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Does anyone know a proof from the first principles that a nilpotent matrix has zero trace. No eigenvalues, no characteristic polynomials, just definition and basic facts about bases and matrices.

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I assume you want the trace of a matrix $A\in M_n(F)$ to be defined as the sum of the diagonal elements and that you take the coefficients in a (commutative) field $F$. Here is an approach using only basic facts about bases and matrices.

1) Recall the trace is commutative $\mathrm{tr}(AB)=\mathrm{tr}(BA)$, as shown by the usual computations. In particular the trace is invariant under similarity (change of basis): $\mathrm{tr}(PAP^{-1})=\mathrm{tr} A$ for every $P$ invertible in $M_n(F)$.

2) If $A$ is nilpotent with degree $k$ (i.e. $A^k=0$ but $A^{k-1}\neq 0$), we have the following flag $$ \{0\}\subseteq \ker A\subseteq \ker A^2\subseteq \ldots\subseteq \ker A^k=F^n $$ where dimensions are strictly increasing. Starting with a basis of $\ker A$, we can complete it into a basis of $\ker A^2$ and so on until we get a basis of $F^n$. If $P$ denotes the corresponding change of basis matrix, then $PAP^{-1}$ is strictly upper-triangular as $A(\ker A^j)\subseteq \ker A^{j-1}$. In particular, the diagonal of $PAP^{-1}$ is zero whence $\mathrm{tr}(PAP^{-1})=0$.

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The trace is not commutative. Even in a commutative field. –  André Caldas Dec 24 '13 at 19:27
    
@AndréCaldas You are right. It is not commutative, but it is cyclic. I.e., Tr$(ABC)=$Tr$(CAB)$=Tr$(BCA)$, so I believe that is all he needs. –  Daniel Montealegre Dec 24 '13 at 19:33
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@julien: I do not have to prove. You have! Show that $\mathrm{tr}(ABC) = \mathrm{tr}(BAC)$. Cheers. –  André Caldas Dec 24 '13 at 19:42
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@JohnMoeller and julien: I prefer to say that it is invariant under xxxx type of permutations. The trace is not a binary operation, and therefore, I find it misleading to say it is commutative. What is associativity, for example? Now, notice that in case of group operations, for example, commutativity plus associativity gives that $abc = bac$. Is the trace associative? In what sense? But anyway... your solution is very nice!!! Cheers. :-) –  André Caldas Dec 28 '13 at 11:53
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@AndréCaldas Yeah, I agree this is sloppy terminology. It just happens I've heard it many times. But hey, I've edited...;-) –  1015 Dec 28 '13 at 14:44

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