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I am attempting to write out three invariants of a general (not necessarily symmetric or Cartesian), 2nd order tensor which we will call $\mathbf{T} = T^i_{\;j}\mathbf{e}_i\mathbf{e}^j \,, \;\; i,j=1,2,3\,$ in this case. In most of the literature I have searched through, invariants are listed as \begin{align} \text{I}_T &= \text{tr}(\mathbf{T}), \\ \text{II}_T &= \frac{1}{2}\left(\text{tr}(\mathbf{T})^2 - \text{tr}(\mathbf{T}^2) \right), \\ \text{III}_T &= \frac{1}{6}\left(\text{tr}(\mathbf{T})^3 - 3\text{tr}(\mathbf{T})\text{tr}(\mathbf{T}^2) + 2\text{tr}(\mathbf{T}^3)\right) = \text{det}(\mathbf{T}). \end{align} However, due to most textbooks in engineering being written for Cartesian analysis, I have a hunch that these expressions for the invariants are strictly for symmetric, Cartesian tensors.

My ultimate goal is to have expressions for the invariants of a general 2nd order tensor in terms of the mixed components like such: \begin{align} \text{I}_T &= T_{\;i}^i, \\ \text{II}_T &= \frac{1}{2}\left(T_{\;i}^i T_{\;j}^j - T^i_{\;j} T_{\;i}^j \right), \\ \text{III}_T &= \frac{1}{6}\left(T_{\;i}^i T_{\;j}^j T_{\;k}^k - 3T_{\;i}^i T_{\;k}^j T_{\;j}^k + 2T_{\;j}^i T_{\;k}^j T_{\;i}^k \right), \end{align} but I am unsure if this is valid for the most general case.

Any info or directions to helpful sources would be appreciated.

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  • $\begingroup$ Aren't these just the coefficients of the characteristic polynomial of $T$ (when considered as a matrix)? $\endgroup$ – Lord Shark the Unknown May 23 '18 at 19:58

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