gives a linear map $S : U \to V$ with $T \circ S = \mathrm{id}_U.$ Thus every surjective linear map with finite-dimensional domain has a right inverse.

59.The restriction of the quotient map to $M$ is an isomorphism $M \cong V/N.$ These identifications depend on the chosen complement, and they give:

$$V \cong N \oplus V/N$$

$$\dim V = \dim N + \dim(V/N)$$

Example

60.Consider the linear map $T : \mathbb{R}^3 \to \mathbb{R}^2$ defined by:

$$T(x, y, z) = (x + y, y + z)$$

61.Each output coordinate is a linear combination of $x, y, z,$ so $T$ is linear. A vector belongs to the kernel when $x + y = 0$ and $y + z = 0.$ Hence:

$$\ker(T) = \mathrm{span}\{\ (-1, 1, -1)\ \}$$

62.For any $(a, b) \in \mathbb{R}^2,$ the vector $(a, 0, b)$ maps to $(a, b),$ so $T$ is surjective. The rank-nullity identity becomes:

$$3 = \dim \ker(T) + \dim \mathrm{im}(T) = 1 + 2$$

63.The plane $M = \{\ (a, 0, b) \mid a, b \in \mathbb{R}\ \}$ meets the kernel only at the origin. Every vector has the decomposition:

$$(x, y, z) = (-y, y, -y) + (x + y, 0, y + z)$$

64.The first term belongs to $\ker(T)$ and the second to $M,$ so $\mathbb{R}^3 = \ker(T) \oplus M.$ The restriction $T|_M$ is the isomorphism $(a, 0, b) \mapsto (a, b),$ and its inverse gives the right inverse $S(a, b) = (a, 0, b).$ The homomorphism theorem identifies the quotient $\mathbb{R}^3/\ker(T)$ with $\mathbb{R}^2.$

65.

Subspaces, bases, dimension and linear maps extend to modules over a ring, but coefficients cannot generally be divided by nonzero scalars. Modules therefore need not have bases, submodules need not have complements, and the dimension arguments used above no longer apply.

66.

For a development of vector spaces from group theory, see Frederick M. Goodman, Algebra: Abstract and Concrete, Section 3.3, listed in the bibliography.