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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.

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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.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)

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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.

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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.

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Can anyone link me a tutorial on tensors?

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