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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.

\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 \(\vec{0}\).

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 \(|\mathbf{x}|\) (or often \(\|\mathbf{x}\|\)), is defined by

\begin{equation*} |\mathbf{x}|^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 \(|\mathbf{x}|=1\text{.}\)

  1. Show that, for any vector \(\mathbf{x}\) and any \(\alpha>0\text{,}\) we have
    \begin{equation*} |\alpha\mathbf{x}|=\alpha|\mathbf{x}|. \end{equation*}
  2. Show that, for any nonzero vector \(\mathbf{x}\text{,}\) the vector \(\frac{\mathbf{x}}{|\mathbf{x}|}\) has norm 1.