Can tensors be equal in all coordinate systems?

[qoqosz]

Homework Statement

Task 1. Show that if components of any tensor of any rank vanish in one particular coordinate system they vanish in all coordinate systems.

Task 2. The components of tensor T are equal to the corresponding components of tensor W in one particular coordinate system; that is, T^0_{ij} = W^0_{ij}.
Show that tensor T is equal to tensor W, T_{ij} = W_{ij} in all coordinate systems.

Homework Equations

The Attempt at a Solution

task 1. I have no idea how to start
task 2. transforming to the any other coordinate system I obtain:

T_{i'j'} = A_{i'}^i A_{j'}^j T^0_{ij} = A_{i'}^i A_{j'}^j W^0_{ij} = W_{i'j'} is it ok?

 

[HallsofIvy]

If you can do 2 you can do 1 by taking W_{ij}= 0]!

But it is easiest to prove 2 by proving 1 first:
You apparently know that if T^0_{ij} are the components of T in one coordinate system, then the coordinates T_{i'j'} in any coordinates system are given by T_{i'j'}= A^i_{i'}A^j_{j'}T^0_{ij}. Okay what if all components of T^0_{ij} are 0?

And if T^0_{ij}= W^0_{ij}[/tex], then T^0_{ij}- W^0_{ij} is also a tensor, with all components 0.

 

[qoqosz]

Well the fact from the 1st task is for me intuitive but I don't understand it's formal proof. Because when some components of T_{ij} are 0 (but not all) than the T_{i'j'} can have no 0 components. So why when all components of T_{ij} are 0 then T_{i'j'} are 0 as well? Is it because changing coordinate system is a linear transformation and so 'A(0)=0'?

 

[qoqosz]

Am I right?

 

[Urmi Roy]

stress is an example of a tensor...If we have a bar (cuboid with a length much greater than other dimensions), and we apply axial forces to it, then there are only normal stresses...however, if we now cut off a piece from it diagonally, then on the inclined surface so obtained, we have both normal and shear stresses...so the same thing appears different in two different perspectives...is this also a basic property of tensors??

 

[Urmi Roy]

Please help!