Section1The vector space $$\R^n$$

Subsection1.1Vectors 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 $$\mathbf{u}_i\text{.}$$

Subsection1.2Standard 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).

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

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

Subsection1.3Inner 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

$$x_i = \mathbf{e}_i\cdot \mathbf{x} \label{ithentry}\tag{1.2}$$

and (1.1) becomes

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

Exercises1.4Exercises

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. $$\displaystyle \mathbf{x}=2\mathbf{e}_1-3\mathbf{e}_2+4\mathbf{e}_4 -(\mathbf{e}_1-\mathbf{e}_2 +\mathbf{e}_3)$$
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

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