Question

Is V \(\otimes\)W = W\(\otimes \)V?
I think yes. How is your opinion?

Answer 1

@wio

Answer 2

what operation is this?

Answer 3

tensor product

Answer 4

oh, outer product...

Answer 5

is this the other name of tensor product?

Answer 6

What happens when you do \(\langle a, b \rangle \otimes \langle c,d\rangle \) and \(\langle c, d \rangle \otimes \langle a,b\rangle \)?

Answer 7

the same solution for both

Answer 8

What about \[
\langle a\rangle \otimes \langle b,c\rangle
\]versus \[
\langle b,c\rangle \otimes \langle a\rangle
\]

Answer 9

the same

Answer 10

Are you sure?

Answer 11

yes for the last one,

Answer 12

I'm getting \[
\langle a\rangle \otimes \langle b,c\rangle = (\langle b,c\rangle \otimes \langle a\rangle)^{T}
\]

Answer 13

\[\langle a\rangle \otimes \langle b,c\rangle = \langle a \otimes b \rangle, \langle a\otimes c \rangle\]

Answer 14

\[
\langle a, b \rangle \otimes \langle c,d\rangle = \begin{bmatrix}
ac&ad \\
bc&bc
\end{bmatrix}
\]
\[
\langle c, d \rangle \otimes \langle a,b\rangle = \begin{bmatrix}
ac&bc \\
ad&bd
\end{bmatrix}
\]

Answer 15

I got you, I don't study this stuff yet. I know matrix way but right now, I am not allowed using it. I have to go from definition. :(

Answer 16

Oh, like the order matter, <a,b> \(\neq <b,a>\)

Answer 17

Thank you very much, I got it and know how to apply.

Answer 18

Is the the same as direct sum - http://en.wikipedia.org/wiki/Direct_sum ?

Answer 19

not sure, just go from the definition : let x =(-1,1) in R^2 and y = (1,0) in R^2
\[x \otimes y = (-1,1)+(-1,0)+(1,1)+(1,0)= (0,2)\\y\otimes x = (1,-1) +(1,1)+(0,-1)+(0,1)=(2,0)\]
so, they are not equal