I am currently working on a subject, in which the dyadic product of two and four vectors result the second rank and fourth rank tensors as follows:
$$ M = a a$$ and $$ \mathbb{M} = a a a a$$ If $a$ is a 3D vector, the above tensors can be stored in a $3\times3$ and $9\times9$ matrix respectively.
How the dyadic product of four vectors is calculated? I could not find the mathematics behind that in the literature. Which kind of product it is?
I found in an article, in a relevant topic, if someone does not want to use the second formula, he can use a model called the quadratic closure, in which the fourth rank tensor equals the dyadic product of second rank tensor $M$ by itself (something like $(aa)(aa)$ ?) and it is exact only if vector $a$ has some specific properties. How can we analyze the error of the model? I mean, how can we elaborate on the error between $(aa)(aa)$ and $aaaa$?
Assume $a$ is an approximated vector and there is some error in that, i.e. $a-a_{exact}=\mathbf h$ and therefore $M$ has an error in itself. What is the difference between the error in $M$ and $\mathbb{M}$? How one can elaborate on that?
Thank you for your help.
It is important to note that for a vector $\mathrm v$ the tensor product $$\mathrm v\otimes \mathrm v$$ Is not a matrix, it is in fact a second order contravariant tensor. As to the product $$\mathrm v \otimes \mathrm v\otimes \mathrm v\otimes \mathrm v$$ The tensor product is associative, so it doesn't matter where you put the parenthesis (this answers your second question). The result will be a fourth order contravariant tensor. To answer your first question (how is it calculated) if $\mathbf{S}, \mathbf T$ are $(p,q)$ and $(r,s)$ tensors respectively their tensor product is a $(p+r,q+s)$ tensor with components that follow a straightforward formula:
$$\left( \mathbf{S\otimes T}\right)^{i_1\dots i_{p+r}}{}_{j_1\dots j_{q+s}}=S^{i_1\dots i_p}{}_{j_1\dots j_q}~T^{i_{p+1}\dots i_{p+r}}{}_{j_{q+1}\dots j_{q+s}}$$
As to your third question, what happens if $\mathrm v$ has some error, i.e if $\mathrm v=\mathrm{u+h}$. Well, since the tensor product is distributive you can just expand:
$$\mathrm{v\otimes v}=(\mathrm{u+h})\otimes(\mathrm{u+h})=\mathrm{u\otimes u}+2\mathrm{u\otimes h}+\mathrm{h\otimes h}$$ So it is just $\mathrm{u\otimes u}$ "plus a bit", where the magnitude of that bit will grow with respect to $\mathrm h$ at a $O(\Vert \mathrm h\Vert^2)$ rate, where I am using the Frobenius norm for a general tensor
$$\Vert\mathbf T\Vert^2=\sum_{ \begin{array}{l} i_{1} ,... ,i_{p}\\ j_{1} ,... ,j_{q} \end{array}} {\left|T^{i_1\dots i_p}{}_{j_1\dots j_q}\right|}^2$$
Similarly for the fourth order product the error grows at a $O(\Vert \mathrm h\Vert^4)$ rate.