Let $k$ be a field of characteristic zero and $A$ be a matrix over $k$. Then the characteristic polynomial of $A$ is $x^n$ if and only if $\text{Tr}(A^i)=0$ for all $1\le i \le n$.

The only proof I can think of is by applying the Jordan normal form to $A$ (considered as a matrix over $\overline{k}$). Is there any slick proof without invoking this theorem?

  • $\begingroup$ For the "if" part, see the proof of Corollary 4.1 in my The trace Cayley-Hamilton theorem, in which I show that $n! c_k = 0$ for every $k \in \left\{1,2,\ldots,n\right\}$ (note that $c_k$ is one of the coefficients of the characteristic polynomial of $A$). $\endgroup$ – darij grinberg Nov 4 '18 at 4:01
  • $\begingroup$ For the "only if" part, it suffices to prove that the trace of a nilpotent matrix is nilpotent (because if $A$ has characteristic polynomial $x^n$, then $A, A^2, \ldots, A^n$ are all nilpotent). This is done in artofproblemsolving.com/community/c7h233169p1288269 . $\endgroup$ – darij grinberg Nov 4 '18 at 4:02
  • 1
    $\begingroup$ Close voters beware: One of the two supposedly duplicate questions is more or less a converse, and the other one assume the traces to be $0$ for all positive powers of $A$ rather than just for the first $n$. Voting to reopen. $\endgroup$ – darij grinberg Feb 9 '19 at 0:47
  • $\begingroup$ @darijgrinberg for all or up to n hardly changes anything though. At least one answer on the dupe is explicit regarding this, and for another it's also quite clear. One might modify the dupe target slightly. $\endgroup$ – quid Feb 9 '19 at 16:52
  • $\begingroup$ @quid: I thought of that, too, but it would require seriously editing a high-rated answer that starts out with a proof of the weaker question. $\endgroup$ – darij grinberg Feb 9 '19 at 22:11

$\newcommand{\Tr}{\operatorname{Tr}}$ This is (now) proven in my note The trace Cayley-Hamilton theorem. More precisely, the "if" part of your claim is Corollary 4.1 (d), while the "only if" part follows from Corollary 4.2. Note that $k$ can be an arbitrary commutative $\mathbb{Q}$-algebra (not necessarily a field); for the "only if" part, it can even be an arbitrary ring.

An alternative way to prove the "only if" part can be obtained from the following theorem:

Theorem 1. Let $R$ be a commutative ring with $1$. Let $ A\in R^{n\times n}$ be a nilpotent matrix. ("Nilpotent" means that there exists an $ m\in\mathbb{N}$ such that $ A^m = 0$; we do not require $m = n$.) Then, $\Tr A$ is a nilpotent element of $ R$.

The following ingenious constructive argument for this theorem was shown to me long ago by Peter Scholze; I had posted it on AoPS, but it's probably of interest to the crowd here just as well.

We shall use the following two facts:

Theorem 2. Let $ f$ be a polynomial in $ R\left[X\right]$, where $ R$ is a commutative ring with $ 1$. Then, the following two statements are equivalent:

Statement 1: The element $ f$ is invertible in $ R\left[X\right]$.

Statement 2: The coefficient of $ f$ before $ X^0$ is invertible in $ R$, and all other coefficients of $ f$ are nilpotent.

Theorem 2 is proven in Characterizing units in polynomial rings and in http://www.mathlinks.ro/Forum/viewtopic.php?t=89417 and in various other places.

Theorem 3. Let $ S$ be a (not necessarily commutative) ring with $ 1$. Let $a\in S$ be nilpotent, and $ x\in S$ be arbitrary such that $xa = ax$. Then, $1 - xa$ is invertible in $ S$.

Proof of Theorem 3. The element $a$ is nilpotent. In other words, there exists some integer $p \geq 0$ such that $a^p = 0$. Consider this $p$.

The elements $x$ and $a$ of $S$ commute (since $xa = ax$). Let $R$ be the subring of $S$ generated by these elements $x$ and $a$. Then, this ring $R$ is generated by two commuting elements (indeed, $x$ and $a$ commute), and thus is commutative. Define $n \in R$ by $n = xa$. Thus, for each nonnegative integer $k$, we have $n^k = \left(xa\right)^k = x^k a^k$ (since we are working in the commutative ring $R$). Applying this to $k = p$, we obtain $n^p = x^p \underbrace{a^p}_{=0} = 0$. Hence, $n$ is nilpotent. Thus, all but finitely many of the elements $n^0, n^1, n^2, \ldots$ are zero. Therefore, the sum $\sum\limits_{i=0}^{\infty} n^i$ is well-defined (more precisely: it converges with respect to the discrete topology). Now, \begin{align} & \left(\sum\limits_{i=0}^{\infty} n^i\right) \left(1-n\right) \\ &= \sum\limits_{i=0}^{\infty} n^i - \left(\sum\limits_{i=0}^{\infty} n^i\right) n = \sum\limits_{i=0}^{\infty} n^i - \sum\limits_{i=0}^{\infty} \underbrace{n^i n}_{=n^{i+1}} \\ &= \sum\limits_{i=0}^{\infty} n^i - \sum\limits_{i=0}^{\infty} n^{i+1} = \sum\limits_{i=0}^{\infty} n^i - \sum\limits_{i=1}^{\infty} n^i \\ & \qquad \left(\text{here, we have substituted $i$ for $i+1$ in the second sum}\right) \\ &= n^0 = 1 . \end{align} Since both $\sum\limits_{i=0}^{\infty} n^i$ and $1-n$ belong to the commutative ring $R$, this equality entails that $\sum\limits_{i=0}^{\infty} n^i$ is an inverse of $1-n$ in $R$. Hence, $\sum\limits_{i=0}^{\infty} n^i$ is an inverse of $1-n$ in $S$ as well. Thus, $1-n$ is invertible in $S$. In other words, $1-xa$ is invertible in $S$ (since $n = xa$). This proves Theorem 3. $\blacksquare$

Proof of Theorem 1. Let $S$ be the ring $\left(R\left[X\right]\right)^{n\times n}$. The unity of this ring $S$ is the identity matrix $I_n$. Via the canonical embedding $ R^{n\times n}\to \left(R\left[X\right]\right)^{n\times n} = S$, we can consider the matrix $A \in R^{n\times n}$ as an element of $S$. It clearly satisfies $XI_n \cdot A = A \cdot XI_n$ (since both sides of this equation equal $XA$). Thus, we can apply Theorem 3 to $a = A$ and $ x = XI_n$, and obtain that $ I_n - XI_n\cdot A$ is invertible in $S$. In other words, $ I_n - XA$ is invertible (since $XI_n \cdot A = XA$).

In other words, there exists $ B\in S$ such that $ \left(I_n - XA\right)B = B\left(I_n - XA\right) = I_n$. Consider this $B$.

Both $I_n - XA$ and $B$ are $n \times n$-matrices over $R\left[X\right]$, and thus their determinants belong to $R\left[X\right]$. We have $\det \left(I_n - XA\right) \cdot \det B = \det\left(\underbrace{\left(I_n - XA\right)B}_{=I_n}\right) = \det\left(I_n\right) = 1$. Therefore, the element $ \det\left(I_n - XA\right)$ is invertible in the commutative ring $R\left[X\right]$.

So $ \det\left(I_n - XA\right)$ is a polynomial in $ R\left[X\right]$ which happens to be invertible in $ R\left[X\right]$. Thus, by Theorem 2 (more specifically, by the "Statement 1 $\Longrightarrow$ Statement 2" direction of this theorem), the coefficient of this polynomial before $ X^0$ is invertible, while all its other coefficients are nilpotent. In particular, the coefficient of $ \det\left(I_n - XA\right)$ before $ X^1$ is nilpotent.

But we claim that the coefficient of $\det\left(I_n - XA\right)$ before $ X^1$ is $- \Tr A$. This may be well-known from linear algebra; if not, the following argument does the trick: The ring $ R\left[X, X^{-1}\right]$ of Laurent polynomials contains the polynomial ring $ R\left[X\right]$ as a subring. Thus, we can consider $S = \left(R\left[X\right]\right)^{n\times n}$ as a subring of the matrix ring $\left(R\left[X, X^{-1}\right]\right)^{n\times n}$. Thus, working over $R\left[X, X^{-1}\right]$, we have $I_n - XA = X\left( X^{-1}I_n - A\right)$, so that \begin{align} \det\left(I_n - XA\right) = \det\left(X\left( X^{-1}I_n - A\right)\right) = X^n\det\left( X^{-1}I_n - A\right), \end{align} and thus \begin{align} & \left(\text{the coefficient of the polynomial $ \det\left(I_n - XA\right)$ before $ X^1$}\right) \\ &= \left(\text{the coefficient of the Laurent polynomial $ \det\left( X^{-1}I_n - A\right)$ before $ X^{1 - n}$}\right) \\ &= \left(\text{the coefficient of the Laurent polynomial $ \det\left( X^{-1}I_n - A\right)$ before $ X^{ - \left(n - 1\right)}$}\right) \\ &= \left(\text{the coefficient of the polynomial $ \det\left(XI_n - A\right)$ before $ X^{n - 1}$}\right) \\ & \qquad \left(\begin{array}{c} \text{here, we have substituted $X$ for $X^{-1}$, using the $R$-algebra automorphism} \\ \text{ of $R\left[X, X^{-1}\right]$ that swaps $X$ with $X^{-1}$} \end{array}\right) \\ &= \left(\text{the coefficient of the characteristic polynomial of the matrix $ A$ before $ X^{n - 1}$}\right) \\ & = - \Tr A \end{align} (where the last equality sign is, e.g., Corollary 3.22 in my The trace Cayley-Hamilton theorem).

Recall that the coefficient of $ \det\left(I_n - XA\right)$ before $ X^1$ is nilpotent. Since we now know that this coefficient is $- \Tr A$, we thus conclude that $- \Tr A$ is nilpotent. Hence, $ \Tr A$ is nilpotent. This proves Theorem 1. $\blacksquare$

Of course, when the commutative ring $R$ is reduced (i.e., has no nilpotents besides $0$), the claim of Theorem 1 can be restated as "$\Tr A = 0$".


Not the answer you're looking for? Browse other questions tagged or ask your own question.