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

Question

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.

Answer 1

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

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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,

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

Answer 2

Late to the party, not sure if this reply adds much. If anything, perhaps the assurance that the first exterior trace is just like the regular trace and - actually - that the $k$-th exterior trace is also like a regular trace for $k>1$.

Suppose $\text{span}\{a_1,\dots,a_n\}=V$. Let $\omega\in\Lambda^n(V^*)$, then $$ x=\sum_{i=1}^n \frac{\omega(a_1\wedge\dots\wedge a_{i-1} \wedge x\wedge a_{i+1}\dots\wedge a_n)}{\omega(a_1\wedge\dots\wedge a_n)}a_i, $$ which essentially is an application of Cramer's rule. Defining
$$ e^i(x)=\frac{\omega(a_1\wedge\dots\wedge a_{i-1} \wedge x\wedge a_{i+1}\dots\wedge a_n)}{\omega(a_1\wedge\dots\wedge a_n)} $$ and $e_i=a_i$, we have $\{e^1,\dots, e^n\}$ as a basis for $V^*$, dual to $\{e_1,\dots e_n\}$, i.e. $\langle e_i,e^j\rangle=\delta^j_i$. (Note: $\langle .,.\rangle:V\times V^*\to \mathbb{R}$ (or whatever), is defined as $(x,f)\mapsto\langle x,f \rangle=f(x)$). We can also write $$ e^i(x)=\frac{\det(a,i\to x)}{\det(a)}, $$ the quotient of the determinant of the matrix $a$ whose columns are formed by $(a_i)_i$, but with column $i$ replaced with $x$ and the determinant of $a$.

Given a basis and dual basis like this, the trace of an operator $A:V\to V$ is computed as $$ \text{tr}(A)=\sum_{1\leq i\leq n}\langle Ae_i,e^i\rangle $$ (Of course, $\text{tr}(A)$ famously does not depend on the chosen basis.)

If we have $A:V\to V$, then $A$ also acts in a linear fashion on higher exterior powers like this $$ A(u_1\wedge \dots \wedge u_k)=A u_1\wedge \dots \wedge A u_k. $$ And those induced linear operators have traces too. To compute their traces we need a dual basis. This is reasonably straightforward. For example, we get a basis for $\Lambda^2(V^*)$, dual to $(e_i\wedge e_j)_{1\leq i<j\leq n}$, in this way: $$ (e^i\wedge e^j)(x,y)=\frac{\det(a,i\to x,j\to y)}{\det(a)}. $$ and $$ \text{tr}_2(A)=\sum_{1\leq i<j\leq n}\langle A(e_i\wedge e_j),e^i\wedge e^j\rangle. $$ Again, we sum the contributions of all basis vectors, this time with a double index.

Next consider $f_k(v_1, \dots\,v_k)$ acted upon by $n$-covector $\omega$. We can express this as $$ \langle f_k(v_1,\dots,v_n),\omega\rangle=\langle v_1\wedge \dots\wedge v_n,\omega\rangle\sum_{1\leq i_1<\dots<i_k\leq n}\frac {\det(v,i_1\to Av_{i_1},\dots,i_k\to Av_{i_k})}{\det(v)}. $$ With $$ \text{tr}_k(A)=\sum_{1\leq i_1<\dots<i_k\leq n}\langle A(e_{i_1}\dots\wedge e_{i_k}),e^{i_1}\dots\wedge e^{i_k}\rangle $$ and using that $\Lambda^n(V)$ is one dimensional, we can conclude that $$ f_k(v_1,\dots,v_n)=\text{tr}_k(A)v_1\wedge \dots \wedge v_n $$ Hence $\text{trace}_k(A)=\text{tr}_k(A)$.

(Sorry about using different symbols like $A$ instead of $T$, $\text{tr}_k(A)$ instead of $\text{trace}(\Lambda^k A)$. I will try to be good next time.)

Answer 3

Here's another proof which uses mixed exterior algebra. The advantage is that it doesn't involve messy computations with bases, sign factors, determinants, etc. but the disadvantage is that it uses a bunch of other machinery. I summarize facts that are needed below, but for details see Chapter 6 of [1].

Background

As above let $V$ be $n$-dimensional, let $V^*$ be dual to $V$, and define $$\textstyle\bigwedge(V^*,V)=\bigwedge V^*\otimes\bigwedge V$$ to be the mixed exterior algebra over $V^*,V$, which is the canonical tensor product of the exterior algebras $\bigwedge V^*$ and $\bigwedge V$. The mixed exterior product (or "dot" product) satisfies $$(u^*\otimes u)\cdot(v^*\otimes v)=(u^*\wedge v^*)\otimes(u\wedge v)$$ for all $u^*,v^*\in\bigwedge V^*$ and $u,v\in\bigwedge V$. Importantly, this product is commutative in the diagonal subalgebra $$\Delta(V^*,V)=\bigoplus_{p=0}^n\Delta_p(V^*,V)\qquad\text{where}\qquad\Delta_p(V^*,V)=\textstyle\bigwedge^p V^*\otimes\bigwedge^p V$$ to which we restrict attention. For $z\in\Delta(V^*,V)$, we write $$z^k=\frac{1}{k!}\underbrace{z\cdots z}_{k\text{ factors}}$$

There is an inner product in $\Delta(V^*,V)$ induced by the scalar product between $V^*$ and $V$ which satisfies $$\langle u^*\otimes u,v^*\otimes v\rangle=\langle u^*,v\rangle\langle v^*,u\rangle$$

There is a canonical linear isomorphism $T:\Delta(V^*,V)\to H(\bigwedge V;\bigwedge V)$ to the space of homogeneous linear transformations of $\bigwedge V$ which satisfies $$T(u^*\otimes u)(v)=\langle u^*,v\rangle u$$ Making appropriate identifications under $T$, we write $$\textstyle H(\bigwedge V;\bigwedge V)=\displaystyle\bigoplus_{p=0}^n L(\textstyle\bigwedge^p V;\bigwedge^p V)$$ so in particular all linear transformations of $V$ and their exterior powers are in there. We make $T$ an algebra isomorphism by defining the (generalized) box product $$\alpha\mathbin{\square}\beta=T(T^{-1}\alpha\cdot T^{-1}\beta)$$ for homogeneous $\alpha,\beta:\bigwedge V\to\bigwedge V$. For $z\in V^*\otimes V$ and $\varphi=T(z):V\to V$, it can be shown that $$T(z^k)=\frac{1}{k!}T(\underbrace{z\cdots z}_{k\text{ factors}})=\frac{1}{k!}\underbrace{\varphi\mathbin{\square}\cdots\mathbin{\square}\varphi}_{k\text{ factors}}=\textstyle\bigwedge^k\varphi$$

$T$ is also an isometry from the inner product above to the trace form $\langle\alpha,\beta\rangle=\mathop{\mathrm{tr}}(\alpha\circ\beta)$. If we define the unit tensor $t=T^{-1}(\iota)$ where $\iota:V\to V$ is the identity map, then for $z\in\Delta_k(V^*,V)$ it follows that $$\langle t^k,z\rangle=\mathop{\mathrm{tr}}(T(z))$$

Finally, it can be shown that $$i(t^k)t^n=t^{n-k}$$ where for $z\in\Delta(V^*,V)$, $i(z)$ is the insertion of $z$, dual to exterior multiplication by $z$: $$\langle i(z)x,y\rangle=\langle x,z\cdot y\rangle=\langle x,y\cdot z\rangle$$

Solution

For $\varphi:V\to V$ with $z=T^{-1}(\varphi)$, we have $$\begin{align*} \textstyle\mathop{\mathrm{tr}}(\bigwedge^k\varphi)&=\langle t^k,z^k\rangle\\ &=\langle i(t^{n-k})t^n,z^k\rangle\\ &=\langle t^n,z^k\cdot t^{n-k}\rangle\\ &=\textstyle\mathop{\mathrm{tr}}(\bigwedge^k\varphi\mathbin{\square}\bigwedge^{n-k}\iota) \end{align*}$$

Let $\Phi=\bigwedge^k\varphi\mathbin{\square}\bigwedge^{n-k}\iota$. Then $\Phi:\bigwedge^n V\to\bigwedge^n V$, and analysis of the box product shows that $$\Phi(v_1\wedge\cdots\wedge v_n)=\frac{1}{k!\,(n-k)!}\sum_{\sigma\in S_n}\varphi_{\sigma(1)}v_1\wedge\cdots\wedge\varphi_{\sigma(n)}v_n\tag{1}$$ where $$\varphi_i=\begin{cases} \varphi&\text{if }1\le i\le k\\ \iota&\text{if }k<i\le n \end{cases}$$

A moment's thought shows that (1) is your desired sum, since for every $1\le i_1<\cdots<i_k\le n$, there will be precisely $k!\,(n-k)!$ terms of form $$v_1\wedge\cdots\wedge\varphi v_{i_1}\wedge\cdots\wedge\varphi v_{i_k}\wedge\cdots\wedge v_n$$ and every term is of one of these forms.

References

1. Greub, W. Multilinear Algebra, 2nd ed. Springer, 1978.