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Let $J := J_r(\lambda)$ be an $r \times r$ Jordan block with complex eigenvalue $\lambda \neq 0$, and consider it acting on $V := \mathbb{C}^{r}$ in the usual way. Then we have the induced map:

$$J^{\otimes k} : V^{\otimes k} \rightarrow V^{\otimes k}$$ $$v_1 \otimes v_2 \otimes \ldots \otimes v_k \mapsto Jv_1 \otimes Jv_2 \otimes \ldots \otimes Jv_k$$

This restricts to an endomorphism of the subspace of alternating tensors $\wedge^k V$ which I will denote by $\wedge^k J$. My question is simply:

What is the Jordan decomposition of $\wedge^k J$?

What I know is that for any two Jordan blocks $J_r(\lambda)$, $J_s(\mu)$, $(r \leq s)$, both $\lambda \neq 0$ and $\mu \neq 0$, that the induced map on the tensor product $\mathbb{C}^r \otimes \mathbb{C}^s$ (whose matrix is simply the Kronecker product $J_r(\lambda) \boxtimes J_s(\mu)$) has Jordan decomposition:

$$J_r(\lambda) \otimes J_s(\mu) \cong J_{r-s+1}(\lambda\mu) \oplus J_{r-s+3}(\lambda\mu) \oplus \ldots \oplus J_{r+s-1}(\lambda\mu)$$

I found a nice paper by Richard Brualdi where he shows this titled "Combinatorial Verification of the Elementary Divisors of Tensor Products". I was hoping to use this formula to pick out the pieces corresponding to the alternating tensors, but that led nowhere. Now I'm hoping to extend his proof to this subspace, but before I spend too much time with that I'd like to make sure that the work hasn't already been done for me!

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