If $A \in M_3(\mathbb C)$ and $\operatorname{tr}(A)=\operatorname{tr}(A^{-1})=0$ and $\det(A)=1$, then $A^3=I_3$ Assume that we have a $3\times 3$ matrix like $A$ in which the coordinates come from the set of complex numbers ($\mathbb C$).
We know that $\operatorname{tr}(A)=\operatorname{tr}(A^{-1})=0$ and $\det(A)=1$ .
Prove that $A^3=I_3$ .  
I don't know what's the relationship between $A^3$ and the trace !
 A: Let $\lambda_i$ be eigenvalues of $A$. Given condition means that
$$
\lambda_1+\lambda_2+\lambda_3=0\quad\text{and}\quad\lambda_1^{-1}+\lambda_2^{-1}+\lambda_3^{-1}=0\quad\text{and}\quad\lambda_1\lambda_2\lambda_3=1
$$
Solve these and we have $\lambda_k=e^{k2\pi/3i}$. Since $A$ has distinct eigenvalues, it is similar to diagonal matrix, i.e
$$
P^{-1}AP=\pmatrix{\lambda_1 \\ & \lambda_2 \\ && \lambda_3}
$$
Thus
$$
A^3=P\pmatrix{\lambda_1^3 \\ & \lambda_2^3 \\ && \lambda_3^3}P^{-1}=PI_3P^{-1}=I_3
$$
A: Expanding on the Hint of Ilisaryus: Use that the trace is the sum of the eigenvalues and the determinant the product and that the eigenvalues of $A^{-1}$ are the inverses of the eigenvalues of $A$. Find the characteristic polynomial and use Cayley-Hamilton.
A: Solving the set of equations:
$$\begin{cases}
\lambda_1+\lambda_2+\lambda_3=0\ \\
\lambda_1^{-1}+\lambda_2^{-1}+\lambda_3^{-1}=0 \\
\lambda_1\lambda_2\lambda_3=1
\end{cases}
$$
you get $6$ solutions, but we can work just on one of these WLOG, that is:
$$\lambda_1 = 1, \lambda_2 = e^{-i\frac{2\pi}{3}} ~ \text{and} ~ \lambda_3 = e^{i\frac{2\pi}{3}}.$$
Indeed, each of the $6$ solution is a permutation of the previous.
Using Cayley-Hamilton theorem, we know that:
$$A^3 - tr(A)A^2 + \frac{1}{2}\left((tr(A))^2 - tr(A^2)\right)A - \det(A)I_3 = 0.$$
By substitution, it is obvious that:
$$A^3 - \frac{1}{2}tr(A^2)A - I_3 = 0.$$
It is clear that $tr(A^2) = \lambda_1^2+\lambda_2^2+\lambda_3^2.$
Then:
$$tr(A^2) = 1^2 + \left(e^{-i\frac{2\pi}{3}}\right)^2 + \left(e^{i\frac{2\pi}{3}}\right)^2 = \\
= 1 + e^{-i\frac{4\pi}{3}} + e^{i\frac{4\pi}{3}} = 1 + 2\cos\left(\frac{4\pi}{3}\right) = \\ = 1 + 2\left(-\frac{1}{2}\right) =
1 - 1 = 0.$$
Finally, you can state that:
$$A^3 - I_3 = 0 \Rightarrow A^3 = I_3.$$
A: This is almost trivial. Suppose $\lambda_1,\lambda_2,\lambda_3$ are the eigenvalues of $A$. The characteristic polynomial of $A^{-1}$ is then
\begin{align*}
&\phantom{=}\left(y-\frac1{\lambda_1}\right) \left(y-\frac1{\lambda_2}\right) \left(y-\frac1{\lambda_3}\right)\\
&= y^3-\left(\frac1{\lambda_1}+\frac1{\lambda_2}+\frac1{\lambda_3}\right)y^2+\left(\frac{\lambda_1+\lambda_2+\lambda_3}{\lambda_1\lambda_2\lambda_3}\right)y-\frac1{\lambda_1\lambda_2\lambda_3}.
\end{align*}
Using the given trace and determinant conditions, you may obtain the values of $\frac1{\lambda_1}+\frac1{\lambda_2}+\frac1{\lambda_3},\ \ \lambda_1+\lambda_2+\lambda_3$ and $\lambda_1\lambda_2\lambda_3$ directly without calculations. Then you may infer that $A^{-3}=I$ and the conclusion follows.
