If $A \in M_3(\mathbb C)$ and $\operatorname{tr}(A)=\operatorname{tr}(A^{-1})=0$ and $\det(A)=1$, then $A^3=I_3$

Question

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 !

Answer 1

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:

$$A^3 - I_3 = 0 \Rightarrow A^3 = I_3.$$

Answer 2

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 $$ A^3=P\pmatrix{\lambda_1^3 \\ & \lambda_2^3 \\ && \lambda_3^3}P^{-1}=PI_3P^{-1}=I_3 $$

Answer 3

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.

Answer 4

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.