Cayley-Hamilton Theorem - Trace of Exterior Power Form

Let $V$ be an $n$-dimensional vector space over a field $F$ (the characteristic of which, for the purpose of this post, may be taken as $0$). Let $T$ be a linear operator on $V$ and $\lambda\in F$.

Sometime ago, somewhere (I can't recall where) I read that

Formula. $\det(T-\lambda I)= \sum_{k=0}^n (-1)^k \text{trace}(\Lambda^k T)\lambda^{n-k}$

I considered a special case to test this out by taking $n=3$. And here is what I got:

Let $\{e_1, e_2, e_3\}$ form a basis for $V$. Then $$\det(T-\lambda I)(e_1\wedge e_2\wedge e_3)=(Te_1-\lambda e_1)\wedge (Te_2-\lambda e_2)\wedge (Te_3- \lambda e_3)$$

whence expanding the RHS we get

$$(\det T)(e_1\wedge e_2\wedge e_3) - \lambda(Te_2\wedge Te_2 \wedge e_3+ Te_1\wedge e_2\wedge Te_3 + e_1\wedge Te_2\wedge Te_3)+\lambda^2(Te_1\wedge e_2\wedge e_3+ e_1\wedge Te_2\wedge e_3+e_1\wedge e_2 \wedge Te_3) - \lambda^3(e_1\wedge e_2\wedge e_3)$$

Since $\det T=\text{trace}(\Lambda^3 T)$, the coefficient of $\lambda^0$ matches with that in the formula. Also, the coefficient of $\lambda^2$ is just $\text{trace}(T)$ by definition so this is also fine.

The Problem. What I am not able to see is how the coefficient of $\lambda$ eaul to $\text{trace}(\Lambda^2T)$.

• Your formula uses $\lambda$ and $x$ to mean the same thing. Also, it may be helpful to first consider the case when the $e_i$ are a basis of eigenvectors for $T$ and recall Vieta's formulas about the roots of polynomials. – Nate Jan 28 '16 at 17:54
• I am sorry about the $x$. I meant $\lambda$. Let me edit. – caffeinemachine Jan 28 '16 at 17:58
• What a beautiful formula! I've never see that particular one before. – Cheerful Parsnip Jan 28 '16 at 18:07

Let us write $Te_i = \sum_j a^j_i e_j$. For $1 \leq i < j \leq n$, we have

$$(\Lambda^2(T))(e_i \wedge e_j) = Te_i \wedge Te_j = \left( \sum_{k_1} a_i^{k_1} e_{k_1} \right) \wedge \left( \sum_{k_2} a_j^{k_2} e_{k_2} \right) = (a_i^i a_j^j - a_i^j a_j^i) (e_i \wedge e_j) + \cdots$$

where the $\cdots$ don't involve $e_i \wedge e_j$ (as the coefficient of $e_i \wedge e_j$ for $i < j$ comes from $k_1 = i, k_2 = j$ or $k_1 = j, k_2 = i$). Hence,

$$\mathrm{tr}(\Lambda^2T) = \sum_{i < j} (a_i^i a_j^j - a_i^j a_j^i).$$

In your case, similar arguments show that

$$Te_1 \wedge Te_2 \wedge e_3 = \left( \sum_{k_1} a_1^{k_1} e_{k_1} \right) \wedge \left( \sum_{k_2} a_2^{k_2} e_{k_2} \right) \wedge e_3 = (a_1^1 a_2^2 - a_1^2 a_2^1) (e_1 \wedge e_2 \wedge e_3), \\ Te_1 \wedge e_2 \wedge Te_3 = \left( \sum_{k_1} a_1^{k_1} e_{k_1} \right) \wedge e_2 \wedge \left( \sum_{k_2} a_3^{k_2} e_{k_2} \right) = (a_1^1 a_3^3 - a_1^3 a_3^1) (e_1 \wedge e_2 \wedge e_3), \\ e_1 \wedge Te_2 \wedge Te_3 = e_1 \wedge \left( \sum_{k_1} a_2^{k_1} e_{k_1} \right) \wedge \left( \sum_{k_2} a_3^{k_2} e_{k_2} \right) = (a_2^2 a_3^3 - a_2^3 a_3^2) (e_1 \wedge e_2 \wedge e_3),$$

and so indeed

$$\mathrm{trace}(\Lambda^2 T)(e_1 \wedge e_2 \wedge e_3) = Te_1 \wedge Te_2 \wedge e_3 + Te_1 \wedge e_2 \wedge Te_3 + e_1 \wedge Te_2 \wedge Te_3.$$

• Thank you. Do you see how to establish the corresponding statement for $\text{trace}(\Lambda^k T)$? – caffeinemachine Jan 31 '16 at 16:34
• For the more general case, it is useful to use the Hodge star construction. Put an inner product on $V$ and choose an orthonormal basis $(e_1, \ldots, e_n)$. Then we have $\mathrm{tr}(\Lambda^kT) (e_1 \wedge \cdots \wedge e_n) = \left( \sum_{I}^{'} \left< (\Lambda^k(T))(e_I), e_I \right> \right) e_1 \wedge \cdots \wedge e_n = \sum_{I}^{'} \Lambda^k(T)(e_I) \wedge (*e_I)$. – levap Jan 31 '16 at 17:24
• Can you please frame this as an answer here math.stackexchange.com/questions/1634778/… – caffeinemachine Jan 31 '16 at 17:26