# Trace of the $k$-th Exterior Power of a Linear Operator

Let $$V$$ be an $$n$$ dimensional vector space over a field $$F$$ and $$T$$ be a linear operator over $$V$$. Assume that the characteristic of $$F$$ is not $$2$$.

Definition. Consider the map $$f_1:V^n\to \Lambda^n V$$ as $$f(v_1, \ldots, v_n)= \sum_{i=1}^n v_1\wedge \cdots \wedge v_{i-1}\wedge Tv_i\wedge v_{i+1} \wedge \cdots \wedge v_n$$This is an alternating multilinear map and thus it induces a unique linear map $$\Lambda^n V\to \Lambda^n V$$. Since $$\dim(\Lambda^n V)=1$$, this linear map is multiplication by a constant which we call the trace of $$T$$.

The above is standard and it naturally calls for the following generalization before which we discuss a notation.

Given an $$n$$ tuple $$(v_1, \ldots, v_n)$$ of vectors in $$V$$ and an increasing $$k$$-tuple $$I=(i_1, \ldots , i_k)$$ of integers between $$1$$ and $$n$$, write $$v_{I, j}$$ to denote $$Tv_j$$ if $$j$$ appears in $$I$$ and simply $$v_j$$ if $$j$$ does not appear in $$I$$. Further write $$v_I$$ to denote $$v_{I, 1}\wedge \cdots \wedge v_{I, n}$$.

Definition. Let $$f_k:V^n\to \Lambda^n V$$ be defined as $$f_k(v_1, \ldots, v_n)= \sum_{I \text{ an increasing }k\text{-tuple}}v_I$$ Then $$f_k$$ is an alternating multilinear map and this induces a unique linear map $$\Lambda^n V\to \Lambda^n V$$. Again, this linear map is multiplication by a constant which we call the $$k$$-th trace of $$T$$ and denote it as $$\text{trace}_k(T)$$.

From this post I have am convinced that the following is true

Statement. $$\text{trace}_k(T)= \text{trace}(\Lambda^k T)$$.

I am unable to prove this.

It is convenient to use the Hodge star to simply the calculations. Choose a non-degenerate symmetric bilinear form $$\left< \cdot, \cdot \right>$$ on $$V$$ that has an orthonormal basis (for example, the one corresponding to the identity matrix) and let $$(e_1, \ldots, e_n)$$ be an orthonormal basis with respect to the chosen bilinear form. We will use $$\sum_{I}$$ to denote summation over increasing multi-indices $$I$$ of size $$k$$. Thus,

$$\mathrm{trace}(\Lambda^k T)(e_1 \wedge \cdots \wedge e_n) = \sum_{I} \left< (\Lambda^kT)(e_I), e_I \right> \left( e_1 \wedge \cdots \wedge e_n \right) = \sum_{I} \Lambda^k T(e_I) \wedge (*e_I) = \sum_{I \coprod J = [n]} \pm \left( \Lambda^k T(e_I) \wedge e_J \right) = \sum_{I \coprod J = [n]} \pm \left(Te_{i_1} \wedge \cdots \wedge Te_{i_k} \wedge e_{j_1} \wedge \cdots \wedge e_{j_{n-k}} \right)$$

where $$J$$ is an increasing multi-index such that $$I \coprod J = [n]$$ and we used the fact that $$*e_I = \pm e_J$$. A sign calculation that uses the definition of the Hodge star shows that in fact the sign is plus which shows that

$$\mathrm{trace}(\Lambda^k T)(e_1 \wedge \cdots \wedge e_n) = f_k(e_1, \cdots, e_n)$$

and thus $$\mathrm{trace}(\Lambda^k T) = \mathrm{trace}_k(T)$$.

One can also show this without using the Hodge star. Choose some basis $$(e_1,\dots,e_n)$$ for $$V$$. The expression for $$f_k(e_1,\dots,e_n)$$ is the sum of $$n \choose k$$ terms where each term is obtained from $$e_1 \wedge \dots \wedge e_n$$ by choosing an increasing tuple $$I = (i_1, \dots, i_k)$$ and applying $$T$$ to each $$e_{i_j}$$ while leaving the rest of the vectors intact and in the same order. Let $$J$$ be the unique increasing tuple $$J$$ such that $$I \coprod J = [n]$$ and then by reordering the vectors in the wedge product, we can write each term as

$$(-1)^{\sigma(I)} Te_{i_1} \wedge \dots \wedge Te_{i_k} \wedge e_{j_1} \wedge \dots \wedge e_{j_{n-k}} = (-1)^{\sigma(I)} \Lambda^k(T)(e_I) \wedge e_J$$

where $$(-1)^{\sigma(I)}$$ is the sign that comes from the reordering. Now,

$$\operatorname{trace}(f_k) = (e^1 \wedge \dots \wedge e^n)(f_k(e_1, \dots, e_n)) = (e^1 \wedge \dots \wedge e^n) \sum_{I} (-1)^{\sigma(I)} \Lambda^k(T)(e_I) \wedge e_J = \sum_{I} (-1)^{\sigma(I)} (e^I \wedge e^J)((-1)^{\sigma(I)} \Lambda^k(e_I) \wedge e_J = \sum_{I} (e^I \wedge e^J)(\Lambda^k(e_I) \wedge e_J).$$

Each $$(e^I \wedge e^J)(\Lambda^k(e_I) \wedge e_J)$$ is the determinant of an upper triangular block matrix whose lower $$(n-k) \times (n-k)$$ block is $$I$$. The vanishing of the rightmost $$k \times (n-k)$$ block comes from "$$e^I(e_J)$$" while the fact that the lower $$(n -k) \times (n-k)$$ block is $$I$$ comes from "$$e^J(e_J)$$". Hence,

$$\operatorname{trace}(f_k) = \sum_{I} e^I(\Lambda^k(e_I)) = \operatorname{trace}(\Lambda^k(T)).$$

• So we are using the fact that if $(e_1, \ldots, e_n)$ is an orhonormal basis of $V$ under the chosen bilinear form, then $(e_I: I\text{ an increasing } k \text{ tuple})$ is an orthonormam basis for the induced bilinear map on $\Lambda^k V$? This is why we can write $\text{trace}(\Lambda^k T)= \sum_I\langle(\Lambda^k T)e_I, e_I\rangle$. Am I right? – caffeinemachine Feb 3 '16 at 10:14
• @caffeinemachine Yep. – levap Feb 3 '16 at 12:09
• Is there a proof without using the Hodge star? Employing an arbitrarily-constructed scalar product seems too ad-hoc here. (Still, this is a great answer already, thank you!) – lisyarus Mar 2 '18 at 9:53
• @lisyarus: Yeah, sure. I've added a proof along those lines. – levap Mar 5 '18 at 11:08