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

\[\mathcal M=(F,V;0_F,1_F,\ldots,0_V,+_V,\ldots)\]

is called a vector space if and only if $$\mathcal M \vDash T_{\mathrm{vs}}$,$ and $v\in V$ is called a vector.

Example 1.1.4. Let $$F$$ be a field. We give $$F^n$$ the operations of coordinate-wise addition and scalar multiplication as follows: For $u=(a_1,a_2,\dots,a_n)\in F^n, v=(b_1,b_2,\dots,b_n)\in F^n,$ and $c\in F,$

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

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

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

Example 1.1.5. An $m\times n$ matrix over $$F$$ is an $$m$-$ tuple of $$n$-$ tuples, which may be written as follows:

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

where $a_{ij}\in F$ for $1\leq i \leq m$ and $1\leq j \leq n.$ Thus the set of all $m\times n$ matrices from a field $$F$$ is the set $$(F^n)^m.$$ We henceforth write ${F}^{m\times n}$ in place of $$(F^n)^m.$$ 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 $$A$$ the zero matrix and denote it by $$O.$$ In addition, $$A$$ is called an upper triangular matrix if $A_{ij}=0$ whenever $$i>j,$$ and is called a lower triangular matrix $A_{ij}=0$ whenever $$i<j.$$ An $m\times n$ matrix is called a square matrix if $$m=n.$$ For a square matrix $A\in F^{n\times n},$ the $$n$-$ tuple $(a_{11},\dots,a_{nn})$ is called the main diagonal of $$A,$$ and the sum of all the entries of the main diagonal is called the trace of $$A.$$ We give ${F}^{m\times n}$ the following operations of matrix addition and scalar multiplication: For $A, B \in {F}^{m\times n}$ and $c\in F,$

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

for $1\leq i \leq m$ and $1\leq j \leq n.$ Then ${F}^{m\times n}$ is a vector space over $$F.$$

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

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

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 $\mathbf 0 (x)=0$ for all $x\in S.$ Then we can see that $$F^S$$ is a vector space over $$F.$$

Example 1.1.7. A polynomial over a field $$F$$ is an expression of the form

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

where $n\in \mathbb N$ and each $$a_k,$$ called the coefficient of $$x^k,$$ is in $$F.$$ 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 from $$F$$ is denoted by $$F[x].$$ 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 $$F[x]$$ the following operations of addition and scalar multiplication:

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

and for $c\in F,$

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

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

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

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

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

Proof. A zero vector was axiomatically defined as follows:

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

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

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

Then $$u+0_V=u+v.$$ By Theorem 1.1.8, $$0_V$$ 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 $$V,$$ then the vector $$-u$$ is unique.

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

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

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

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

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 $$V,$$ the following statements are true:

$$0u=0_V$$ for all $u\in V.$
$$(-a)u=-(au)=a(-u)$$ for each $a\in F$ and each $u\in V.$
$$a0_V=0_V$$ for all $a\in F.$

Proof. (1) The zero vector is the unique vector that behaves in the way $\forall u \in V \ (u + 0_V = u)$ (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 $$au+(-(au))=0_V$$ (the additive inverse of $$au$).$ But we know that $$au+(-a)u=(a+(-a))u=0u=0_V.$$ Thus $$(-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+a0_V = a(u+0_V)=au.$$ Thus $$a0_V$$ is the zero vector. $\square$

Exercises

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

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

Claim. Let $$a$$ be an element of a field $$F,$$ and $$u$$ be an element of a vector space $$V$$ over $$F.$$ If $a\neq 0$ and $u\neq 0_V,$ then $au\neq 0_V.$

Subproof. Suppose, for contradiction, that $$au=0_V.$$ Since $a\neq 0,$ there exists $a^{-1}\in F$ such that $a^{-1}a=1.$ Then $u=1u=(a^{-1}a)u=a^{-1}(au)=0_V.$ This contradicts $u\neq 0_V.$ Therefore $au\neq 0_V.$ $\blacksquare$

By the contrapositive of this Claim, $$a+(-b)=0$$ because $u\neq 0_V.$ 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 $a^{-1}\in F$ 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 $$V$$ denote the set of all differentiable real-valued functions defined on the real line ${\mathbb{R}}.$ Prove that $$V$$ is a vector space.

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

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

The additive inverse of $f\in V$ 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 $f,g\in V, c\in\mathbb R,$ and $$x\in\mathbb R$.$ If $f,g\in V,$ then $$D(f+g)=Df + Dg.$$ Thus $f+g\in V.$ Similarly, if $f\in V$ and $c\in \mathbb R,$ then $$D(cf)=c(Df).$$ Thus $cf\in V.$ Therefore closure is guaranteed so that the operations work well on $$V.$$ We now prove that the vector space axioms hold on $$V$:$ (1) Since $$(f+g)(x)=f(x)+g(x)=g(x)+f(x)=(g+f)(x)$$ for all $x\in \mathbb R,$ $$f+g=g+f.$$ (2) 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 $x\in \mathbb R,$ $$(f+g)+h=f+(g+h).$$ (3) Since $(f+\mathbf 0)(x)=f(x)+\mathbf 0(x)=f(x)+0=f(x)$ for all $x\in \mathbb R,$ $f+\mathbf 0=f.$ (4) Since $(f+(-f))(x)=f(x)+(-f(x))=0=\mathbf 0(x)$ for all $x\in \mathbb R,$ $f+(-f)=\mathbf 0.$ (5) Since $$(1f)(x)=1f(x)=f(x)$$ for all $x\in \mathbb R,$ $$1f=f.$$ (6) Since $$((ab)f)(x)=(ab)f(x)=a(bf(x))=a(bf)(x)$$ for all $x\in \mathbb R,$ $$(ab)f=a(bf).$$ (7) 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 $x\in \mathbb R,$ $$a(f+g)=af+ag.$$ (8) 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 $$V$$ is a vector space over $\mathbb R.$ $\square$

Exercise 4. Let $V=\lbrace 0_V \rbrace$ and define two operations by $$0_V + 0_V = 0_V$$ and $$c0_V = 0_V$$ for all scalar $c\in F.$ Prove that $$V$$ is a vector space over $$F.$$ ( $$V$$ is called the zero vector space.)

Proof. The zero vector is $$0_V,$$ and the additive inverse of $$0_V$$ is $$0_V.$$ Then the following hold: (1) $$0_V + 0_V = 0_V + 0_V.$$ (2) $$(0_V + 0_V) + 0_V = 0_V + 0_V = 0_V + (0_V+ 0_V).$$ (3) $$0_V + 0_V= 0_V.$$ (4) $$0_V + (-0_V)=0_V + 0_V=0_V.$$ (5) $$10_V=0_V.$$ (6) $$(ab)0_V=0_V=a0_V=a(b0_V).$$ (7) $$a(0_V + 0_V)=a0_V=0_V=0_V+0_V=a0_V+a0_V.$$ (8) $$(a+b)0_V=0_V=0_V+0_V=a0_V+b0_V.$$ Thus all vector space axioms hold. Therefore $$V$$ is a vector space over $$F.$$ $\square$

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

Proof. Let

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

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

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

Since $\mathbf 0(-x)=0=\mathbf 0(x)$ for all $x\in \mathbb R,$ $\mathbf 0 \in V.$ The additive inverse of $f\in V$ 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 $x\in \mathbb R,$ $-f \in V.$ The operations are pointwise addition and multiplication defined by

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

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

External

1.1. Vector Spaces

Exercises

References