${A_1}^k + {A_2}^k + \cdots +{A_n}^k = 0$ for all $k \in \mathbb N_{>0}$ $\implies$ $A_i$ are all nilpotent. Does the following statement hold true?
Given $n$ real matrices $A_1, A_2, \cdots A_n$, if their $k$-th power sum is zero for all $k \in \mathbb N_{>0}$ then they are all nilpotent.
Given $n$ real variables $x_1, x_2, \cdots x_n$, if their $k$-th power sum is zero for all $k \in \mathbb N_{>0}$ then they are all identical to zero. This can be shown by gathering up the same values and using the well known fact that the Vandermonte matrix is invertible. I tried the same method. The entry ${x_i}^j$ turns into ${A_i}^j \otimes I$ so the determinant is given by the product of ${(\det(A_i -A_j))}^n$ which is not necessarily nonzero. I got stuck here.
 A: It is true for all $n$, for matrices with coefficients in any field $K$ of characteristic $0$
Indeed, one can obviously assume that $K$ is algebraically closed (because every field has an algebraic closure - if you're only concerned with real matrices, consider $K=\mathbb{C}$).
Then $tr(A^k) = \displaystyle\sum_{\lambda\in Spec(A)\setminus\{0\}} m_\lambda^A\lambda^k$ (where $m_\lambda^A$ is the multiplicity of $\lambda$ in $A$, this is easily seen by trigonalizing)
Therefore, with your hypothesis, $tr(\displaystyle\sum_i A_i^k) = 0, \displaystyle\sum_i \sum_{\lambda\in Spec(A_i)\setminus\{0\}} m_\lambda^{A_i}\lambda^k =0$.
Now if you regroup the terms together, this gives you some nonzero complex numbers $\lambda_1,...,\lambda_l$ and some nonzero integers $m_1,...,m_l$ such that $\displaystyle\sum_j m_j \lambda_j^k =0$ for all $k$.
If you stop $k$ at the right point this gives you a VanderMonde matrix which should therefore be invertible but has a nonzero vector in its kernel ($(m_j)_{j\in \{1,...,l\}}$), which is a contradiction.
What did we contradict though ? Well the fact that this vector was nonzero, in other words, that $l>0$ : so the dimension is zero, but if we gi back up what we did, this means that all of the $Spec(A_i)\setminus\{0\}$ are empty, in other words $Spec(A_i) \subset \{0\}$, which implies by the Cayley-Hamilton theorem that $A_i$ is nilpotent.
Notice that the necessary hypothesis was not "for all $k>0$" but rather "for all $k\in \{1,..., na\}$" where $a$ is the dimension of the matrices for instance.
Note : as noted in the comments, this argument does not work (at least not without working a bit more) in positive characteristic !
Edit 2 : In fact, Jyrki Lahtonen's comment makes it easy to come up with counterexamples in characteristic $p$ : put $n=p$ and $A_i = I_n$. Then clearly $A_i$ is not nilpotent, but $\displaystyle\sum_i A_i^k = \sum_i A_i = pI_p = 0$. Of course you can come up with many more examples this way
