Identity tensor as a tensor product of two vectors Any second order tensor in a given basis can be expressed as a matrix. Also, as any second order tensor can be expressed a tensor product of two first order tensors (or vectors), I would like to find $u$ and $v$ that satisfy:
$$u \otimes v = \left[ { \begin{array}{ccc} 
1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}}  \right] $$
 A: As pointed out, you need to take the sum of products of first order tensors to be able to obtain any second order tensor. To achieve this, write out the matrix corresponding to $e_i\otimes e_j$ and see what you get; you should be able to find the decomposition from here.
A: We have a correspondence
$$\begin{pmatrix}\color{Magenta}{a} \\ \color{Magenta}{b} \\ \color{Magenta}{c}\end{pmatrix} \otimes \begin{pmatrix}\color{Blue}{x} & \color{Blue}{y} & \color{Blue}{z}\end{pmatrix} \quad\longleftrightarrow\quad\begin{pmatrix} \color{Magenta}{a} \color{Blue}{x} & \color{Magenta}{a} \color{Blue}{y} & \color{Magenta}{a} \color{Blue}{z} \\ \color{Magenta}{b} \color{Blue}{x} & \color{Magenta}{b} \color{Blue}{y} & \color{Magenta}{b} \color{Blue}{z} \\ \color{Magenta}{c} \color{Blue}{x} & \color{Magenta}{c} \color{Blue}{y} & \color{Magenta}{c} \color{Blue}{z} \end{pmatrix} $$
As you can see, each column is a multiple of the pink column vector (by $x$, $y$ and $x$ respectively) and each row is a multiple of the blue row vector (by $a$, $b$ and $c$ respectively). As the columns and rows of the matrix are linearly dependent the matrix is always singular.
If $U$ and $V$ are two vector spaces, $U\otimes V$ does contain elements of the form $u\otimes v$ but these elements (pure tensors) are not closed under addition; there are sums of pure tensors that cannot be "factorized" into the form of a single pure tensor. The space $U\otimes V$ is the span of pure tensors.
If ${\cal B}_U$ and ${\cal B}_V$ are bases of $U$ and $V$ respectively then the pure tensors $u\otimes v$ ($u\in U, v\in V$) form a basis for $U\otimes V$. In particular $e_i\otimes e_j$ corresponds to what matrices? Now try to write the identity $3\times3$ matrix as a linear combination of them.
