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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.

Rodrigo de Azevedo
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Norbert
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  • See also the proof of Theorem 1 in my answer https://math.stackexchange.com/questions/2983730/the-characteristic-polynomial-of-a-is-xn-if-and-only-if-texttrai-0/2983792#2983792 . – darij grinberg Feb 04 '19 at 04:38

2 Answers2

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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 $\mathrm{tr}(PAP^{-1})=0$.

Julien
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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
  • Nice argument! By the way, the trace is invariant under cyclic permutations. :-) – André Caldas Dec 24 '13 at 19:33
  • @DanielMontealegre: I know. Sorry if I was not of much help... I could have been more didactic. – André Caldas Dec 24 '13 at 19:34
  • I knew this solution, but may be there is something more simple. I doubt whether it exist but anyway let's wait... – Norbert Dec 24 '13 at 19:34
  • @AndréCaldas Guys, if you managed to find a counter-example to this, you should publish this without waiting any longer... ;-) – Julien Dec 24 '13 at 19:39
  • @Norbert Yes, I assumed you were aware of this argument. But since you did not mention it... it seemed to qualify as basic and elementary. – Julien Dec 24 '13 at 19:40
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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
  • @AndréCaldas As much I like jokes, you lost me. Of course $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ (the trace is commutative if the ground field is commutative). While the last identity you mentioned is false in general (this permutation is not cyclic). What is this all about? – Julien Dec 24 '13 at 19:51
  • @julien André isn't saying that $\mathrm{tr}(AB)\neq\mathrm{tr}(BA)$. But it's misleading to say that the trace is commutative when $\mathrm{tr}(ABC)\neq\mathrm{tr}(BAC)$ in general. It's better to say that it's cyclic. – Josephine Moeller Dec 24 '13 at 19:55
  • @JohnMoeller Ah...thanks for explaining. For (lots of?) people, the trace is commutative simply means $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ as it refers to the product as a binary operation. – Julien Dec 24 '13 at 20:01
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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...;-) – Julien Dec 28 '13 at 14:44
  • In step 2) when you say "If P denotes the corresponding change of basis matrix", what change of basis are you referring to? – gen May 01 '19 at 15:31
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If you're working over a field of characteristic $p$, you can view this as a consequence of the identity $Tr(X^p) = Tr(X)^p$, which has a short combinatorial proof by expanding out $Tr(X)^p$. Even though this doesn't apply in generality (i.e. characteristic 0), I still think it is worth mentioning.

Henry Bosch
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