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: enter image description here

(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?

  • $\begingroup$ I'm going to delete my answer.. Ignore it because I read your initial question wrong $\endgroup$
    – John M
    Commented Jan 19, 2015 at 22:34
  • $\begingroup$ @John M No problem! Perhaps you've deleted it already, but I don't think I see your answer. $\endgroup$
    – Guest
    Commented Jan 19, 2015 at 22:44

2 Answers 2


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.

  • $\begingroup$ I see! Just to check that I understand, so a matrix always has these parts; meaning, these parts added together define a matrix? There are no more parts to it? Then each part can be analyzed separately under certain symmetries, as you say? Thank you! $\endgroup$
    – Guest
    Commented Jan 20, 2015 at 2:06
  • $\begingroup$ That's basically it $\endgroup$
    – Phoenix87
    Commented Jan 20, 2015 at 23:32
  • $\begingroup$ How to find the decomposition for other symmetries? e.g. not SO(3) or the more general SO(N) like here but, say, SO(3,1). $\endgroup$
    – Quillo
    Commented Jan 17, 2023 at 10:28

They can be broken up in many ways, it's like choosing a base for a vectorial system, however in physics the symmetries play an important role, that's why we use this system to reduce tensors, because it contains physical information. The matrix can be broken into these three parts and in other ways too, this is just the most convenient.

And yes, it is a factor $\frac{2}{3}$ in the second $\delta$.

  • $\begingroup$ If possible, can you explain a little bit more how this relates to bases? Such as, in this case there is a trace, a symmetric part, and an anti-symmetric part, but why is there no inverse or transpose of the matrix, for example? Perhaps the symmetric and anti-symmetric parts are there because of anti-commutating and commutating relation of the matrices (such as what was talked about here: physics.stackexchange.com/q/146565), but why the trace? Sorry, I'm still a bit confused. Also, thank you for commenting about the 2/3! $\endgroup$
    – Guest
    Commented Jan 19, 2015 at 22:43
  • $\begingroup$ 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). $\endgroup$
    – manuel91
    Commented Jan 19, 2015 at 22:49
  • $\begingroup$ Symmetric and anti-symmetric parts are there because they are important in physics, they are related to commutation or to fluid vortexes, etc. The trace is there because it accounts for scalar quantities, a good example of it is the inertia moment, which is the trace of the inertia tensor. In quantum mechanics the diagonal component is related to the eigenstates associated to the matrix operator for example. $\endgroup$
    – manuel91
    Commented Jan 19, 2015 at 22:51

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