Simple proof that if $A^n=I$ then $\mathrm{tr}(A^{-1})=\overline{\mathrm{tr}(A)}$ Let $A$ be a linear map from a finite dimensional complex vector space to itself. If $A$ has finite order then the trace of its inverse is the conjugate of its trace.
I know two proofs of this fact, but they both require linear algebra facts whose proofs are themselves quite involved.


*

*Since $A^n=I$, the eigenvalues of $A$ are roots of unity. Hence they have unit norm, and so their reciprocals are their conjugates. Then the result follows from following facts: (a) The eigenvalues of $A^{-1}$ are the reciprocals of the eigenvalues of $A$, (b) the dimensions of the eigenspaces of $A^{-1}$ are equal to the dimensions of the corresponding eigenspaces of $A$, (c) the trace is equal to the sum of the (generalised) eigenvalues. The proof of (a) is relatively easy, but (b) and (c) seem to require the existence of Jordan Normal Form, which requires a lot of work.

*By Weyl's Unitary Trick, there's a inner product for which $A$ is unitary (this proof is itself a fair amount of work). So in an orthonormal basis (which we must construct with the Gram-Schmidt procedure) the inverse of $A$ is given by its conjugate transpose (one must also prove this). So the trace of the inverse is the conjugate of the trace.
Since  the condition $A^n=I$ and the consequence $\mathrm{tr}(A^{-1})=\overline{\mathrm{tr}(A)}$ are both elementary statements, I'm wondering if there's a short proof from first principles (ideally without quoting any big linear algebra Theorems). Can anyone find one?

If the trouble is just a simple proof for the fact that
$$
\text{tr}\,A=\sum_{i=1}^n\lambda_i
$$
you can try the following approach instead of JNF.


*

*In the field $\Bbb C$ we can factorize $\det(\lambda I-A)=\prod_{i=1}^n(\lambda-\lambda_i)$ where $\lambda_i$ are all  eigenvalues (possibly repeated with multiplicities). The coefficient for $\lambda^{n-1}$ is $\color{red}{-\sum_{i=1}^n\lambda_i}$. 

*Prove (e.g. expanding the determinant along the first column + induction) that the coefficients for $\lambda^n$ and $\lambda^{n-1}$ are build from the main diagonal product only
\begin{align}
\det(\lambda I-A)&=\begin{vmatrix}\lambda-a_{11} & -a_{12} & \ldots & -a_{1n}\\-a_{21} & \lambda-a_{22} & \ldots & -a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
-a_{n1} & -a_{n2} & \ldots & \lambda-a_{nn}
\end{vmatrix}=\prod_{i=1}^n(\lambda-a_{ii})+<\text{terms of $\deg\le n-2$}>=\\
&=\lambda^n\color{red}{-\text{tr}\,A}\,\lambda^{n-1}+<\text{terms of $\deg\le n-2$}>.
\end{align}
It is because for any cofactor $A_{j1}$, $j>1$,  in the first column you remove two $\lambda-a_{ii}$ elements: one from the first column and one from the $j$th row.

*Compare the red coefficients to conclude.

