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If $A\in M_n(\mathbb{R})$, and satisfies $$A^3+A=0$$how to show that$$\text{Trace}(A)=0$$

thanks

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You should edit your question so that it's, well, a question, instead of a statement. –  Max Nov 19 '12 at 8:20
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I have edited. maybe it looks like a question. –  Laura Nov 19 '12 at 8:54

1 Answer 1

up vote 12 down vote accepted

$$\text{Trace}(A) = \sum_{k=1}^{n} \lambda_k$$ 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.

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