# Cam's Blog

## November 14, 2009

### Coefficients of characteristic polynomials

Filed under: Math — cfranc @ 7:43 pm

If you’ve taken linear algebra and noticed that the trace and determinant of a matrix appear, up to sign, as coefficients of the characteristic polynomial of the matrix, then you may have wondered if the other coefficients have a similarly nice description. This turns out to be the case, but to explain we’ll need to assume knowledge about exterior powers of a vector space. Let $K$ be a field, let $V$ denote a finite dimensional vector space over $K$ and let $f \colon V \to V$ denote a $K$-linear map. Recall that $\Lambda^0V = K$ and $\Lambda^1V = V$. Recall furthermore that the exterior power is a functor, where $\Lambda^0f$ is the identity map of $K$ and for $k \geq 1$ one has

$(\Lambda^kf)(v_1 \wedge \cdots \wedge v_k) = f(v_i)\wedge \cdots \wedge f(v_k)$.

Let $\chi_f(X) = X^n + a_1X^{n-1} + \cdots + a_n$ be the characteristic polynomial of the linear operator $f$. Then it turns out that $a_k = (-1)^k\text{Tr}(\Lambda^k f)$. Our class was asked to prove this when I was a fourth year undergraduate at Queen’s, and I did so via a messy computation involving a choice of basis. I’d like a more sophisticated proof of this fact, and while I don’t have one yet, I have managed to give a nice proof of the following fact: let $\lambda$ be an eigenvalue of $f$. Then $\sum_{k = 0}^n (-1)^k\text{Tr}(\lambda^k f)\lambda^{n-k} = 0$. This fact follows if you accept my description of the coefficients of $\chi_f(X)$ simply by substituting $\lambda$ into the characteristic polynomial. We’ll give an independent proof using the following cool lemma:

Let

$0 \longrightarrow V_0 \longrightarrow V_1 \longrightarrow \cdots \longrightarrow V_n \longrightarrow 0$

be an exact sequence of $K$-vector spaces, and let $f_i \in \text{End}_K(V_i)$ for $i = 0, \ldots, n$ be endormorphisms commuting with the arrows of the exact sequence above. Then $\sum_{k = 0}^n (-1)^k \text{Tr}(f_i) = 0$.

This lemma is not difficult to prove. For a five term short exact sequence, build a nice basis for $V_1$ which induces bases on $V_0$ and $V_2$ such that $f_0$ and $f_2$ are upper triangular in these bases. Then $f_1$ will also be upper triangular, and you can compute it’s trace to be the sum of the traces of $f_0$ and $f_2$. Note that one should really base-change to an algebraically closed field, so that the matrices can be brought to upper-triangular form. To deduce the general result from the case of a five term short exact sequence, simply break the longer sequence up into $n+1$ five term short exact sequences; see also Proposition $2.11$ of Atiyah and MacDonald’s Introduction to Commutative Algebra.

With the lemma out of the way, recall that we now want to use it to prove that if $\lambda$ is an eigenvalue for our operator $f \colon V \to V$, then

$\sum_{k = 0}^n (-1)^k\text{Tr}(\Lambda^k f)\lambda^{n-k} = 0$.

When $\lambda = 0$ this amounts to proving simply that $\text{Tr}(\Lambda^n f) = 0$. Recall that $\text{Det}(f) = \text{Tr}(\Lambda^n f)$, and in this case $\text{Det}(f) = 0$ since $f$ is not invertible. So we may now suppose that $\lambda \neq 0$, and in this case the lined formula may be rearranged to read

$\sum_{k = 0}^n (-1)^k\text{Tr}(\Lambda^k (f/\lambda)) = 0$.

When expressed in this form, it is clear that we should apply the homological lemma proved above taking $V_k = \Lambda^k V$ and $f_k = \Lambda^k (f/\lambda)$ for $k = 0, \ldots, n$. The exact sequence

$0 \longrightarrow \Lambda^0 V \longrightarrow \Lambda^1 V \longrightarrow \Lambda^2 V \longrightarrow \cdots \longrightarrow \Lambda^n V \longrightarrow 0$

is defined in the following way: let $v \in V$ denote an eigenvector for $f$ of eigenvalue $\lambda$. Let $\Lambda^0 V \to \Lambda^1V$ be defined by mapping $\alpha \mapsto \alpha v$. Then for $k \geq 1$ define $\Lambda^k V \to \Lambda^{k+1}V$ by sending $w \mapsto w \wedge v$. Completing $\{v\}$ to a basis for $V$ allows one to easily show that the sequence above is exact. Now one simply needs to check that the sequence of maps $\Lambda^k(f/\lambda)$ commutes with the maps of the exact sequence, which is a simple calculation boiling down to the fact that $f/\lambda$ fixes the vector $v$. This completes the proof.

There are several things that I’m unhappy with in this post. Firstly, I haven’t shown that $p(X) = \sum_{k = 0}^n (-1)^k\text{Tr}(\Lambda^k f)X^{n - k}$ is the characteristic polynomial of $f$. All I’ve shown is that $p(X)$ has all of the eigenvalues of $f$ as roots. This isn’t even enough to prove that the minimal polynomial of $f$ divides my polynomial! What I’d like to prove is that

$\sum_{k = 0}^n (-1)^k\text{Tr}(\Lambda^k f)f^{n - k} = 0$

via some kind of “homological” argument which avoids the use of the Cayley-Hamilton theorem (from which this result does follow). This fact has always looked to me like some sort of Lefshetz fixed-point formula, and I think it’s intriguing that the proof of the partial result above relied on considering the operator $f/\lambda$ and its fixed point $v$.

Anyway, I’ll end with one other thing that I’d like to see proved in an elegant way. For this we’ll need to assume that the characteristic of our base field $K$ is zero. It’s not too hard to show that $f$ is nilpotent if and only if $\text{Tr}(f^k) = 0$ for all $k \geq 1$. Since $f$ is nilpotent if and only if its characteristic polynomial is of the form $X^n$, the description of the coefficients of the characteristic polynomial above suggests that one might be able to express $\text{Tr}(\Lambda^k f)$ as a polynomial expression of the various $\text{Tr}(f^j)$. Without too much difficulty I was able to prove that

$\text{Tr}(\Lambda^2f) = (1/2)(\text{Tr}(f^2) - \text{Tr}(f)^2)$.

I quickly gave up trying to deduce a general formula, since it was getting late, and instead turned to the internet for help. During my brief search I was only able to find the following link describing “Bocher’s formula” for the coefficients of the characteristic polynomial. In our notation it reads

$\text{Tr}(\Lambda^k f) = \frac{1}{k} \left(\sum_{j = 1}^{k} (-1)^{k - j}\text{Tr}(\Lambda^{k - j} f)\text{Tr}(f^j)\right)$

for $k \geq 1$. I’ve never seen this proved, so don’t quote this result from me. However, when $k = 2$ it recovers the formula I found, and when $k > 2$ it gives a recursive formula for $\text{Tr}(\Lambda^k f)$ as a polynomial expression of $\text{Tr}(f^j)$ for $j = 0, \ldots k$. So it’d be nice to have an elegant proof to this version of “Bocher’s formula” as well.