# Proving that the coefficients of the characteristic polynomial are the traces of the exterior powers

Let $T$ be an endomorphism of a finite-dimensional vector space $V$. Let $$f(x)=x^n+c_1x^{n-1}+ \dots + c_n$$ be the characteristic polynomial of $T$. It is well known that $c_m=(-1)^m\text{tr}(\bigwedge^mT)$.

If the base field is $\mathbf{C}$, then we can prove it using a density argument. The statement is true for diagonalizable matrices, which are dense in $M_n(\mathbf{C})$. This actually enough to prove it general, but I don't find it very illuminating. I would like to see an abstract proof of this result.

Thank you!

• This makes me happy. I'll give it a though. – Alex Youcis Apr 15 '12 at 0:42
• Consider the standard basis for the exterior products, and how your matrix acts on it. Now think of how the coefficients of the characteristic polynomial can be written in terms of the eigenvalues. This is enough because we can always pass to an algebraic closure. – user641 Apr 15 '12 at 1:24
• Dear @SteveD, I'm not sure I understand what you are suggesting. What do you suggest we do with the eigenvalues? In general $V$ will not have a basis of eigenvectors of $T$, even over an algebraically closed field. – Bruno Joyal Apr 15 '12 at 4:28
• Sorry I think I misunderstood the question. I thought you were concerned about invertible matrices, but extending the result to other fields. I now see you are more concerned with non-invertible matrices. – user641 Apr 15 '12 at 6:37
• No need to diagonalize. Write down the action of $A$ on $\bigwedge^m V$ in the standard basis. What is the trace of that matrix? And what is $c_m$? – David E Speyer Apr 15 '12 at 14:46

## 2 Answers

$\newcommand{\tr}{\operatorname{tr}}$

First, let us give an alternative description of $\tr {\bigwedge}^m T$.

Let $e_1,e_2,\dots,e_n$ be a fixed basis for $V$, and for $J \subset \{1,2,\dots,n\}$ with $|J| = m$, let $T_J$ denote the $m \times m$ matrix acquired by keeping the entries $T_{i,j}$ with $i,j \in J$. Alternatively, if $P_J$ is the projection onto $\operatorname{span} \{e_i\}_{i \in J}$ then $T_J = P_J T P_J$. My claim is that: $$\tr {\bigwedge}^m T = \sum_J \det T_J$$ where the sum runs over all $J$ with $|J| = m$. More precisely, if $e_J := \bigwedge_{i \in J} e_i$ (in any order, but fixed once and for all), then ${\bigwedge}^m T e_J = \det T_J e_J + \sum_{I \neq J} \alpha_I e_I$. This can be showed by brute-force application of the definition on exterior power and the determinant (the one with sum over permutations). Another possibility is to say that: $${\bigwedge}^m P_J {\bigwedge}^m T e_J = {\bigwedge}^m P_J {\bigwedge}^m T {\bigwedge}^m P_J e_J = {\bigwedge}^m T_J e_J = \det T_J e_J$$ where the last equality holds because $m$ is the dimension of the space on which $T_J$ acts (Wiki lists this property as one of the possible definitions of $\det$).

Now, because $\tr {\bigwedge}^m T$ is by definition the sum of coeffs at $e_J$ in ${\bigwedge}^m T e_J$ (note that $e_J$ form the basis), it turns out that the formula with the sum works.

Thus, it remains to see that $c_m = (-1)^m \sum_J \det T_J$. This can be done by looking at how the characteristic polynomial is computed from the definition, using the matrix form of $T$. I think this is known, and has been asked on MSE (except the sign differs). The gist of the proof I know is that to get $(-x)^{n-m}$ from $\det(T - xI)$, you need to choose the term $-x$ in $n-m$ places, and from what remains you get the coefficient $\det T_J$, where $J$ is the set of indices where you did not take the $-x$.

• I really enjoyed your answer! Just a quick comment : the map $P_J$ has a kernel, so technically speaking $\det T_J = 0$. You probably want to replace it by the inclusion map $I_J : \mathrm{span}(\{e_j \}_{j \in J}) \to V$, the rest of the proof still works fine (if you replace $P_J$ by $I_J$ at the appropriate places). – Patrick Da Silva Feb 16 '17 at 21:21

I know this question is quite old, but still I was asking myself the same question recently so I was trying to find the solution ; finding this question (and Feanor's answer) wasn't completely enlightening, although it was helpful. I took the time to type my solution up, hence I thought I should add it here.

There is a little intro to exterior algebra because I admit I wasn't totally familiar with it myself and I had to read back through my notes. I assume not every reader will be familiar with it so I thought it was a good idea. It's on my academia.edu page, which I will link here.

https://www.academia.edu/5901122/On_The_Characteristic_Polynomial

Note : the question suggests it holds only for endomorphisms of finite dimensional vector spaces, but in fact it works for any endomorphism of $R^n$ where $R$ is a commutative unital ring. There's really nothing special about $\mathbb C$ in the statement.

Hope that helps,

• Can you please tell me where did you read this material? – caffeinemachine Jul 21 '15 at 12:19
• @caffeinemachine : I couldn't find this material anywhere either, so I just sat down and thought about this. I am pretty happy about this "injection sign" thing, it took me a while to get it right. – Patrick Da Silva Jul 21 '15 at 12:47