Identity matrix

Let $A \in \mathcal{M}_3 \left( \mathbb{R} \right)$ such that $\det A =1$ and ${\rm tr} (A)= {\rm tr} (A^{-1})=0$. Prove that $A^3=\mathbb{I}_{3 \times 3}$.

Solution

Analyzing the data and recalling some formulae we get that, if $\lambda_1,\; \lambda_2, \; \lambda_3 $ are the eigenvalues of the matrix,

• $\lambda_1 \lambda_2 \lambda_3 =1$
• $\displaystyle  \lambda_1 +\lambda_2 + \lambda_3 = \frac{1}{\lambda _1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3} =0$

Applying Cayley - Hamilton we get the result.