If $A$ be an invertible matrix $M_3(\mathbb{R})$ and we have $\operatorname{tr}(A)=\operatorname{tr}(A^2)=0$ and $\det(A)=1$, then what is the characteristic polynomial of $A$?


If $\lambda_1,\lambda_2,\lambda_3$ are the eigenvalues of $A$, then the characteristic polynomial is:

\begin{align*}\det(xI-A) &= (x-\lambda_1)(x-\lambda_2)(x-\lambda_3) \\ &= x^3 - (\lambda_1+\lambda_2+\lambda_3)x^2+(\lambda_1\lambda_2+\lambda_2\lambda_3+\lambda_3\lambda_1)x-\lambda_1\lambda_2\lambda_3\end{align*}

The determinant of a matrix is the product of its eigenvalues. Hence $\lambda_1\lambda_2\lambda_3 = \det A = 1$.

The trace of a matrix is the sum of its eigenvalues. Hence, $\lambda_1+\lambda_2+\lambda_3 = \text{tr}(A) = 0$.

Also, the eigenvalues of $A^2$ are $\lambda_1^2,\lambda_2^2,\lambda_3^2$. Hence, $\lambda_1^2+\lambda_2^2+\lambda_3^2 = \text{tr}(A^2) = 0$.

Now, use the fact that $(\lambda_1+\lambda_2+\lambda_3)^2 = \lambda_1^2+\lambda_2^2+\lambda_3^2+2(\lambda_1\lambda_2+\lambda_2\lambda_3+\lambda_3\lambda_1)$ to get $\lambda_1\lambda_2+\lambda_2\lambda_3+\lambda_3\lambda_1 = 0$.

Therefore, the characteristic polynomial is $x^3-1$.


By Cayley-Hamilton, the characteristic formula is:

$$x^3 − \text{tr}(A)x^2+\frac{1}{2}[\text{tr}^2(A)−\text{tr}(A^2)]x−\det A =0$$

In particular, the characteristic polynomial becomes $x^3-1$.


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