If $A\in M_n(\mathbb{R})$, and satisfies $A^3+A=0$how to show that$\text{Trace}(A)=0$
thanks
If $A\in M_n(\mathbb{R})$, and satisfies $A^3+A=0$how to show that$\text{Trace}(A)=0$
thanks
$\text{Trace}(A) = \sum_{k=1}^{n} \lambda_k$ Since $A^3 + A = 0$, this implies that the eigenvalues are $\lambda = 0, \pm i$. Since $A \in M_n(\mathbb{R})$, the imaginary eigenvalues occur in conjugate pairs i.e. if $m$ eigenvalues are $+i$, then there exists exactly $m$ eigenvalues $-i$. The remaining $n-2m$ eigenvalues are $0$.
Hence, the sum of the eigenvalues is $m \times i + m \times (-i) + (n-2m) \times 0 = 0$ This implies that the trace of $A$ is zero.