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Let $A_{(2n+1)\times(2n+1)}$ be a symmetric matrix of Rank $2n$. Then does $\operatorname{tr}A=0$ imply $\operatorname{tr}A^3=0$? If not, Under what condition?

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The answer is NO. For a counterexample, let

$$A=\operatorname{diag}(1,3,-2,-2,0)$$ We have $\operatorname{Tr}(A)=0$ and $\operatorname{Tr}(A^3)=12\ne0$.

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Hint

$1)$ If $A$ is symmetric then we can write:

$$A=P^TDP$$ where $D$ is a diagonal matrix and $P$ is orthogonal. So,

$$A^3=P^TD^3P$$

$2)$ Use that $Trace(XY)=Trace(YX)$

So,

$$Trace(A)=Trace(D)=0$$

and,

$$Trace(A^3)=Trace(D^3)$$

Now compare $Trace(D)$ and $Trace (D^3)$.

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  • $\begingroup$ What's your point ? $\endgroup$ – Gabriel Romon Sep 5 '17 at 17:26
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    $\begingroup$ @LeGrandDODOM: That is a hint. My point is to help the OP be able to go further. $\endgroup$ – Arnaldo Sep 5 '17 at 17:27
  • $\begingroup$ A hint towards what ? $\endgroup$ – Gabriel Romon Sep 5 '17 at 17:27
  • $\begingroup$ @LeGrandDODOM The condition under which $tr(A)=tr(A^3)=0$, I think. $\endgroup$ – Duchamp Gérard H. E. Sep 5 '17 at 17:28
  • $\begingroup$ @LeGrandDODOM: I think is clear now. $\endgroup$ – Arnaldo Sep 5 '17 at 17:31
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The fact that $\operatorname{Rank}(A) = 2n$ implies $\operatorname{Null}(A) = 1$, so $0$ is an eigenvalue with multiplicity $1$. Other than that, the eigenvalues are free to range over other non-zero values, provided they do sum to $0$. Certainly if $n = 1$, then the other two eigenvalues must be negatives of each other and non-zero, giving us distinct eigenvalues and hence a diagonalisable matrix. If $\lambda > 0$ is the positive eigenvalue, it follows therefore that $$\operatorname{tr}(A^3) = 0^3 + \lambda^3 + (-\lambda)^3 = 0.$$ When we shed the assumption that $n = 1$, things get trickier. Even in (presumably) the nicest case where $A$ is diagonalisable, things start to fall down. For example, $$A = \operatorname{diag}(0, 1, 2, 3, -6)$$ has a $0$ trace and rank $4$, but $$A^3 = \operatorname{diag}(0, 1, 8, 27, -216),$$ which has a trace of $-180$.

Fundamentally, you need to navigate this issue of the so-called "Freshman's Dream", where the sum of powers is not the power of the sum. Steering away from diagonalisable, symmetric, normal, or even towards defective matrices won't help in the end, as the eigenvalues will still cube, and that's what you're trying to force to be $0$.

I think limiting $n$ to $1$, or forcing eigenvalues to be specific values (as someone suggested, maybe $\lambda = \pm 1$, or possibly other roots of unity?) is your best bet for sufficient conditions, but I don't think it's going to be particularly helpful for anything non-trivial.

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