Vector Spaces

Definition 1.1.1. Let $\mathcal L_{\mathrm{vs}}$ be the two-sorted first-order language with sorts $$F$$ and $$V.$$ The language contains constant symbols $$0_F, 1_F$$ of sort $$F,$$ function symbols $+_F, \cdot_F: F \times F \to F,$ a unary function symbol $-_F: F \to F,$ a constant symbol $$0_V$$ of sort $$V,$$ a function symbol $+_V: V \times V \to V,$ a unary function symbol $-_V: V \to V,$ and a function symbol $\cdot_{FV}: F \times V \to V$ for scalar multiplication.

Definition 1.1.2. Let $T_{\mathrm{vs}}$ be the $\mathcal L_{\mathrm{vs}}$ -theory consisting of the following axioms. First, the $$F$$ -sort is a field under $0_F, 1_F, +_F, -_F, \cdot_F.$ Second, the $$V$$ -sort is an abelian group under $$0_V, +_V, -_V.$$ That is, the following axioms hold:

$\forall u, v \in V \ (u +_V v = v +_V u),$
$\forall u, v, w \in V \ ((u +_V v) +_V w = u +_V (v +_V w)),$
$\forall v \in V \ (v +_V 0_V = v),$
$\forall v \in V \ (v +_V (-_V v) = 0_V).$

Finally, the scalar multiplication symbol $\cdot_{FV}$ satisfies the following axioms:

$\forall v \in V \ (1_F \cdot_{FV} v = v),$
$\forall a, b \in F \ \forall v \in V \ ((a \cdot_F b)\cdot_{FV} v = a \cdot_{FV} (b \cdot_{FV} v)),$
$\forall a \in F \ \forall u, v \in V \ (a \cdot_{FV} (u +_V v) = (a \cdot_{FV} u) +_V (a \cdot_{FV} v)),$
$\forall a, b \in F \ \forall v \in V \ ((a +_F b) \cdot_{FV} v = (a \cdot_{FV} v) +_V (b \cdot_{FV} v)).$

Definition 1.1.3. An $\mathcal L_{\mathrm{vs}}$ -structure $\mathbb V$ is called a vector space if $\mathbb V \vDash T_{\mathrm{vs}},$ and $$v$$ is called a vector if $v\in \mathbb V.$

Example 1.1.4. An object of the form $(a_1,a_2,\dots,a_n),$ where $a_1,a_2,\dots,a_n$ are scalars of a field $\mathbb F,$ is called an $$n$$ -tuple. We say that $a_1,a_2,\dots,a_n$ are entries of the $$n$$ -tuple. Two $$n$$ -tuples $(a_1,a_2,\dots,a_n)$ and $(b_1,b_2,\dots,b_n)$ are equal if and only if $$a_i=b_i$$ for all $i\in \lbrace 1,2,\dots,n\rbrace.$ The set of all $$n$$ -tuples with entries from $\mathbb F$ is denoted by $\mathbb F^n.$
We give $\mathbb F^n$ the operations of coordinate-wise addition and scalar multiplication as follows: For $u=(a_1,a_2,\dots,a_n)\in \mathbb F^n, v=(b_1,b_2,\dots,b_n)\in \mathbb F^n,$ and $c\in \mathbb F,$

\[u+v=(a_1+b_1, a_2+b_2,\dots, a_n+b_n) \quad \text{and} \quad cu=(ca_1,ca_2,\dots,ca_n).\]

Then, $\mathbb F^n$ is a vector space over $\mathbb F.$ A vector in $\mathbb F^n$ may be written as

\[\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix}.\]

Example 1.1.5. An $m\times n$ matrix with scalars of $\mathbb F$ is a rectangular array of the form

\[\begin{pmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{pmatrix}.\]

We say that $a_{ij}\ (1\leq i \leq m, 1\leq j \leq n)$ is an entry of the $m\times n$ matrix, and we call the entries $a_{ij}$ with $$i=j$$ the diagonal entries of the matrix. If each entry equals zero we call the matrix the zero matrix and denote it by $$O.$$ In addition, the matrix is called square if $$m=n.$$ Two $m\times n$ matrices are equal if and only if all their corresponding entries are equal. And the set of all $m\times n$ matrices from a field $\mathbb F$ is denoted by ${\mathbb F}^{m\times n}.$
We give ${\mathbb F}^{m\times n}$ the following operations of matrix addition and scalar multiplication: For $A, B \in {\mathbb F}^{m\times n}$ and $c\in \mathbb F,$

\[(A+B)_{ij}=A_{ij}+B_{ij} \quad \text{and} \quad (cA)_{ij}=cA_{ij}\]

for $1\leq i \leq m$ and $1\leq j \leq n.$ Then ${\mathbb F}^{m\times n}$ is a vector space over $\mathbb F.$
Let $(\mathbb F^n)^m$ denote the set of all $$m$$ -tuples of vectors in $\mathbb F^n$ . That is,

\[(\mathbb F^n)^m = \lbrace (v_1, \dots, v_m) : v_i \in \mathbb F^n \text{ for each } i = 1, \dots, m\rbrace .\]

We can regard $(\mathbb F^n)^m$ as a vector space over $\mathbb F$ by defining addition and scalar multiplication componentwise:

\[(v_1, \dots, v_m) + (w_1, \dots, w_m) = (v_1 + w_1, \dots, v_m + w_m),\]

and

\[c(v_1, \dots, v_m) = (cv_1, \dots, cv_m).\]

There is a natural map $\Phi:{\mathbb F}^{m\times n}\to(\mathbb F^n)^m$ defined by $\Phi(A)=(r_1(A),\dots,r_m(A)),$ where $$r_i(A)$$ denotes the $$i$$ th row vector of $$A.$$ Indeed, it is bijective, since every $$m$$ -tuple of vectors in $\mathbb F^n$ uniquely determines an $m\times n$ matrix by placing those vectors as rows. Moreover, $\Phi$ preserves addition and scalar multiplication:

\[\begin{aligned} \Phi(A+B) &= (r_1(A+B),\dots,r_m(A+B))\\ &= (r_1(A)+r_1(B),\dots,r_m(A)+r_m(B))\\ &= \Phi(A)+\Phi(B), \end{aligned}\]

and

\[\begin{aligned} \Phi(cA) &= (r_1(cA),\dots,r_m(cA))\\ &= (cr_1(A),\dots,cr_m(A))\\ &= c\Phi(A). \end{aligned}\]

Hence $\Phi$ is a vector space isomorphism, and ${\mathbb F}^{m\times n}$ and $(\mathbb F^n)^m$ are isomorphic as vector spaces. That is,

\[{\mathbb F}^{m\times n} \cong (\mathbb F^n)^m.\]

Likewise, we can see that ${\mathbb F}^{m\times n}$ and $(\mathbb F^m)^n$ are isomorphic as vector spaces.

Example 1.1.6. Let $$S$$ be a nonempty set and $\mathbb F$ be a field, and let $\mathbb F^S$ denote the set of all functions from $$S$$ to $\mathbb F.$ We give $\mathbb F^S$ the operations of point-wise addition and scalar multiplication: For $f,g\in \mathbb F^S$ and $c\in \mathbb F,$

\[(f+g)(x)=f(x)+g(x) \quad \text{and} \quad (cf)(x)=c(f(x))\]

for all $x\in S.$ Incidentally, We know that the definition of the equality of two functions $$f$$ and $$g$$ is that $$f(x)=g(x)$$ for all $x\in S,$ and the definition of the zero function is that $\mathit 0(x)=0$ for all $x\in S.$ Then we can see that $\mathbb F^S$ is a vector space over $\mathbb F.$

Example 1.1.7. A polynomial from a field $\mathbb F$ is an expression of the form

\[a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0,\]

where $$n$$ is a nonnegative integer and each $$a_k,$$ called the coefficient of $$x^k,$$ is in $\mathbb F.$ If $a_n=a_{n-1}=\cdots =a_0=0,$ then the polynomial is called the zero polynomial and, its degree is defined to be $-\infty;$ otherwise, the degree of a polynomial is defined to be the largest exponent of $$x$$ that appears in the expression with a nonzero coefficient. Two polynomials,

\[P(x)=a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0,\]

and

\[Q(x)=b_m x^m + b_{m-1} x^{m-1} + \cdots + b_1 x + b_0,\]

are equal if and only if $$m=n$$ and $$a_i=b_i$$ for all $i\in \lbrace 0,1,\dots, n\rbrace.$ The set of all polynomials with coefficients from $\mathbb F$ is denoted by $\mathbb F[x].$
If $m\leq n,$ then $$Q(x)$$ can be written as

\[Q(x)=b_n x^n + b_{n-1} x^{n-1} + \cdots + b_1 x + b_0,\]

where $b_{m+1}=b_{m+2}=\cdots =b_n=0.$ And we give $\mathbb F[x]$ the following operations of addition and scalar multiplication:

\[(P+Q)(x)=(a_n+b_n)x^n+(a_{n-1}+b_{n-1})x^{n-1}+\cdots + (a_1+b_1)x+(a_0+b_0)\]

and for $c\in \mathbb F,$

\[(cP)(x)=ca_n x^n + ca_{n-1} x^{n-1} + \cdots + ca_1 x + ca_0.\]

Then $\mathbb F[x]$ is a vector space over $\mathbb F.$

Theorem 1.1.8. If $$u,v,$$ and $$w$$ are vectors in a vector space $\mathbb V$ such that $$u+w=v+w,$$ then $$u=v.$$

Proof. We have $u=u+\mathit 0=u+(w+(-w))=(u+w)+(-w),$ and $v=v+\mathit 0=v+(w+(-w))=(v+w)+(-w).$ But by the condition, $$(u+w)+(-w)=(v+w)+(-w).$$ Thus $$u=v.$$ $\square$

Corollary 1.1.9. In a vector space $\mathbb V,$ the zero vector is unique.

Proof. A zero vector was axiomatically defined as follows:

\[\forall u \in \mathbb V \ (u + \mathit 0 = u).\]

The question is if there can be more than one vector that behaves like $\mathit 0.$ Suppose that

\[u+\mathit 0=u \quad \text{and} \quad u+v=u.\]

Then $u+\mathit 0=u+v.$ By Theorem 1.1.8, $\mathit 0$ and $$v$$ are the same. Thus there is only one zero vector. $\square$

Corollary 1.1.10. If $$u$$ is a vector in a vector space $\mathbb V,$ then the vector $$-u$$ is unique.

Proof. An additive inverse of $$u$$ was axiomatically defined as follows:

\[\forall u \in \mathbb V \ (u + (-u) = \mathit 0).\]

Suppose that $$v$$ and $$w$$ are additive inverses of $$u,$$ that is,

\[u+v=\mathit 0 \quad \text{and} \quad u+w=\mathit 0.\]

Then $$u+v=u+w.$$ By Theorem 1.1.8, $$v$$ and $$w$$ are the same. Thus there is only one additive inverse of $$u.$$ $\square$

Theorem 1.1.11. In a vector space $\mathbb V,$ the following statements are true:

$0u=\mathit 0$ for all $u\in \mathbb V.$
$$(-a)u=-(au)=a(-u)$$ for each $a\in \mathbb F$ and each $u\in \mathbb V.$
$a\mathit 0=\mathit 0$ for all $a\in \mathbb F.$

Proof. (1) The zero vector is the unique vector that behaves in the way $\forall u \in \mathbb V \ (u + \mathit 0 = u)$ (the additive identity). But for a given vector $$au,$$ $$au + 0u = (a+0)u=au.$$ Thus $$0u$$ is the zero vector.

(2) The vector $$-(au)$$ is the unique element such that $au+(-(au))=\mathit 0$ (the additive inverse of $$au$$ ). But we know that $au+(-a)u=(a+(-a))u=0u=\mathit 0.$ Thus $$(-a)u=-(au).$$ From this, we can see that $$-(au)=(-a)u=((-1)a)u=(-1)(au).$$ Thus we have $(-1)\cdot u=-u.$ Finally we get $$(-a)u=(a(-1))u=a((-1)u)=a(-u).$$

(3) $au+a\mathit 0 = a(u+\mathit 0)=au.$ Thus $a\mathit 0$ is the zero vector. $\square$

Exercises

Exercise 1. Prove that, in a vector space, if $u\neq \mathit 0$ and $$au=bu$$ then $$a=b.$$

Proof. $au+(-(au))=\mathit 0.$ Since $$au=bu,$$ $au+(-(bu))=\mathit 0.$ But by Theorem 1.1.11, $$-(bu)=(-b)u.$$ Thus $au+(-b)u=\mathit 0.$ It follows that $(a+(-b))u=\mathit 0.$ We now prove a lemma.

Claim. Let $$a$$ be an element of a field $\mathbb F,$ and $$u$$ be an element of a vector space $\mathbb V$ over $\mathbb F.$ If $a\neq 0$ and $u\neq \mathit 0,$ then $au\neq \mathit 0.$
Subproof. Suppose, for contradiction, that $au=\mathit 0.$ Since $a\neq 0,$ there exists $a^{-1}\in \mathbb F$ such that $a^{-1}a=1.$ Then $u=1u=(a^{-1}a)u=a^{-1}(au)=\mathit 0.$ This contradicts $u\neq \mathit 0.$ Therefore $au\neq \mathit 0.$ $\Diamond$

By the contrapositive of this Claim, $$a+(-b)=0$$ because $u\neq \mathit 0.$ Since the additive inverse of a scalar is unique, $$a=b.$$ $\square$

Exercise 2. Prove that, in a vector space, if $a \neq 0$ and $$au=av$$ then $$u=v.$$
Proof. Since $a\neq 0,$ there exists $a^{-1}\in \mathbb F$ such that $a^{-1}a=1.$ Thus $u=1u=(a^{-1}a)u=a^{-1}(au)=a^{-1}(av)=(a^{-1}a)v=1v=v.$ $\square$

Exercise 3. Let $\mathbb V$ denote the set of all differentiable real-valued functions defined on the real line ${\mathbb{R}}.$ Prove that $\mathbb V$ is a vector space.

Proof. The zero vector is the zero function $\mathit 0:\mathbb R\to \mathbb R,$ defined by

\[\forall x\in \mathbb R \ (\mathit 0(x)=0).\]

The additive inverse of $f\in \mathbb V$ is the function $$-f,$$ defined by

\[\forall x\in \mathbb R \ ((-f)(x)=-f(x)).\]

The operations are pointwise addition and multiplication defined by

\[(f+g)(x)=f(x)+g(x), \qquad (cf)(x)=c f(x)\]

for $f,g\in \mathbb V, c\in\mathbb R,$ and $x\in\mathbb R$ . If $f,g\in \mathbb V,$ then $$D(f+g)=Df + Dg.$$ Thus $f+g\in \mathbb V.$ Similarly, if $f\in \mathbb V$ and $c\in \mathbb R,$ then $$D(cf)=c(Df).$$ Thus $cf\in \mathbb V.$ Therefore closure is guaranteed so that the operations work well on $\mathbb V.$
We now prove that the vector space axioms hold on $\mathbb V.$

Since $$(f+g)(x)=f(x)+g(x)=g(x)+f(x)=(g+f)(x)$$ for all $x\in \mathbb R,$ $$f+g=g+f.$$
Since $$((f+g)+h)(x)=(f+g)(x)+h(x)=(f(x)+g(x))+h(x)=f(x)+(g(x)+h(x))=f(x)+(g+h)(x)=(f+(g+h))(x)$$ for all $x\in \mathbb R,$ $$(f+g)+h=f+(g+h).$$
Since $(f+\mathit 0)(x)=f(x)+\mathit 0(x)=f(x)+0=f(x)$ for all $x\in \mathbb R,$ $f+\mathit 0=f.$
Since $(f+(-f))(x)=f(x)+(-f(x))=0=\mathit 0(x)$ for all $x\in \mathbb R,$ $f+(-f)=\mathit 0.$
Since $$(1f)(x)=1f(x)=f(x)$$ for all $x\in \mathbb R,$ $$1f=f.$$
Since $$((ab)f)(x)=(ab)f(x)=a(bf(x))=a(bf)(x)$$ for all $x\in \mathbb R,$ $$(ab)f=a(bf).$$
Since $$(a(f+g))(x)=a(f+g)(x)=a(f(x)+g(x))=af(x)+ag(x)=(af)(x)+(ag)(x)=(af+ag)(x)$$ for all $x\in \mathbb R,$ $$a(f+g)=af+ag.$$
Since $$((a+b)f)(x)=(a+b)f(x)=af(x)+bf(x)=(af)(x)+(bf)(x)=(af+bf)(x)$$ for all $x\in \mathbb R,$ $$(a+b)f=af+bf.$$

Thus all vector space axioms hold. Therefore $\mathbb V$ is a vector space over $\mathbb R.$ $\square$

Exercise 4. Let $\mathbb V=\lbrace \mathit 0 \rbrace$ and define two operations by $\mathit 0 + \mathit 0 = \mathit 0$ and $c\mathit 0 = \mathit 0$ for all scalar $c\in \mathbb F.$ Prove that $\mathbb V$ is a vector space over $\mathbb F.$ ( $\mathbb V$ is called the zero vector space.)

Proof. The zero vector is $\mathit 0,$ and the additive inverse of $\mathit 0$ is $\mathit 0.$ Then the following holds:

$\mathit 0 + \mathit 0 = \mathit 0 + \mathit 0.$
$(\mathit 0 + \mathit 0) + \mathit 0 = \mathit 0 + \mathit 0 = \mathit 0 + (\mathit 0+ \mathit 0).$
$\mathit 0 + \mathit 0= \mathit 0.$
$\mathit 0 + (-\mathit 0)=\mathit 0 + \mathit 0=\mathit 0.$
$1\mathit 0=\mathit 0.$
$(ab)\mathit 0=\mathit 0=a\mathit 0=a(b\mathit 0).$
$a(\mathit 0 + \mathit 0)=a\mathit 0=\mathit 0=\mathit 0+\mathit 0=a\mathit 0+a\mathit 0.$
$(a+b)\mathit 0=\mathit 0=\mathit 0+\mathit 0=a\mathit 0+b\mathit 0.$

Thus all vector space axioms hold. Therefore $\mathbb V$ is a vector space over $\mathbb F.$ $\square$

Exercise 5. Prove that the set of even functions defined on the real line is a vector space.

Proof. Let

\[\mathbb V=\lbrace f:\mathbb R\to \mathbb R : \forall x \in \mathbb R\ ( f(-x)=f(x)\rbrace .\]

The zero vector is the zero function $\mathit 0:\mathbb R\to \mathbb R,$ defined by

\[\forall x\in \mathbb R \ (\mathit 0(x)=0).\]

Since $\mathit 0(-x)=0=\mathit 0(x)$ for all $x\in \mathbb R,$ $\mathit 0 \in \mathbb V.$ The additive inverse of $f\in \mathbb V$ is the function $$-f,$$ defined by

\[\forall x\in \mathbb R \ ((-f)(x)=-f(x)).\]

Since $$(-f)(-x)=-f(-x)=-f(x)=(-f)(x)$$ for all $x\in \mathbb R,$ $-f \in \mathbb V.$ The operations are pointwise addition and multiplication defined by

\[(f+g)(x)=f(x)+g(x), \qquad (cf)(x)=c f(x)\]

for $f,g\in \mathbb V, c\in\mathbb R,$ and $x\in\mathbb R$ . If $f,g\in \mathbb V,$ then $$(f+g)(-x)=f(-x)+g(-x)=f(x)+g(x)=(f+g)(x).$$ Thus $f+g\in \mathbb V.$ Similarly, if $f\in \mathbb V$ and $c\in \mathbb R,$ then $$(cf)(-x)=cf(-x)=cf(x)=(cf)(x).$$ Thus $cf\in \mathbb V.$ Therefore closure is guaranteed so that the operations work well on $\mathbb V.$
In exactly the same way done in Exercise 3, one proves that the vector axioms hold on $\mathbb V.$ Therefore $\mathbb V$ is a vector space over $\mathbb R.$ $\square$

References

Friedberg, S. H., Insel, A. J., & Spence, L. E. (2025). Linear algebra (5th ed.). Pearson Education South Asia Pte Ltd.

Semantics

1.1. Vector Spaces

Exercises

References