Are each two triangles in $\mathbb R^n, \mathbb R^m$ affinely equivalent?

Question

I am wondering whether each two non-degenerate triangles (spanned by three affinely independent vertices) $T_1 \subseteq \mathbb R^n$, $T_2 \subseteq \mathbb R^m, n,m \geq 2,$ are affinely equivalent, i.e. there exists an affine map $f:\mathbb R^n \to \mathbb R^m$ with $f(x_i) = f(y_i)$ for each $i$, where $x_i, y_i$ are the vertices of the resp. triangle. And if so, is it possible to specify such a map?

Answer 1

Assuming that $T_1$ is not degenerate, $x_2-x_1$ and $x_3-x_1$ are linearly independent. Extend these two vectors to a basis for $\mathbb R^n$. You can then construct a linear transformation that maps $x_2-x_1$ to $y_2-y_1$, maps $x_3-x_1$ to $y_3-y_1$, and maps the other basis vectors to whatever you like.

Combine this with an appropriate translation such that your affine map sends $x_1$ to $y_1$ and you're done.