What is the tensor product of a vector with itself? And another tensor question Thank you for the good comments, I will show you what I am trying to work with, and perhaps somebody can tell me how to work with it:
$$T=I+\frac{a\otimes b + b\otimes a - a\otimes a - b\otimes b}{(1-a\cdot b)}\tag{1}$$
Where a and b are unit vectors and I is the identity tensor and T is a tensor. Can I do anything to simplify this? Eventually I need to solve Ta but I don't want the answer, I want to know what to do with this type of thing. Thank y'all so much.
 A: It sounds kind of like you are working in the tensor algebra $T(V)$ of a vector space $V$. The way to think of $T(V)$ is that it is the "freest" associative algebra "generated" by $V$. The quotes are in there because they need a lot more explaining, but they can be accepted at face value for now.
Since $V$ sits inside $T(V)$, and $T(V)$ has a product $\otimes$, $v\otimes v$ is the product of $v$ with itself in this algebra. Actually, you can multiply $v\otimes w$ for whatever vectors you want, and you just wind up with another element of $T(V)$.
The trick about the tensor algebra is to not get sucked into believing everything in $T(V)$ looks like $v\otimes w$ (or, for that matter, finite products of more than two vectors). For instance, $(v\otimes v) +w$ is definitely not of that form for nonzero $w$ in $V$, and even $(v\otimes w)+(a\otimes b)$ may be inexpressible that way. Furthermore, you have to deal with things like $v\otimes w\otimes a$ and products with even more elements! General elements of $T(V)$ look like linear combinations of finite tensor products of things in $V$.

Update:
No matter if you are working in $V\otimes V$ or $T(V)$, you can reduce the expression you gave slightly:
$$\frac{a\otimes b + b\otimes a - a\otimes a - b\otimes b}{(1-a\cdot b)}$$ 
At least $a\otimes b-a\otimes a=a\otimes(b-a)$, and likewise $b\otimes a-b\otimes b=b\otimes(a-b)$. This is just the bilinearity axiom of tensor products.
Then, since $a\otimes(b-a)=-a\otimes(a-b)$, we have this denominator:
$-a\otimes(a-b)+b\otimes(a-b)=(b-a)\otimes(a-b)$
