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The characteristic equation of matrix A is $$\lambda ^3 - I_1\lambda^2 + I_2\lambda-I_3 = 0 $$
For orthogonal matrix $$I_3 = det(A) = \pm1$$ $$I_1 = tr(A)$$
Taking examples of orthogonal matrices, it looks like $I_1 = I_2$. Is this true always for an orthogonal matrix? Is there some proof?
This isn't true. E.g. when $A$ is the negative of the identity matrix, its characteristic polynomial is $(x+1)^3=x^3+3x^2+3x+1$.
In general, the characteristic polynomial of a $3\times3$ nonsingular matrix $A$ is in the form of \begin{aligned} &x^3-\sum_i\lambda_i(A)x^2+\sum_{i\ne j}\lambda_i(A)\lambda_j(A)x-\prod_i\lambda_i(A)\\ =\,&x^3-\operatorname{tr}(A)x^2+\det(A)\operatorname{tr}(A^{-1})x-\det(A). \end{aligned} You are essentially asking whether $$ \operatorname{tr}(A)=\det(A)\operatorname{tr}(A^{-1}) $$ when $A$ is orthogonal. Since $A^{-1}=A^T$ in this case, the above equality is true if and only if $\operatorname{tr}(A)=0$ or $\det(A)=1$.
