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Let $A$ = {triangles in $\mathbb{R^2}$}. We can let $(x_1,y_1)$,$(x_2,y_2)$,$(x_3,y_3)$ be the vertices of the triangle. The group $GL(2,\mathbb{R})$ acts on $A$ by acting on the vectors of the vertices which form the triangle. $GL(2,\mathbb{R})$ acts on $\mathbb{R}$ via the action: $y \in \mathbb{R}$, $B \in GL(2,\mathbb{R})$ then $B(y) = |\det(B)|y$. Now we let $f: A \rightarrow \mathbb{R}$ which maps a triangle in $A$ to its area. We want to show that $f$ is equivariant with respect to these group actions.

Solution attempt: Let $t \in X$ and $A \in GL(2, \mathbb{R})$. $A(f(t))=|\det(A)|f(t) $ where $f(t)$ is the area of triangle $t$. We can let the area of $t$ be represented by a 3x3 determinant from linear algebra. From here I thought of use determinant properties, but we cannot multiply a 2x2 to a 3x3 matrix.

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Hint: we may compute the area of a triangle via a $2\times2$ determinant of two displacement vectors.

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  • $\begingroup$ I think I know what your trying to say. We can split the 3x3 determinant of the area into a sum of three separate 2x2 determinants and then go from there? $\endgroup$
    – user77404
    Aug 6, 2013 at 0:51
  • $\begingroup$ No. Form two displacement vectors to represent two sides of the triangle. Put them together to form a $2\times2$ matrix. Halve the absolute value of the determinant and you have the area of the triangle. $\endgroup$
    – anon
    Aug 6, 2013 at 0:52
  • $\begingroup$ I think I remember this from linear algebra. When we have a parallelogram we can find the area of it by use the two displacement vectors. For triangle maybe we could divide by 2, it is along those lines, could I have a link to maybe a general formula? $\endgroup$
    – user77404
    Aug 6, 2013 at 0:56
  • $\begingroup$ @user77404 Yes (note that "halve" means "divide by 2"). You've just said the general formula yourself in your comment, but here's a link to some mathematics behind it. Although by 'general formula' perhaps you're referring to the fact that the determinant of a $n\times n$ matrix is the hypervolume of the parallelotope formed by the $n$ column vectors? In this case too, the general formula is right there in the previous sentence. $\endgroup$
    – anon
    Aug 6, 2013 at 0:59
  • $\begingroup$ Great, and from that we could simplify what we have above into 1/2(det(AT)) where T is the matrix that determined area of T and from this we can conclude this also equals f(A(t)) for equivariant. But the only thing is we are not told what action $Gl(2,R)$ has on the vertices, so will we be sure that f(A(t)) = 1/2(det(AT))? $\endgroup$
    – user77404
    Aug 6, 2013 at 1:08

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