6

Suppose $V$ is a vector space over $k$ and $\dim(V) = N$. Let $A \in\operatorname{End}(V)$. Let $\wedge^n A \in \operatorname{End}(\wedge^n V)$ where $\wedge^n$ is the $n$-th exterior power. I am having trouble finding an expression for $\operatorname{tr}(\wedge^n A)$. The case when $A$ is diagonalizable is solved here: Symmetric and exterior power of representation

My attempts so far:

  1. Use $\phi: \wedge^n V \to V^{\otimes n}$ where $\left[v\right] \overset{\phi}{\mapsto} \sum_{\sigma \in S_n}\operatorname{sign}(\sigma) \sigma(v)$ (where $\left[v\right]$ denotes the congruence class of $v \in V^{\otimes n}$ in $\wedge^n V$).
  2. Try to write down the restriction of the map $A^{\otimes n}$ to $\phi(\wedge^n A)$ as a matrix. This seems tricky and I am stuck on figuring out all the indices.

Is there a more natural way to find $\operatorname{tr}(\wedge^n A)$ without writing down the matrix?

Community
  • 1
beeflavor
  • 881

1 Answers1

5

Recall that if $\beta = \{e_1,\dotsc,e_N\}$ is a basis for $V$, then $\wedge^n \beta := \{e_{i_1} \wedge \cdots \wedge e_{i_n}\}_{i_1 < \cdots < i_n}$ is a basis for $\wedge^n V$. Then, if we write $$ A e_j = \sum_{i=1}^N A_{ij}e_i, $$ we have that $$ (\wedge^n A)(e_{j_1} \wedge \cdots \wedge e_{j_n}) = Ae_{j_1} \wedge \cdots \wedge Ae_{j_n}\\ = \left(\sum_{i_1=1}^N A_{i_1 j_1} e_{i_1} \right) \wedge \cdots \wedge \left(\sum_{i_n=1}^N A_{i_n j_n} e_{i_n} \right)\\ = \sum_{i_1 < \cdots < i_n} \left( \sum_{\pi \in S_n} (-1)^\pi A_{i_{1}j_{\pi(1)}} \cdots A_{i_{n}j_{\pi(n)}}\right) \; e_{i_1} \wedge \cdots \wedge e_{i_n}\\ = \sum_{i_1 < \cdots < i_n} [A]_\beta(i_1,\dotsc,i_n;j_1,\dotsc,j_n) \; e_{i_1} \wedge \cdots \wedge e_{i_n}, $$ where $T(i_1,\dotsc,i_n;j_1,\dotsc,j_n)$ denotes the $n$-th order minor of $T \in M_N(F)$ corresponding to the $n$ rows $i_1 < \dotsc < i_n$ and $n$ columns $j_1 < \dotsc < j_n$. Hence, $[\wedge^n A]_{\wedge^n\beta}$ is precisely the matrix of $n$-th order minors of $[A]_\beta$, and, in particular, $$ \operatorname{Tr}(\wedge^n A) = \sum_{i_1 < \cdots < i_n} [A]_\beta(i_1,\dotsc,i_n;i_1,\dotsc,i_n) = \sum_{i_1 < \cdots < i_n} \sum_{\pi \in S_n} (-1)^\pi A_{i_{1}i_{\pi(1)}} \cdots A_{i_{n}i_{\pi(n)}}. $$

The theoretical upshot of all this is that $\wedge^n A$ is the coordinate-free, exterior-algebraic avatar of the matrix of $n$-th order minors, just as $\wedge^N A$, in particular, is the coordinate-free, exterior-algebraic avatar of the determinant. Indeed, the adjugate of $A$ can be defined, in coordinate-free, exterior-algebraic terms, as the unique operator $A^{\text{adj}} \in \operatorname{End}(V)$ such that $$ \forall v \in V, \; \omega \in \wedge^{N-1}V, \quad A^{\text{adj}}v \wedge \omega = v \wedge (\wedge^{N-1}A)(\omega), $$ and you can then prove that $$ A^{\text{adj}}A = \det(A)1_V, $$ giving the translation into coordinate-free terms of the formula for the inverse of an invertible matrix.

Branimir Ćaćić
  • 13,684