The $-1$-eigenspace of the exchange map is $V \wedge V$

Question

Let $V$ be a vector space, and $T(\mathbf{v}_0 \otimes \mathbf{v}_1) = \mathbf{v}_1 \otimes \mathbf{v}_0: V \otimes V \to V \otimes V$ be a linear map.

Theorem:

The eigenspace of $T$ with eigenvalue $-1$ is $V \wedge V$.

How to prove this theorem? I am able to prove that tensors of the form$\mathbf{v}_0 \otimes \mathbf{v}_1 - \mathbf{v}_1 \otimes \mathbf{v}_0$ are eigenvectors of $T$, but I could not prove that they are the only tensors with such property.

Answer 1

1. Fix a basis $e_i$ for $V$.

2. Let $v = \Sigma c_{ij} e_i \otimes e_j$ be a tensor in your space. If $v$ is an -1 eigenvector, then $v + Tv = 0$. But $v + Tv = \Sigma (c_{ij} + c_{ji}) e_i \otimes e_j$, which implies that $c_{ij} = - c_{ji}$. Hence $v = \Sigma c_{ij} (e_i \otimes e_j - e_j \otimes e_i)$.