# Why can tensors be broken up into parts?

I have found these notes: http://www.physics.usu.edu/Wheeler/QuantumMechanics/QMWignerEckartTheorem.pdf

Which state on page two that a matrix (M) can be broken up into rotationally independent pieces like so:

(ps: I believe the last term should have 2/3 and not 1/3 in front of the delta_ij, but please tell me if I am wrong).

My question is: why is it that a matrix can be broken up into these three parts? Is it always these three parts (trace, symmetric, and anti-symmetric parts)? I am new to the concept of irreducible tensors and I think this relates to them.

I have tried reading previous threads about this on here: Irreducible tensors concept and here: Irreducible decomposition of higher order tensors however, none answer the question of why it is even possible to break up a matrix like so. Where does this concept come from?

• I'm going to delete my answer.. Ignore it because I read your initial question wrong – John M Jan 19 '15 at 22:34
• @John M No problem! Perhaps you've deleted it already, but I don't think I see your answer. – Guest Jan 19 '15 at 22:44

An order-2 tensor is usually decomposed in that way because each part behaves differently under a certain symmetry. For example, if the symmetry is just rotation, then the term with the trace transforms like a scalar; the anti-symmetric part $M_{ij}-M_{ji}$ of the tensor transforms like a pseudo-vector, while the traceless symmetric part (the last term) transforms like an ordinary 2-tensor.
And yes, it is a factor $\frac{2}{3}$ in the second $\delta$.
• I was just making an analogy in the sense that the choice of a basis can be done in many ways. You have the transpose of the matrix there ($M_{ij}^T = M_{ji}$), the inverse is not there because there's no trivial way to put its value in terms of indexes ($|M|$ has a lot of terms when decomposed as a function of matrix elements). – manuel91 Jan 19 '15 at 22:49