Shear mapping defined through triangle ABC with A, B fixed points

Question

I am stuck on the following problem:

Let $\mathcal{A}^2$ be an affine plane with the associated vector space $V$ and the points $A, \: B=\tau_u A, \: C=\tau_v A \in \mathcal{A}^2$ be the corners of a non-degenerate triangle. Show that $$(A,B,C) \xrightarrow{\sigma} (A,B,\tau_u C)$$ defines a shear mapping.

I guess we have to show that the linear part of $\sigma$ is equal to $id_V + w\omega$, where $\omega \in V^\ast$ and $w\in ker(\omega)$, but how to do that?

I would be really grateful for any help (both for the formal proof and the geometric intuition behind).

Answer 1

A drawing should convince you that, with respect to affine basis $(A,\vec{AB},\vec{AC})$, the matrix of the transformation is:

$$\begin{pmatrix}1&1\\0&1\end{pmatrix}$$

which has the characteristic form of a shear (also called transvection) matrix:

$$\begin{pmatrix}1&a\\0&1\end{pmatrix}$$