1.2  Vector Spaces

Tuples, matrices, functions, and polynomials are different kinds of objects, yet each can be added to another of its kind and can be multiplied by a scalar. A vector space captures the rules the operations share: addition forms an abelian group, and scalar multiplication by a field works consistently with addition and field arithmetic. This common structure lets us study the fundamental principles such as linear independence, transformation and dimension without depending on what the vectors themselves look like.

Definition 1.2.1. Let \(F\) be a field. Then, a vector space over \(F\) consists of an abelian group \(V\) under addition, together with a scalar multiplication \(F\times V \to V\) such that, for all \(a,b\in F\) and \(u,v\in V\),

  1. \(1v = v\),
  2. \((ab)v = a(bv)\),
  3. \(a(u + v) = au + av\), and
  4. \((a + b)v = av + bv\).

The elements of a vector space are called vectors.

Example 1.2.2. Let \(F\) be a field. An object of the form \((a_1,a_2,\dots,a_n)\) with \(a_1,a_2,\ldots,a_n\in F\) is called an \(n\)-tuple with entries from \(F\). The set of all \(n\)-tuples from \(F\) is denoted by \(F^n\). We give \(F^n\) the operations of coordinate-wise addition and scalar multiplication. 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 row vector \((a_1,a_2,\dots,a_n)\) may be written as a column vector

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

Example 1.2.3. An \(m\times n\) matrix with entries from a field \(F\) is a rectangular array of the form

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

where each \(a_{ij}\) is an element of \(F\), called an entry of the matrix. We call the entries \(a_{ij}\) with \(i=j\) the diagonal entries. The diagonal entries compose the main diagonal of the matrix. If every entry equals zero, the matrix is called the zero matrix, denoted by \(O\). The entries \(a_{i1},\ldots,a_{in}\) compose the \(i\)th row of the matrix, and the entries \(a_{1j},\ldots,a_{mj}\) compose the \(j\)th column of the matrix. If the number of rows and columns are equal, the matrix is called a square matrix. The set of all \(m\times n\) matrices from \(F\) is denoted by \(\operatorname{Mat}_{m\times n}(F)\). We give \(\operatorname{Mat}_{m\times n}(F)\) the following operations of matrix addition and scalar multiplication. For \(A, B \in \operatorname{Mat}_{m\times n}(F)\) 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 \(\operatorname{Mat}_{m\times n}(F)\) is a vector space over \(F.\)

Example 1.2.4. 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 \(x\in S.\) Two functions \(f\) and \(g\) are equal if \(f(x)=g(x)\) for all \(x\in S,\) and the zero function is defined by \(\mathbf 0 (x)=0\) for all \(x\in S\). We can see that \(F^S\) is a vector space over \(F.\)

Example 1.2.5. 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 \(n\in \mathbb Z_{\ge 0}\) 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 their corresponding coefficients agree after padding with zero coefficients to the same length. The set of all polynomials over \(F\) 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 1.2.6. Let \(V\) be a vector space over a field \(F\).

  1. For all \(u\in V,\) we have \(0u=0_V\).
  2. For each \(a\in F\) and each \(u\in V,\) we have \((-a)u=-(au)=a(-u)\).
  3. For all \(a\in F,\) we have \(a0_V=0_V\).

Proof. (1) The zero vector is unique. Given a vector \(au\), we have \(au + 0u = (a+0)u=au.\) 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 \((-a)u=-(au).\) It follows 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) We have \(au+a0_V = a(u+0_V)=au.\) Thus \(a0_V\) is the zero vector. \(\square\)

Exercises

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

Proof. We have \(au+(-(au))=0_V.\) Since \(au=bu,\) \(au+(-(bu))=0_V.\) However, by Theorem 1.2.6, \(-(bu)=(-b)u.\) Thus \(au+(-b)u=0_V.\) It follows that \((a+(-b))u=0_V.\) We now prove a lemma.

Lemma. 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 lemma, \(a+(-b)=0\) because \(u\neq 0_V.\) Since the additive inverse of a scalar is unique, \(a=b.\) \(\square\)

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 \(u=1u=(a^{-1}a)u=a^{-1}(au)=a^{-1}(av)=(a^{-1}a)v=1v=v.\) \(\square\)