# 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

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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

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