linear algebratensor-productsvectors
So it's easy to take the tensor product of two vectors,
$$\begin{bmatrix}a\\b\\\end{bmatrix}\otimes\begin{bmatrix}c\\d\\\end{bmatrix}=\begin{bmatrix}ac\\ad\\bc\\bd\\\end{bmatrix}$$
but it seems much more difficult to go backwards, to "factor" the vector. Is there a method to find two vectors whose tensor product is a given vector?
In my specific problem, the magnitude of the resultant vector is 1, if that's relevant.
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.
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
To find a solution, assume your tensor is $(x,y,z,w)^\top.$ We must have $xw=abcd=yz$ for there to be any solution.
Assume $x\neq 0.$
Then $$\frac{b}{a}=\frac zx\\\frac dc=\frac yx.$$
So you get that $$\begin{pmatrix}1\\z/x \end{pmatrix}\otimes \begin{pmatrix}1\\y/x\end{pmatrix}= \begin{pmatrix}1\\y/x\\z/x\\yz/x^2 \end{pmatrix}= \begin{pmatrix}1\\y/x\\z/x\\w/x \end{pmatrix}= \frac1x\begin{pmatrix}x\\y\\z\\w \end{pmatrix},$$ With one step because $yz=xw.$
So we can choose $a=1,b=z/x,c=x,d=y.$
More generally, if $uv=x,$ we can choose:
$$a=u,b=uz/x=z/v,c=v,d=vy/x=y/u.$$
If $x=0,$ you can do the same with any other non-zero coordinate.
If they are all zero, you can set $a=b=0$ and $c,d$ to anything, or $c=d=0$ and $a,b$ anything.