Question

If V is a finite dimension vector space, x, y are in V, is it true that \(x\otimes y\) = \(y\otimes x\)
Please, help

Answer 1

Do I have to separate it into cases:
1/ x, y are in the same basis, then x \(\otimes ~~y\) = \( y ~~\otimes~~x\)
2/ if they are in different basis, then x \(\otimes~~y\neq x\otimes y\)

Answer 2

If I remember correctly, cross products are anticommunitive, that is
x⊗y = -y⊗x
It's been a while since I did vector spaces though.

Answer 3

but I need proof for whatever I say, Pleeeeeeeeeease

Answer 4

because orientation matters in vector spaces, the only case where x⊗y = y⊗x is true, would be the case where x,y are null vectors (x=0,y=0). For proof you can do an example, pick any two not null vectors and do both cross products, they will have different answers.

Answer 5

how do u define \(\otimes\) ?

Answer 6

I did, but I confused. Let x, y are in R^2. {(1,0) , (0,1)} is the basis of R^2
let x = (1,2) and y = (2,0 ) in R^2 ,
so, x = ( e1 + 2e2) , y = (2e1)
x \(\otimes y\) = (e1 +2e2) \(\otimes~~ 2e1\)

Answer 7

=(e1\(\otimes 2e1)+ (2e2 ~~\otimes ~2e1)\)

Answer 8

\(y~\otimes x = 2e1\otimes (e1+2e2)=(2e1\otimes e1 )+ (2e1\otimes 2e2)\)
are they the same? I see they are the same but they are supposedly not.

Answer 9

@helder_edwin it's notation for cross product.
\[\large\vec{x}\otimes \vec{y} =\left[ \begin{matrix}
\textbf{i}& \textbf{j} & \textbf{k} \\
x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3\\
\end{matrix} \right]
= \left[\begin{matrix}
x_2& x_3\\
y_2&y_3\\
\end{matrix} \right]\textbf{i}-\left[\begin{matrix}
x_1& x_3\\
y_1&y_3\\
\end{matrix} \right]\textbf{j}+\left[\begin{matrix}
x_1& x_2\\
y_1&y_2\\
\end{matrix} \right]\textbf{i}\]

Answer 10

OK. But this operation is only defined in R^3. Not in any finite dimensional space vector.

Answer 11

And u were right. It is not commutative. This can be seen from the definition u just posted.

(The last i should be k, though.)

Answer 12

indeed it should, as it happens that's a lot of latex to type out, lol. Also, the question didn't mention what degree space we were in, so i assumed R^3.

Answer 13

I agree: a lot to type.

Answer 14

Question: tensor product = cross product?

Answer 15

NO !!

Answer 16

why? my lecture uses \(\otimes \) to indicate tensor product. I am sorry for that. I mean the tensor product

Answer 17

that is the correct notation for tensor product, I assumed you meant cross product for a couple reasons, mostly relating to the fact that tensor products don't show up on here nearly as often as cross products :X sorry about that.

Answer 18

if it is tensor product, then?? the conclusion??

Answer 19

yes. that was strange. that's why I asked for the definition. The symbol \(\otimes\) is (as far as I know) solely used to denote the tensor product. While the cross product is denoted by \(\times\) and is only defined in R^3.

Answer 20

when it comes to modules, \(\otimes\) is not exactly an operation so it is not commutative. this comes from the definition of the tensor product of modules. (U can check any book in homological algebra.)

but when it comes to vector spaces, i don't really know. but I guess it is not true. again from the definition (construction) of a tensor product.

Answer 21

Thank you

Answer 22

u r welcome