## 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.