# Invariants of a tensor [closed]

I asked (almost) the same question in the math exchange.

I'm teaching a course, and I need a simple and intuitive proof that the invariants of a matrix ($3\times3$, but it doesn't matter) can be expressed as linear combinations of traces of its power. When I say "invariance" I mean under orthonormal transformation of the axes $A\to Q A Q^{-1}$ for orthonormal $Q$.

For example, that only linear invariant scalar is the trace, that every quadratic invariant scalar is a combination of $\operatorname{tr}(A)^2$ and $\operatorname{tr}(A^2)$, and that every cubic invariant is a combination of $\operatorname{tr}(A)^3$, $\operatorname{tr}(A^2)\operatorname{tr}(A)$, and $\operatorname{tr}(A^3)$.

The proof for the linear case is trivial (and intuitive) but I can't find a generalization for the quadratic case: say that f(A) is a scalar invariant that is linear in the entries of A. That means $$f=\sum_{ij}C_{ij}A_{ij}$$ where $C$ is some matrix. $C$ should be unchanged when applying an infinitesimal rotation to $A$. This means (here there're two lines of algebra) that $C$ commutes with the generators of $SO(n)$. The only $C$ that does that is the identity. QED

-
 Hi yohBS, and welcome to Physics Stack Exchange! We strongly discourage cross-posting questions unless you have had the question at one site for some time without getting any response. In any case, this is really a mathematical question and accordingly seems to be off topic here. – David Zaslavsky♦ Dec 15 '11 at 16:33 Agreed with David. This really contains no physics, which would be the chief reason for not putting it on Physics.SE in my book. – Mark S. Everitt Dec 15 '11 at 16:38

## closed as off topic by David Zaslavsky♦Dec 15 '11 at 16:33

Questions on Physics Stack Exchange are expected to relate to physics within the scope defined in the FAQ. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about closed questions here.