Eigenvalues of the second exterior power of a linear operator Let $K$ be a field of characteristic zero and $V=K^n$. Let $T: V \to V$ be a linear map with eigenvalues $\lambda_1,...,\lambda_n \in K$ , not necessarily all distinct. Let $\wedge^2 V$ be the second exterior power of $V$ and $\wedge^2T: \wedge^2 V \to \wedge^2 V$ is the linear map defined as $\wedge^2T(x\wedge y)=T(x)\wedge T(y),\forall x,y\in V$ and extend it to whole of $\wedge ^2V$ linearly. 
What are all the eigenvalues of $\wedge^2T$ ? 
I can easily show that  all $\lambda_i\lambda_j$ s , with $i\ne j$ , are eigenvalues of $\wedge^2T$; my question is: are there any other eigenvalues ?
EDIT: I would accept an answer even if just for the $K=\mathbb C$ case  
 A: I guess that by saying that "$T$ has eigenvalues $\lambda_1, \dots, \lambda_n \in K$, not necessarily all distinct", you mean that the characteristic polynomial of $T$ splits over $K$ with roots $\lambda_1, \dots, \lambda_n$ which might have multiplicities. Since the characteristic polynomial of $T$ splits over $K$, you can find an ordered basis $(e_1, \dots, e_n)$ of $V$ with respect to which $T$ is given by an upper triangular matrix $A$ with the $\lambda_i$'s on the diagonal. That is, we have for each $1 \leq i \leq n$
$$ Te_i = \lambda_i e_i \mod \operatorname{span} \{ e_j \}_{j < i}$$
Let me introduce some notation. Denote by $\mathcal{P}_2(n)$ the collection of subsets of size two of $\{ 1, \dots, n \}$. We can order the elements of $\mathcal{P}_2(n)$ according to the lexicographical order. Namely, if $\alpha = \{ i, j \}$ with $i < j$ and $\beta = \{ k, l \}$ with $k < l$ then
$$ \alpha < \beta \iff (i < k) \textrm{ or } (i = k \textrm{ and } j < l) $$
so for example, in $\mathcal{P}_2(3)$ we have
$$ \{ 1, 2 \} < \{ 1, 3 \} < \{ 2, 3 \}. $$
Given $\alpha \in \mathcal{P}_2(n)$ we can write $\alpha$ uniquely as $\alpha = \{ i, j \}$ with $1 \leq i < j \leq n$. Set $e_{\alpha} = e_i \wedge e_j$. Then $(e_{\alpha})_{\alpha \in \mathcal{P}_2(n)}$ is an ordered basis for $\Lambda^2(V)$. For example, for $n = 3$ we have the ordered basis
$$ (e_1 \wedge e_2, e_1 \wedge e_3, e_2 \wedge e_3). $$
Now let $\alpha \in \mathcal{P}_2(n)$ with $\alpha = \{ i, j \}$ for $i < j$. Write
$$ Te_i = \lambda_i e_i + \sum_{k < i} c_k e_k,
\\ Te_j = \lambda_j e_j + \sum_{l < j} d_l e_l. $$
Then
$$ \Lambda^2(T)(e_{\alpha}) = Te_i \wedge Te_j = \left( \lambda_i e_i + \sum_{k < i} c_k e_k \right) \wedge \left( \lambda_j e_j + \sum_{l < j} d_l e_l \right) \\
= \lambda_i \lambda_j e_i \wedge e_j + \\
 \sum_{i < l < j} \lambda_i d_l e_i \wedge e_l - \sum_{l < i} \lambda_i d_l e_l \wedge e_i +
\\ \sum_{k < i} \lambda_j c_k e_k \wedge e_j + \\
 \sum_{k < i, l < j, k < l} c_k d_l e_k \wedge e_l - \sum_{k < i, l < j, k > l} c_k d_l e_l \wedge e_k\\
= \lambda_i \lambda_j e_i \wedge e_j \mod \operatorname{span} \{ e_{\beta} \}_{\beta < \alpha}. $$
Hence, $\Lambda^2(T)$ is represented with respect to the ordered basis $(e_{\alpha})$ by an upper triangular matrix whose diagonal elements are $\lambda_i \lambda_j$ with $i < j$. This means that the roots of the characteristic polynomial of $\Lambda^2(T)$ are precisely $\lambda_i \lambda_j$ for $i < j$ and those are the eigenvalues of $\Lambda^2(T)$.

Remark: This argument works over any field (no need to require that the characteristic of $K$ is zero) and generalizes in a straightforward way for higher exterior powers $\Lambda^k(T)$.
