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I'm trying to prove that the trace of a nilpotent matrix is $0$. Here's what I have:

Let $A \in M_n(k)$, where $k$ is a field. If $A$ is nilpotent then, in fact, $A^n=0$, (this result was proven in the previous problem.) By Proposition 20 on page 428 (Dummit & Foote) the characteristic polynomial of $A$ divides some power of the minimal polynomial of $A$, and so is of the form $x^m$ for some $m \ge t$. Thus, the eigenvalues of $A$ are all equal to $0$ and the only solution to the equation $xv-Av=0$ is the trivial solution. Since it is assumed that $v \neq 0$, then it must be that $A=0$.

This is a much stronger conclusion than the conclusion I'm aiming for and so I am suspicious. I probably made an error. Please help.

user26857
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Sam
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    Since $\chi_A(X)=X^n-{\rm tr};(A) X^{n-1}+\cdots+(-1)^n\det(A)$, you get your answer. – Pedro Apr 05 '15 at 00:43
  • It is not clear what you means by "the only solution to the equation $xv−Av=0$ is the trivial solution", notably what you are solving for. Normally one would use such a phrase for the zero solution to a linear homogeneous equation (system). Here that would be the case if you are solving for (the vector) $v$, not for (the scalar) $x$. Indeed if $x$ is any nonzero constant then the equation only has the trivial solution for $v$. But then your following sentence makes little sense; it would seem like you are interpreting it as solving for $A$! – Marc van Leeuwen Apr 05 '15 at 00:59
  • @MarcvanLeeuwen Thanks for pointing that out! – Sam Apr 05 '15 at 01:07

3 Answers3

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To expand on the comment by Pedro Tamaroff, I can add the following.

The initial part of your argument seems to aim at the (correct) conclusion that the characteristic polynomial must be $X^n$ (remember that unlike the minimal polynomial, you know the degree of the characteristic polynomial beforehand). But you never really make that conclusion, which you could reach as follows: since some $X^k$ annihilates $A$, the minimal polynomial divides $X^k$ and must be a power of $X$, but then the characteristic polynomial, which divides a power of the minimal polynomial, must also be a power of $X$, necessarily $X^n$. (Note that this starts with the order $k$ of nilpotency of$~A$; there is no need to argue for $k=n$ initially, although the conclusion of the argument shows, by Cayley-Hamilton, that $X^n$ annihilates $A$ as well.)

Once you know the characteristic polynomial, you can simply read off the trace as minus its subdominant coefficient, which is zero for $X^n$.

A more low-tech argument is possible as well, provided you know that the trace is invariant under change of basis. The chain of subspaces $\{0\}=\ker A^0\subseteq\ker A^1\subseteq\ker A^2\subseteq\cdots$ grows up to the whole space for nilpotent$~A$. Choosing a basis of $\ker A^1$, then extending to a basis of the next space $\ker A^2$, and so on, eventually gives a basis of the whole space. By construction change of basis of $A$ to this new basis leads to a strictly upper triangular matrix; its trace is obviously$~0$.

Marc van Leeuwen
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Even if $A\ne 0$ has all of its eigenvalues $0$ it doesn't mean $Av=xv$ has no nontrivial solutions; consider for instance $A=(\begin{smallmatrix}0&1\\0&0\end{smallmatrix})$ and $v=(\begin{smallmatrix}1\\0\end{smallmatrix})$.

Anyway, to do the problem, know/use the fact that the trace is the sum of eigenvalues.

anon
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Let $A \in M_n(k)$, where $k$ is a field.

$A$ is similar to its Jordan Canonical form. That is, for some $P \in GL_n(k)$, we have $PAP^{-1}=J_A$. Since $Tr(AB)=Tr(BA)$, we then have $Tr(PAP^{-1})=Tr(A)$.

If $A$ is nilpotent then $A^k=0$ for some $k$. Thus the minimal polynomial of $A$ is a power of $x$. We know the characteristic polynomial of $A$ divides some power of the minimal polynomial of $A$ and so is $x^n$. Thus, all eigenvalues of $A$ are equal to $0$ and the trace of $J_A$ is $0$.

Sam
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    The Jordan form is not defined over arbitrary fields. Your argument could work after extending to an algebraically closed field, but as it stands the claim $P\in GL_n(k)$ is simply wrong. Also, your argument does not actually need the Jordan form: once you know the characteristic polynomial$~\chi_A$, you know the trace of $A$ itself (from the coefficient of $X^{n-1}$ in$~\chi_A$). – Marc van Leeuwen Aug 27 '16 at 06:51