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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.

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There are many possible answers to this question. Can you give some background? – k.stm Jan 26 '13 at 17:29
Thanks K. Stm., I didn't realize what I was asking was so ambiguous. Hope this helps. – Marcus Jan 26 '13 at 17:59
Over which field are you working? Which space are $a$ and $b$ from and what dimension does it have? – k.stm Jan 26 '13 at 18:02
Now those questions I do not have the answers to, as I was only given as much information as I put down. Is there any way of working with this problem without knowing that? – Marcus Jan 26 '13 at 18:05
Your question, and your comment to @rschwieb’s answer, do not yet make it clear what space you are thinking of. Is it $V\otimes V$? Or is it the direct sum of all spaces $V^{\otimes\,n}$, over all nonnegative $n$? I have to confess that I still don’t know what you mean by “the identity tensor”, which I would have thought was in $V\otimes(V^*)$. – Lubin Jan 26 '13 at 19:53

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:


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Thanks especially rschwieb for the interesting answer, I'm not sure I understand it completely but I do understand how my question didn't have a specific solution. Hopefully my edit makes things clearer. – Marcus Jan 26 '13 at 18:00
@Marcus Can you tell us yet if you are working in the tensor algebra $T(V)$, or if you are just working in $V\otimes V$? Or might you be working in a Clifford algebra or symmetric algebra?? I'm adding a bit to address your edits. – rschwieb Jan 27 '13 at 1:24
That makes the numerator $(a-b)\otimes(a-b)$, right? – Rahul Jan 27 '13 at 1:32
@RahulNarain Ah yes, you just factor out a -1 to obtain that huh... I had been thinking you needed antisymmetry to do that but that was obviously a mirage. Thanks for pointing it out. – rschwieb Jan 27 '13 at 1:35
@RahulNarain Although I think it's $(b-a)\otimes(a-b)$? – rschwieb Jan 27 '13 at 1:44

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