- #1

- 383

- 5

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Superposed_Cat
- Start date

- #1

- 383

- 5

- #2

- 18,517

- 8,418

However, all tensors are not rank 2 and those that are not cannot be represented as a matrix (you would have to use a matrix with more than 2 dimensions). Also, not all matrices are tensors. There are non-square matrices, matrices not transforming in the proper way (a matrix is a priori only a rectangular array of numbers) to represent a tensor, etc. For many applications, you will only encounter tensors of rank 2 or lower and then representation with matrices is very convenient.

- #3

FactChecker

Science Advisor

Gold Member

- 6,973

- 2,895

- #4

- 18,517

- 8,418

Example: The identity matrix is a diagonal matrix of 1's. If the coordinate system is in feet or inches, the diagonals are still 1's. So the identity matrix is a math concept that does not transform correctly (from coordinates of feet to coordinates of inches) to represent a physical entity. For the same matrix to represent a tensor, it would have to be defined in a way that its diagonal 1's in the coordinates of feet would transform to either 12's or 1/12's diagonal elements in coordinates of inches (there are covarient and contravarient tensors)

While I agree that transformation properties of tensors are important, I think the unit matrix is not a very illuminating (and somewhat misleading) example. In particular, consider the (1,1)-tensor ##\delta^\alpha_\beta## such that ##\delta^\alpha_\beta V^\beta = V^\alpha##, where ##V## is a vector. This tensor will be represented by the unit matrix in all frames (the unit matrix is a transformation from the vector space of column matrices to itself and therefore naturally represents a (1,1)-tensor, you can fiddle around to make a square matrix represent an arbitrary rank-2 tensor, but I would say it is slightly less natural). The tensor transformation properties follow trivially from the chain rule.

- #5

FactChecker

Science Advisor

Gold Member

- 6,973

- 2,895

Ok. I retract my statement and will stay out of this discussion.While I agree that transformation properties of tensors are important, I think the unit matrix is not a very illuminating (and somewhat misleading) example. In particular, consider the (1,1)-tensor ##\delta^\alpha_\beta## such that ##\delta^\alpha_\beta V^\beta = V^\alpha##, where ##V## is a vector. This tensor will be represented by the unit matrix in all frames (the unit matrix is a transformation from the vector space of column matrices to itself and therefore naturally represents a (1,1)-tensor, you can fiddle around to make a square matrix represent an arbitrary rank-2 tensor, but I would say it is slightly less natural). The tensor transformation properties follow trivially from the chain rule.

- #6

- 83

- 10

When ##V## is finite dimensional, ##V^{**}=V##, and a rank ##(p,q)## tensor is in ##V^{*p}\otimes V^{q}##. A linear transformation from ##V## to itself can be represented by an element ##\omega \in V\otimes V^*##. If we pick bases ##\epsilon_j## for ##V## and ##\epsilon_k^*## for ##V^*## (with ##\epsilon_k^*(\epsilon_j)=\delta_{kj}##), then we can expand ##\omega## as ##\omega = \sum_{j,k} \omega_{jk}\epsilon_j\otimes \epsilon_k^*##, and the components ##\omega_{jk}## can be interpreted as elements of a matrix.

Matrices have a different kind of structure from tensors. While matrices can be used to represent tensors in a wide range of settings, matrix multiplication (say between square matrices) is only meaningful in the tensor context when the tensors have the form ##V\otimes V^{*}## for some vector space ##V##, or when there is a linear map between ##V## and ##V^*## (i.e. an inner product or metric) and ##p+q## is even (or if you consider multiplication between special families of tensors). Tensors have more structure than matrices, but questions about matrices have a very different flavor from questions about tensors.

- #7

- 383

- 5

Can anyone link me a tutorial on tensors?

- #8

- 83

- 10

Share: