Skip to main content

Section 1 The vector space \(\R^n\)

Subsection 1.1 Vectors and vector operations

A list \((x_1,x_2,\ldots,x_n)\) of real numbers, separated by commas and delimited by parentheses, denotes an \(n\)-tuple of real numbers. The set of all \(n\)-tuples of real numbers, denoted \(\R^n\), is called \(n\)-dimensional (real) space.

\begin{equation*} \R^n=\{(x_1,x_2,\ldots,x_n)\colon x_i\in \R, 1\leq i\leq n\} \end{equation*}

The operation of vector addition of two \(n\)-tuples is defined by

\begin{equation*} (x_1,x_2,\ldots,x_n) + (y_1,y_2,\ldots,y_n) = (x_1+y_1,x_2+y_2,\ldots,x_n+y_n) \end{equation*}

and scalar multiplication of an \(n\)-tuple \((x_1,x_2,\ldots,x_n)\) by a real number (or scalar) \(\alpha\) is defined by

\begin{equation*} \alpha(x_1,x_2,\ldots,x_n) = (\alpha x_1,\alpha x_2,\ldots,\alpha x_n). \end{equation*}

With the operations of addition and scalar multiplication, called vector operations, the set \(\R^n\) is a vector space and \(n\)-tuples are vectors.

Vectors are often denoted by lowercase letters in bold Roman font (better for typing), and sometimes italic font decorated with arrows (better for handwriting). Here are examples. 1 

\begin{align*} \mathbf{x} \amp= (x_1,x_2,\ldots,x_n)\\ \vec{x} \amp = (x_1,x_2,\ldots,x_n) \end{align*}

The vector \((0,0,\ldots,0)\) with all entries equal to 0 is called the zero vector and is denoted \(\mathbf{0}\) or \(\vec{0}\).

In many multivariable calculus and introductory physics texts, vectors are denoted using angle brackets, rather than parentheses, for delimiters. That is, these texts write \(\langle x_1,x_2,\ldots,x_n\rangle\) instead of \((x_1,x_2,\ldots,x_n)\text{.}\) In these notes, we do not observe this convention.

A sum of the form

\begin{equation*} \sum_{i=1}^r c_i \mathbf{u}_i = c_1\mathbf{u}_1 + c_2\mathbf{u}_2 + \cdots + c_r\mathbf{u}_r \end{equation*}

where \(c_1,c_2,\ldots,c_r\) are scalars and \(\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_r\) are vectors is called a linear combination of the vectors \(\mathbf{u}_i\text{.}\)

Subsection 1.2 Standard basis vectors

The vector \(\mathbf{e}_i=(0,\ldots,0,1,0,\ldots,0)\) with a single 1 in the \(i\)th position and zeroes in all other positions is called the \(i\)th standard basis vector in \(\R^n\text{.}\) In \(\R^2\text{,}\) the standard basis vectors \(\mathbf{e}_1=(1,0)\) and \(\mathbf{e}_2=(0,1)\) are also called \({\mathbf i}\) and \(\mathbf{j}\text{,}\) respectively. In \(\R^3\text{,}\) the standard basis vectors \(\mathbf{e}_1=(1,0,0)\text{,}\) \(\mathbf{e}_2=(0,1,0)\) and \(\mathbf{e}_3=(0,0,1)\) are also called \({\mathbf i}\text{,}\) \(\mathbf{j}\text{,}\) and \(\mathbf{k}\text{,}\) respectively. Given a vector \(\mathbf{x}=(x_1,\ldots,x_n)\text{,}\) we have the following representation of \(\mathbf{x}\) as a sum of scalar multiples of the standard basis vectors (note that the summation sign indicates vector addition).

\begin{equation} \mathbf{x} = \sum_{i=1}^n x_i\mathbf{e}_i\label{sumofcomps}\tag{1.1} \end{equation}

Equation (1.1) says that every vector is a linear combination of standard basis vectors.

Subsection 1.3 Inner product

The inner product or dot product of two vectors \(\mathbf{x}=(x_1,\ldots,x_n)\) and \(\mathbf{y}=(y_1,\ldots,y_n)\) is defined to be the real number

\begin{equation*} \mathbf{x}\cdot \mathbf{y} = x_1y_1+x_2y_2 + \cdots +x_ny_n. \end{equation*}

In terms of inner product, the \(i\)th coordinate \(x_i\) of the vector \(\mathbf{x}=(x_1,\ldots,x_n)\) is given by

\begin{equation} x_i = \mathbf{e}_i\cdot \mathbf{x} \label{ithentry}\tag{1.2} \end{equation}

and (1.1) becomes

\begin{equation} \mathbf{x} = \sum_{i=1}^n \left(\mathbf{e}_i\cdot \mathbf{x}\right)\mathbf{e}_i. \label{sumofstandprojs}\tag{1.3} \end{equation}

Exercises 1.4 Exercises

1.

Write each of the following vectors \(\mathbf{x}\) in the form \((x_1,x_2,\ldots,x_n)\) and \(\sum_i x_i\mathbf{e}_i\text{.}\) For \(n=2,3\text{,}\) also write \(\mathbf{x}\) using \(\mathbf{i},\mathbf{j},\mathbf{k}\) notation. Example: Given \(\mathbf{x}=3(2,4,-1)\text{,}\) write \(x=(6,12,-3)=6\mathbf{e}_1+12\mathbf{e}_2-3\mathbf{e}_3=6{\mathbf i}+12{\mathbf j}-3{\mathbf k}\text{.}\)

  1. \(\displaystyle \mathbf{x}=(3,2)-(5,-2)\)
  2. \(\displaystyle \mathbf{x}=2(-1,2,1) + 3(2,-2,0)\)
  3. \(\displaystyle \mathbf{x}=2\mathbf{e}_1-3\mathbf{e}_2+4\mathbf{e}_4 -(\mathbf{e}_1-\mathbf{e}_2 +\mathbf{e}_3)\)
Answer
  1. \(\displaystyle \mathbf{x}=(-2,4)=-2\mathbf{e}_1 + 4\mathbf{e}_2 = -2\mathbf{i} + 4\mathbf{j}\)
  2. \(\displaystyle \mathbf{x}=(4,-2,2)= 4\mathbf{e}_1 - 2\mathbf{e}_2 + 2\mathbf{e}_3 = 4\mathbf{i} - 2\mathbf{j} +2 \mathbf{k}\)
  3. \(\displaystyle \mathbf{x}=(1, -2, -1, 4) = \mathbf{e}_1 - 2\mathbf{e}_2- \mathbf{e}_3 + 4\mathbf{e}_4\)
2.

Verify equation (1.1).

3.

Verify equation (1.2).

4.

Show that

\begin{equation*} \mathbf{e}_i\cdot \mathbf{e}_j = \delta_{ij} \end{equation*}

where \(\delta_{ij}\), called the Kronecker delta, is given by

\begin{equation} \delta_{ij}=\left\{ \begin{array}{cc} 1 \amp \mbox{if $i=j$}\\ 0 \amp \mbox{if $i\neq j$} \end{array} \right.\tag{1.4} \end{equation}
5.

The length or norm of vector \(\mathbf{x}=(x_1,x_2,\ldots,x_n)\text{,}\) denoted \(\left\|\mathbf{x}\right\|\), is defined by

\begin{equation*} \left\|\mathbf{x}\right\|^2 = \mathbf{x}\cdot \mathbf{x} = x_1^2 + x_2^2 + \cdots + x_n^2. \end{equation*}

A vector \(\mathbf{x}\) is said be be normalized or have norm 1 if \(\left\|\mathbf{x}\right\|=1\text{.}\)

  1. Show that, for any vector \(\mathbf{x}\) and any real \(\alpha\text{,}\) we have
    \begin{equation*} \left\|\alpha\mathbf{x}\right\|=|\alpha| \left\|\mathbf{x}\right\|. \end{equation*}
  2. Show that, for any nonzero vector \(\mathbf{x}\text{,}\) the vector \(\frac{\mathbf{x}}{\left\|\mathbf{x}\right\|}\) has norm 1.