Vector Spaces, Linear Subspaces, Span and Direct Sums

Let $F$ be a field (Field), with additive identity $0_F$, multiplicative identity $1_F$, and the field axioms as stated there. A vector space over $F$, also called an $F$-vector space, consists of

• a set $V$, whose elements are called vectors;
• a binary operation $+ : V \times V \to V$ on $V$ (Binary operation on a set; associativity, commutativity, and a subset closed under the operation), the vector addition;
• an element $0_V \in V$, the zero vector;
• a map $\cdot \,:\, F \times V \to V$, the scalar multiplication, written $\lambda v := \cdot(\lambda, v)$;

subject to the following axioms, in which $u, v \in V$ and $\lambda, \mu \in F$ are arbitrary.

• (V1) $(V, +, 0_V)$ is an abelian group (Group and abelian group): $+$ is associative and commutative, $0_V$ is a two-sided identity for $+$ (Left identity, right identity, and two-sided identity for a binary operation), and every $v \in V$ has an additive inverse (Left inverse, right inverse, and invertible element of a monoid).
• (V2) $\lambda(u + v) = \lambda u + \lambda v$.
• (V3) $(\lambda + \mu)v = \lambda v + \mu v$.
• (V4) $(\lambda\mu)v = \lambda(\mu v)$.
• (V5) $1_F v = v$.

The elements of $F$ are called scalars. When several vector spaces are in play we write $0_V$ for the zero of $V$, and we write $-v$ for the additive inverse of $v$ and $u - v := u + (-v)$.

The notation $0_V$ and $-v$ is legitimate. Axiom (V1) asserts only that some two-sided identity and some additive inverses exist. That there is at most one two-sided identity for $+$ is A left identity and a right identity for the same binary operation are equal; hence there is at most one two-sided identity, and that an invertible element of a monoid has exactly one inverse is In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided; both are proved before Group and abelian group and are inherited here with the group structure. So $0_V$ and $-v$ denote well-defined elements, and nothing below re-derives them.

What (V1) buys, and why it is not restated. Associativity, commutativity, the identity law $v + 0_V = v = 0_V + v$, the inverse law $v + (-v) = 0_V$, cancellation (Cancellation in a group: $gx = gy$ or $xg = yg$ forces $x = y$; equivalently left and right translation by $g$ are bijections of $G$, so $gx = h$ and $xg = h$ each have exactly one solution) and the inverse identities (In a group $e^{-1} = e$, $(g^{-1})^{-1} = g$ and $(gh)^{-1} = h^{-1}g^{-1}$, the order of the last product being essential) are facts about abelian groups. They are quoted from the group page wherever they are used and are never proved again for vectors.

• Scalar multiplication is not a binary operation on a set. It is a map $F \times V \to V$ with arguments from two different sets, so Binary operation on a set; associativity, commutativity, and a subset closed under the operation, which is about a map $S \times S \to S$, does not apply to it and is never cited for it. The definition above cites that item for the vector addition only. In particular "closed under scalar multiplication" below always means $\lambda v \in W$ for $\lambda \in F$ and $v \in W$, which is not an instance of the closure condition defined there.

• (V5) is an axiom, not a consequence of (V2)–(V4). Take any abelian group $(V,+,0_V)$ and define $\lambda v := 0_V$ for every $\lambda$ and $v$. Then (V2), (V3) and (V4) all hold, both sides of each being $0_V$, while (V5) fails as soon as $V \ne \{0_V\}$. So (V5) has to be imposed, and it is what ties the scalar action to the identity of $F$.

• Two structures, one set. A vector space is data: the set $V$, the addition, the zero, and the scalar multiplication, over a fixed field $F$. The same set may carry vector-space structures over different fields, and the field is part of the statement of every result below. A field is a vector space over itself, and over any subfield $K \subseteq F$ every $F$-vector space is a $K$-vector space by restricting the scalars is the first place where that matters.

• The field is the published one. No field axiom is restated here; $F$ is a field in the sense of Field, whose axiom (A) already says that $(F,+)$ is an abelian group and whose axiom (M) says the same of $(F \setminus \{0_F\}, \cdot)$. In particular every field is a vector space over itself, which is A field is a vector space over itself, and over any subfield $K \subseteq F$ every $F$-vector space is a $K$-vector space by restricting the scalars.

Let $V$ be a vector space over a field $F$ (Vector space over a field). For all $\lambda \in F$ and $v \in V$:

1. $0_F v = 0_V$;
2. $\lambda 0_V = 0_V$;
3. $(-\lambda)v = -(\lambda v)$, and also $\lambda(-v) = -(\lambda v)$;
4. $(-1_F)v = -v$;
5. if $\lambda v = 0_V$ then $\lambda = 0_F$ or $v = 0_V$.

Here $0_F$ and $1_F$ are the additive and multiplicative identities of $F$, $0_V$ is the zero vector, $-\lambda$ is the additive inverse of $\lambda$ in $F$, and $-v$ is the additive inverse of $v$ in the abelian group $(V,+,0_V)$.

• None of this is an axiom. The scalar axioms (V2)–(V5) say nothing directly about $0_F$, $0_V$ or negatives; each claim above is extracted by writing one element in two ways and cancelling in the abelian group $(V,+,0_V)$. That is the same device that gives $0 \cdot a = 0$ in a field (Multiplication by zero: $0 \cdot a = 0$), and the proofs are deliberately parallel.

• Claim 5 is what makes $\{0_V\}$ the only "degenerate" scalar multiple. It is used below to compute $\operatorname{span}\{v\}$ ($\operatorname{span}\{v\} = \{\, \lambda v : \lambda \in F \,\}$, which is $\{0_V\}$ when $v = 0_V$, and when $v \ne 0_V$ contains $0_V$ only as the multiple $0_F v$) and, in that form, is the vector-space analogue of a field having no zero divisors (A field has no zero divisors: $ab = 0 \Rightarrow a = 0$ or $b = 0$). Its converse directions, claims 1 and 2, say that both degenerate products really are $0_V$.

• Claim 4 is the bridge to the additive group. It says the additive inverse of a vector is a scalar multiple of it, which is why closure under scalar multiplication already forces closure under negatives; that is what makes a linear subspace a subgroup of $(V,+)$ without a separate axiom.

Let $F$ be a field (Field) and let $X$ be any set. Write

$F^{X} \;:=\; \{\, f \;:\; f \text{ is a function } X \to F \,\},$

and for $f \in F^{X}$ write $f(x)$ for its value at $x \in X$. Two elements of $F^{X}$ are equal exactly when they agree at every point of $X$. Define

$(f + g)(x) \;:=\; f(x) + g(x), \qquad (\lambda f)(x) \;:=\; \lambda\, f(x), \qquad 0_{F^{X}}(x) \;:=\; 0_F,$

for $f, g \in F^{X}$, $\lambda \in F$ and $x \in X$, the operations on the right being those of $F$. These are the pointwise operations.

These rules really are the required data. For $f, g \in F^{X}$ the assignment $x \mapsto f(x) + g(x)$ is a function $X \to F$, so $+$ is a binary operation $F^{X} \times F^{X} \to F^{X}$ (Binary operation on a set; associativity, commutativity, and a subset closed under the operation); for $\lambda \in F$ and $f \in F^{X}$ the assignment $x \mapsto \lambda f(x)$ is a function $X \to F$, so scalar multiplication is a map $F \times F^{X} \to F^{X}$; and $0_{F^{X}}$, the constant function at $0_F$, is an element of $F^{X}$.

$F^{X}$ is a vector space over $F$ (Vector space over a field). Each axiom is an equation between elements of $F^{X}$, hence holds exactly when it holds at every $x \in X$ after evaluation, and there it is the corresponding field axiom applied to the values $f(x), g(x), \lambda, \mu$:

• associativity and commutativity of $+$, and $f + 0_{F^{X}} = f$, come from the same laws for $+$ in $F$; the additive inverse of $f$ is $x \mapsto -f(x)$, which lies in $F^{X}$ and satisfies $f + (-f) = 0_{F^{X}}$ pointwise. This is axiom (V1);
• $\bigl(\lambda(f+g)\bigr)(x) = \lambda(f(x)+g(x)) = \lambda f(x) + \lambda g(x)$ is (V2), by distributivity in $F$;
• $\bigl((\lambda+\mu)f\bigr)(x) = (\lambda+\mu)f(x) = \lambda f(x) + \mu f(x)$ is (V3), by distributivity in $F$;
• $\bigl((\lambda\mu)f\bigr)(x) = (\lambda\mu)f(x) = \lambda(\mu f(x))$ is (V4), by associativity of multiplication in $F$;
• $(1_F f)(x) = 1_F f(x) = f(x)$ is (V5), by the multiplicative identity law in $F$.

The case $X = n$

A natural number is a von Neumann natural (The natural numbers $\mathbb{N}$ (von Neumann)), that is a set, and $n = \{\, m \in \mathbb{N} : m < n \,\} = \{0, 1, \dots, n-1\}$ (On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$). Taking $X = n$ therefore gives

$F^{n} \;=\; \{\, x \;:\; x \text{ is a function } n \to F \,\},$

whose elements are written $x = (x_0, x_1, \dots, x_{n-1})$ with $x_i := x(i)$ for $i < n$. The coordinates are indexed from $0$, because $0 \in n$ whenever $n \ne 0$ and $n \notin n$ always. The operations read

$(x + y)_i = x_i + y_i, \qquad (\lambda x)_i = \lambda x_i \qquad (i < n),$

and the zero of $F^{n}$ is the tuple all of whose coordinates are $0_F$.

The two boundary cases. $\mathbb{N}$ contains $0$, so $n = 0$ is a genuine case. Since $0 = \varnothing$, the set $F^{0} = F^{\varnothing}$ has exactly one element, the empty function; that element is $0_{F^{0}}$, so $F^{0}$ is the zero space $\{0_{F^{0}}\}$, not the empty set. For $n = 1$ we have $1 = \{0\}$, and the map $F^{1} \to F$ sending $x$ to its single coordinate $x_0$ is a bijection satisfying $(x+y)_0 = x_0 + y_0$ and $(\lambda x)_0 = \lambda x_0$; we use it to read $F^{1}$ as $F$ where convenient. (No general notion of isomorphism of vector spaces is available on this page, and none is claimed here: what is asserted is exactly the displayed bijection and the two displayed equations.)

• The same set $F^{X}$ also carries a ring structure, and the two must not be conflated. For a ring $R$ and a set $X$, The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations equips the set $R^{X}$ of all functions $X \to R$ with pointwise addition and pointwise multiplication. Taking $R = F$, the underlying set is literally the same set $F^{X}$ as here, and the addition is literally the same operation, $(f+g)(x) = f(x) + g(x)$ in both. What differs is the second operation:

  	second operation	type
  ring of functions	$(fg)(x) = f(x)\,g(x)$	$F^{X} \times F^{X} \to F^{X}$
  vector space (here)	$(\lambda f)(x) = \lambda\, f(x)$	$F \times F^{X} \to F^{X}$

  Neither is a special case of the other, since they do not even have the same domain: one multiplies two functions, the other multiplies a function by a scalar. They agree in the following sense, and this is the whole of the relation between them: for $\lambda \in F$ let $c_\lambda \in F^{X}$ be the constant function at $\lambda$; then $\lambda f$ and the ring product $c_\lambda f$ have the same value $\lambda f(x)$ at every $x$, so they are equal. Both structures are present on $F^{X}$ at once, and nothing on this page uses the ring product.

• Why $F^{n}$ is defined this way rather than as "$n$-tuples". An $n$-tuple is already a function on an index set, and taking that index set to be the natural number $n$ itself makes the coordinates, the finite sums of The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n = 0$) equal to the identity and the induction arguments below all run over the same object. The price is that every index starts at $0$, and that $F^{0}$ is a one-element space; both are recorded above so that no statement on this page or its companion is quietly restricted to $n \ge 1$.

• $X$ is arbitrary. Nothing above assumes $X$ finite, countable, or nonempty. The case $X = \mathbb{N}$ gives the space of all families of scalars indexed by $\mathbb{N}$, and the case $X = m \times n$ gives the matrices of The vector space $M_{m \times n}(F) := F^{\,m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations.

Let $F$ be a field (Field) and let $m, n \in \mathbb{N}$. Recall that a natural number is a von Neumann natural (The natural numbers $\mathbb{N}$ (von Neumann)), so $m = \{0, \dots, m-1\}$ and $n = \{0, \dots, n-1\}$ (On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$), and let $m \times n$ be their cartesian product. An $m$ by $n$ matrix over $F$ is an element of the function space $F^{\,m \times n}$ (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$), that is a function $a : m \times n \to F$; we write

$a_{ij} \;:=\; a(i,j) \qquad (i < m,\; j < n)$

for its entries, $i$ being the row index and $j$ the column index. Write

$M_{m \times n}(F) \;:=\; F^{\,m \times n},$

and $M_n(F) := M_{n \times n}(F)$ for the square case.

Since $M_{m \times n}(F)$ is the function space $F^{\,m \times n}$, it is a vector space over $F$ with the pointwise operations of The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$, which read entrywise:

$(a + b)_{ij} = a_{ij} + b_{ij}, \qquad (\lambda a)_{ij} = \lambda\, a_{ij} \qquad (i < m,\; j < n),$

and the zero of $M_{m \times n}(F)$ is the matrix all of whose entries are $0_F$. No verification is needed beyond that already carried out in The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$ for an arbitrary index set: this is the case $X = m \times n$.

Both indices start at $0$. The rows are indexed by $i \in m = \{0,\dots,m-1\}$ and the columns by $j \in n = \{0,\dots,n-1\}$, so the entries of a $2$ by $3$ matrix are $a_{00}, a_{01}, a_{02}, a_{10}, a_{11}, a_{12}$.

The degenerate shapes. $\mathbb{N}$ contains $0$, so $m = 0$ and $n = 0$ are genuine cases. If $m = 0$ or $n = 0$ then $m \times n = \varnothing$, so $M_{m \times n}(F)$ has exactly one element, the empty function, and is the zero space; there is no matrix of shape $0$ by $n$ other than that one.

• This item introduces the vector-space structure and the notation, and nothing else. There is no matrix product here, no identity matrix, no ring $M_n(F)$, no determinant and no matrix of a linear map. Those belong to a later page, which must add the product to this object rather than introduce a second notion of matrix; the addition and the scalar multiplication used there are the entrywise ones defined above.

• A matrix is a function, so equality of matrices is equality of functions: $a = b$ exactly when $a_{ij} = b_{ij}$ for all $i < m$ and $j < n$. Nothing below relies on a matrix being written as a rectangular array; the array is a way of displaying the function.

• Why the shape is a pair of naturals rather than a pair of finite index sets. Taking the index set to be $m \times n$ with $m$ and $n$ natural numbers keeps the coordinates of a matrix, of a tuple in $F^{n}$ (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$) and the finite sums of The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n = 0$) equal to the identity all indexed by the same kind of object, which is what makes the boundary cases above statable at all.

Let $F$ be a field (Field).

1. $F$ is a vector space over itself (Vector space over a field): take the set to be $F$, the vector addition to be the field addition, the zero vector to be $0_F$, and the scalar multiplication $F \times F \to F$ to be the field multiplication.
2. Let $K \subseteq F$ be a subfield (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations) and let $V$ be a vector space over $F$. Then $V$, with the same addition and the same zero vector and with the scalar multiplication restricted to $K \times V$, is a vector space over $K$. This is called restricting the scalars from $F$ to $K$.
3. In particular $F$ is a vector space over every subfield $K \subseteq F$, with the field multiplication restricted to $K \times F$ as scalar multiplication.