A fact about the kernel of a linear transformation on an F[x]-vector space

Let V be an n-dimensional vector space over a field F with basis B = \{v_1, \ldots, v_n\}, let T be a linear transformation on V with matrix A, and make V into an F[x]-module in the usual way. That is, we have T(v_j) = \sum a_{i,j} v_i for each v_j and A = [a_{i,j}]. Let F[x]^n be the free module of rank n over F[x] and let E = \{ \xi_1, \ldots, \xi_n \} be a basis. Let \varphi : F[x]^n \rightarrow V be the (surjective) F[x]-module homomorphism given by defining \varphi(\xi_i) = v_i and extending linearly.

As demonstrated in the series of exercises beginning here, once we have a generating set for \mathsf{ker}\ \varphi, we can compute the invariant factors of A. In the next few exercises, we will find such a generating set.

Show that the elements \omega_j = \sum_{i=1}^n (\delta_{i,j}x - a_{i,j}) \xi_i = x \xi_j - \sum_{i=1}^n a_{i,j} \xi_i are in \mathsf{ker}\ \varphi, where \delta_{i,j} is the Kronecker delta.

Note that \varphi(\omega_j) = \varphi(x \xi_j - \sum a_{i,j} \xi_i) = x \varphi(\xi_j) - \sum a_{i,j} \varphi(\xi_i) = x \cdot v_j - \sum a_{i,j} v_i = T(v_j) - \sum a_{i,j} v_i = 0 as desired.

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