# 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.

## 1 Answer

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