The characteristic polynomial of $A$ is $x^n$ if and only if $\text{Tr}(A^i)=0$ for all $1\le i \le n$. [duplicate]

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?

marked as duplicate by user26857 abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 9 at 0:30

• 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 $A$). – darij grinberg Nov 4 '18 at 4:01
• 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 . – darij grinberg Nov 4 '18 at 4:02
• 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. – darij grinberg Feb 9 at 0:47
• @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. – quid Feb 9 at 16:52
• @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. – darij grinberg Feb 9 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 "$$\Tr A = 0$$".