The other product that can be taken between two vectors can be represented as follows.

(1)
\begin{align} \left[ \begin{array}{c} a_1\\a_2\\a_3 \end{array} \right] \cdot \left[ \begin{array}{ccc} b_1& b_2 &b_3 \end{array} \right]= \left[\begin{array}{ccc}a_1b_1&a_1b_2&a_1b_3\\a_2b_1&a_2b_2&a_2b_3\\a_3b_1&a_3b_2&a_3b_3 \end{array}\right]. \end{align}

This operation is called dyadic product and expressed as

(2)
\begin{align} \boldsymbol{a}\otimes\boldsymbol{b}= \left[\begin{array}{ccc}a_1b_1&a_1b_2&a_1b_3\\a_2b_1&a_2b_2&a_2b_3\\a_3b_1&a_3b_2&a_3b_3 \end{array}\right]. \end{align}

$\otimes$ is called dyadic and $\boldsymbol{a}\otimes\boldsymbol{b}$ is called a dyad.
Dyadic products between basis can be calculated as

(3)
\begin{align} \boldsymbol{e}_1\otimes\boldsymbol{e}_1= \left[ \begin{array}{c} 1\\0\\0 \end{array} \right] \cdot \left[ \begin{array}{ccc} 1& 0 &0 \end{array} \right]= \left[\begin{array}{ccc}1&0&0\\0&0&0\\0&0&0 \end{array}\right], \end{align}
(4)
\begin{align} \boldsymbol{e}_1\otimes\boldsymbol{e}_2= \left[ \begin{array}{c} 1\\0\\0 \end{array} \right] \cdot \left[ \begin{array}{ccc} 0& 1 &0 \end{array} \right]= \left[\begin{array}{ccc}0&1&0\\0&0&0\\0&0&0 \end{array}\right], \end{align}

and so on.
It's likely that dyadic products between arbitrary vectors can be represented by linear combinations of dyadic products between basis (and it is true). In this sense, 9 combinations of dyadic products between basis (in 3-dim. case) become basis of rank-2 tensors.
Only from the properties of multiplications, dyadic products between arbitrary vectors can be calculated as

(5)
\begin{align} \boldsymbol{a}\otimes\boldsymbol{b}=\left(a_i\boldsymbol{e}_i\right)\otimes\left(b_i\boldsymbol{e}_i\right)=a_ib_j\boldsymbol{e}_i\otimes\boldsymbol{e}_j \end{align}
(6)
\begin{align} =a_1b_1\left[\begin{array}{ccc}1&0&0\\0&0&0\\0&0&0 \end{array}\right]+a_1b_2\left[\begin{array}{ccc}0&1&0\\0&0&0\\0&0&0 \end{array}\right]\cdots= \left[\begin{array}{ccc}a_1b_1&a_1b_2&a_1b_3\\a_2b_1&a_2b_2&a_2b_3\\a_3b_1&a_3b_2&a_3b_3 \end{array}\right]. \end{align}

It seems good to understand that $\boldsymbol{e}_i\otimes\boldsymbol{e}_j$ are 9 independent basis and $a_ib_j\boldsymbol{e}_i\otimes\boldsymbol{e}_j$ is a linear combination of them.
However, there is one thing one should pay attention, that is, rank-2 tensors are similar to 9-dim. vectors indeed but they are not completely same.
Let's check the following two calculations:

(7)
\begin{align} \left(\left[ \begin{array}{c} a_1\\a_2\\a_3 \end{array} \right] \cdot \left[ \begin{array}{ccc} b_1& b_2 &b_3 \end{array} \right]\right)\cdot\left[ \begin{array}{c} c_1\\c_2\\c_3 \end{array} \right] = \left[\begin{array}{c} a_1\\a_2\\a_3 \end{array} \right]\left(b_1c_1+b_2c_2+b_3c_3\right) \end{align}

and

(8)
\begin{align} \left[ \begin{array}{ccc} c_1&c_2&c_3 \end{array} \right] \cdot \left(\left[ \begin{array}{c} a_1\\a_2\\a_3 \end{array} \right] \cdot \left[ \begin{array}{ccc} b_1& b_2 &b_3 \end{array} \right]\right)=\left(a_1c_1+a_2c_2+a_3c_3\right)\left[ \begin{array}{ccc} b_1& b_2 &b_3 \end{array} \right]. \end{align}

Thus, sometimes the dyadic product is defined by

(9)
\begin{align} \boldsymbol{a}\otimes\boldsymbol{b}\cdot\boldsymbol{c}=\boldsymbol{a}\left(\boldsymbol{b}\cdot\boldsymbol{c}\right) \end{align}

and

(10)
\begin{align} \boldsymbol{c}\cdot\boldsymbol{a}\otimes\boldsymbol{b}=\left(\boldsymbol{c}\cdot\boldsymbol{a}\right)\boldsymbol{b}. \end{align}

These definitions are dependent on only inner products between vectors; hence it is more general and practical than Eq. (2). Various tensor calculus can be derived from this definition.
Moreover, this definition is valid for general case in which base vectors are not orthonormal; hence this definition plays an important role in tensor calculus.
However, in most textbooks, this definition is posed in the very beginning of the introduction of tensors and makes beginners terrified. Hence, in this document, this definition was posed after a long introduction.

page revision: 16, last edited: 08 May 2015 04:29