Is Qij=AiBj a Tensor of Rank 2?

flintbox
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Homework Statement


Suppose A and B are vectors. Show that the object Q with nine components Qij=AiBj is a tensor of rank 2.

Homework Equations


A tensor transforms under rotations (R) as a vector:
Tij'=RinRjmTnm

The Attempt at a Solution


I wanted to just create the matrix, but I don't know how to prove that this is also a tensor.
 
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flintbox said:

Homework Statement


Suppose A and B are vectors. Show that the object Q with nine components Qij=AiBj is a tensor of rank 2.

Homework Equations


A tensor transforms under rotations (R) as a vector:
Tij'=RinRjmTnm

The Attempt at a Solution


I wanted to just create the matrix, but I don't know how to prove that this is also a tensor.
Well, how do the vectors ##A## and ##B## transform under rotations?
 
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I think I understand it now:
$$A_i' B_j' = R_{in}A_n R_{jm}A_m$$
$$A_i' B_j' = R_{in} R_{jm} (A_n A_m)$$
$$A_i' B_j' = R_{in} R_{jm} Q_{nm}$$
$$A_i' B_j' = Q'_{nm}$$
So we for proving something is a tensor, we just apply some transformations to it, right?
 
flintbox said:
Thanks a lot!
I think I understand it now:
$$A_i' B_j' = R_{in}A_n R_{jm}A_m$$
$$A_i' B_j' = R_{in} R_{jm} (A_n A_m)$$
$$A_i' B_j' = R_{in} R_{jm} Q_{nm}$$
$$A_i' B_j' = Q'_{nm}$$
So we for proving something is a tensor, we just apply some transformations to it, right?

Yes, something is a tensor if it transforms like a tensor.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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