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:
$$u \otimes v = \left[ { \begin{array}{ccc}
1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}} \right] $$
[Math] Identity tensor as a tensor product of two vectors
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The equation $(v_1\otimes w_1)(v_2\otimes w_2)=(v_1v_1\otimes w_1w_2)$ tells us how to multiply pure tensors but does not immediately say how to multiply sums of pure tensors. But there's only one possible way to extend the multiplication operation to sums of pure tensors in a way that the distributive property holds, and that is through the distributive property itself: when multiplying sums of pure tensors, first distribute and then use the given rule. It is very often in situations like these that we only define multiplication for a set of things and then "extend linearly" to all else via distributivity.
Let's assume everything has finite dimension here.
1) I don't know exactly to what property $(f\otimes g)(...) = f(...)g(...)$ refer to exactly in the second definition of tensor. For example if we have $f$ to be a (0,2) tensor and $g$ a (0,2) tensor. the product would be a (0,4) tensor so $(f\otimes g)(v_1,v_2,v_3,v_4) = f(v_1,v_2)g(v_3,v_4)$. To what kind of tensor in definition 2 would this be isomorphic ?
A tensor product of two vector spaces $V$ and $W$ is a pair $(\mathsf{T} ,t)$, where $\mathsf{T} $ is a vector space and $t\colon V \times W \to \mathsf{T} $ is bilinear, such that if $\{{\bf v}_i\}$ and $\{{\bf w}_j\}$ are bases for $V$ and $W$, then $\{t({\bf v}_i,{\bf w}_j)\}$ spans $\mathsf{T} $, and if given $b\colon V \times W \to Z$ is any bilinear map (arriving at an arbitrary vector space $Z$), there is an unique linear map $\overline{b}\colon \mathsf{T} \to Z$ such that $\overline{b}\circ t = b$. Meaning that bilinear maps $b$ factor through $\mathsf{T} $ and we have all the information needed in a single linear map $\overline{b}$. One then proves that all tensor products of $V$ and $W$ are isomorphic, and so we put the usual notations $\mathsf{T} \equiv V \otimes W$, $t \equiv \otimes$, and write ${\bf v} \otimes {\bf w}$ for $t({\bf v},{\bf w})$.
An explicit construction is to take the free vector space with basis $V \times W$, and take its quotient by the subspace spanned by the elements of the form \begin{align} &({\bf v}_1+{\bf v}_2,{\bf w}) - ({\bf v}_1,{\bf w})-({\bf v}_2,{\bf w}), \\ & ({\bf v},{\bf w}_1+{\bf w}_2)-({\bf v},{\bf w}_1)-({\bf v},{\bf w}_2), \\ & (\lambda{\bf v},{\bf w}) - ({\bf v},\lambda{\bf w}).\end{align} We then denote the class of $({\bf v},{\bf w})$ by ${\bf v} \otimes {\bf w}$.
One generalizes all of this by considering more spaces, writing "multilinear maps" instead of "bilinear maps", and so on. The space $${\frak T}^{(r,s)}(V) = \{ f\colon (V^\ast)^r \times V^s \to \Bbb R \mid f \text{ is multilinear} \}$$is isomorphic to $V^{\otimes r}\otimes (V^\ast)^{\otimes s}$, and that isomorphism does not depend on a choice of basis (so it is better than your average run-of-the-mill isomorphism). Well, being more honest, we use a basis to define the isomorphism, but then we check that it would be same if we started with another basis. We say that $T \in {\frak T}^{(r,s)}(V)$ is a $r-$times contravariant and $s-$times covariant tensor. We of course have an operation $$\otimes \colon {\frak T}^{(r,s)}(V) \times {\frak T}^{(r',s')}(V) \to {\frak T}^{(r+r',s+s')}(V).$$
Now we hopefully understand a little better what a tensor product is, we can simply note that ${\frak T}^{(0,4)}(V) \cong (V^\ast)^{\otimes 4}$, and if $\{{\bf v}_i\}$ is a basis for $V$, and $\{{\bf v}^i\}$ is the dual basis, then $$f \otimes g = \sum_{i,j,k,\ell} f_{ijk\ell} {\bf v}^i \otimes {\bf v}^j\otimes {\bf v}^k \otimes {\bf v}^\ell,$$where $f_{ijkl} = f({\bf v}_i,{\bf v}_j,{\bf v}_k,{\bf v}_\ell)$ and ${\bf v}^i \otimes {\bf v}^j\otimes {\bf v}^k \otimes {\bf v}^\ell \in {\frak T}^{(0,4)}(V)$. It corresponds to that same expression seen as a linear combination of ${\bf v}^i \otimes {\bf v}^j \otimes {\bf v}^k \otimes {\bf v}^\ell$, the class of $({\bf v}^i,{\bf v}^j,{\bf v}^k,{\bf v}^\ell)$ in that quotient - that expression is an element of $(V^\ast)^{\otimes 4}$. Maybe I shouldn't have been lazy and used another notation for the classes until now - I'll gladly explain it all again if you have trouble following.
2) I am confused about what is a tensor component. As I understand tensor components are the scalars that form a linear combination of basis tensors. In definition 1, I see books defining the tensor components as $T^{a_1,...,a_k}_{b_1,...,b_k} = T(a_1,...,a_k,b_1,...,b_k)$. Where $a_i$ and $\{b_i\}$ are a basis of the vector space and covector space. For definition 2 I see tensor component written as : \begin{align} \sum\sum A_{ij}a_i\otimes b_j \end{align} How do these components relate ?
Components do depend on a choice of basis. The choice of notation used in your textbook was bad, we only keep indices on $T$, not the vectors. I mean one would write $$T^{i_1...i_r}_{\qquad j_1...j_s} \stackrel{\rm def.}{=} T({\bf v}^{i_1},...,{\bf v}^{i_r},{\bf v}_{j_1},...,{\bf v}_{j_s})$$instead. And with this notation, we'd have $$T = \sum_{i_1,...,i_r,j_1,...,j_s} T^{i_1...i_r}_{\qquad j_1...j_s} {\bf v}_{i_1}\otimes \cdots \otimes {\bf v}_{i_r}\otimes {\bf v}^{j_1}\otimes \cdots \otimes {\bf v}^{j_s}.$$With Einstein's summation convention, we'd only write $$T = T^{i_1...i_r}_{\qquad j_1...j_s} {\bf v}_{i_1}\otimes \cdots \otimes {\bf v}_{i_r}\otimes {\bf v}^{j_1}\otimes \cdots \otimes {\bf v}^{j_s}$$ with all the summations implied (and that's why index balance is good - if the same index appears twice, up and below, sum over it).
When you ask how these things relate, the reasonable thing to consider is another basis $\{{{\bf v}_i}'\}$, the corresponding dual basis $\{{{\bf v}^i}'\}$, write $${T^{i_1...i_r}_{\qquad j_1...j_s}}' = T({{\bf v}^{i_1}}',...,{{\bf v}^{i_r}}',{{\bf v}_{j_1}}',...,{{\bf v}_{j_s}}')$$and see how we can express this in terms of the "old" components $T^{i_1...i_r}_{\qquad j_1...j_s}$.
If you're still alive after all that index juggling, you'll certainly pardon me by ilustrating the relation only in the $(1,1)$ case. Write ${{\bf v}_j}' = \sum_i \alpha_{ij} v_i$. It is an easy linear algebra exercise to check that ${{\bf v}^j}' = \sum_i \beta_{ij} {\bf v}^i$, where $(\beta_{ij})$ is the inverse matrix of $(\alpha_{ij})$. Then $${T^i_{\hspace{1ex} j}}'=T({{\bf v}^i}',{{\bf v}_j}') = \sum_{k, \ell }\beta_{ki}\alpha_{\ell j}T({{\bf v}^k}',{{\bf v}^\ell}') = \sum_{k,\ell} \beta_{ki}\alpha_{\ell j}T^k_{\hspace{1ex}\ell}.$$
If we had more entries, then more $\alpha$'s and $\beta$'s would pop out. You can see that in any physics book, for instance. I like A Short Course in General Relativity, by Foster & Nightingale. By the way, it is costumary to denote the entries of the inverse matrix by writing indices upstairs. In this notation, and using Einstein's convention, we'd have simply $${T^i_{\hspace{1ex} j}}' = \alpha^{ki}\alpha_{\ell j}T^k_{\hspace{1ex} \ell}.$$
3) I was trying to work out a basic example of an endomorphism $\mathbb{R}^2 \longrightarrow \mathbb{R}^2$ by using the two definitions but couldn't end up with the same set of components...
When we write components in a basis, they're real numbers, and our discussion does not quite apply if the codomain of the bilinear map isn't $\Bbb R$.
Best Answer
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.