Weight of dual Galois representation Let $K$ be a number field. We say that an $\ell$-adic Galois representation is of pure weight $n$ if for almost all primes $\mathfrak{p}$ the eigenvalues of $\text{Frob}_{\mathfrak{p}}$ are of absolute value $p^n$. A good source of these is the cohomology of smooth proper $K$-schemes, where $H^n(X_{\bar{K}},\mathbb{Q}_\ell)$ is of weight $n$. My question is how does the weight behave with respect to the dual? Since $\mathbb{Q}_\ell(1):=H^2(\mathbb{P}^1)^*$ is of weight $2$, you would expect that the weight of the dual $\rho^*$ is minus the weight of $\rho$. However I don't know how to take an eigenvector of $\rho$ and get an eigenvector of $\rho^*$....
 A: In any finite-dimensional $K$-vector space $V$, an endomorphism $u$ has the same characteristic polynomial as its dual. It’s easy to see that, for instance, by considering the matrix of $u$ in a certain basis $\mathcal{B}$, and the matrix of the dual of $u$ In the basis dual to $\mathcal{B}$ – and see that they are transposes one of the other.
However, as David Loeffler points out, the dual representation is considered by taking the inverse of the dual of the group action (indeed, $u \in End(V) \longmapsto u^* \in End(V^*)$ is contravariant, so to restore the correct fonctoriality we need to take the inverse as well).
In particular, this means that the dual of a representation of weight $n$ is of weight $-n$.
However, this doesn’t mean that it’s easy to construct an eigenvector of the dual representation directly from an eigenvector of the original representation. Instead, we need to use a full diagonalization (or at least a trigonalization) of the Frobenius, and then take the dual basis to really “see” the weights acting on the dual representation.
