Is this a simple change of variables, or is there something else going on? This is a question from a past statistics class, and I'm a bit puzzled by the first few steps in its solution.
Suppose $(X,Y)$ is uniformly distributed on the triangle with vertices $(0,0), (1,1), (-1,1)$. 
Suppose $(U,V)$ is uniformly distributed on the triangle with vertices $(0,0), (1,0), (0,1)$. 
At this point, it is just asserted that
$\begin{bmatrix}X \\Y \end{bmatrix} = \begin{bmatrix}1 & -1\\1 & 1\end{bmatrix}\begin{bmatrix}U \\ V \end{bmatrix}$
I don't see the justification in this. I think it's some kind of change of coordinate, from the first triangle to the second triangle, but I don't really know what's going on.
There is a result in this class that if $X$ is uniformly distributed over a region $R$ in $\mathbb{R}^{n}$, then given a linear isomorphic $\phi: \mathbb{R}^{n} \to \mathbb{R}^{n}$, then $\phi{X}$ is uniformly distributed on $\phi{R}$.
But I just can't recognize this as being such a coordinate change. Can anyone help me see it, or explain what's going on?
 A: The matrix:
$\begin{bmatrix}1 & -1 \\ 1 & 1\end{bmatrix}$ is a transformation matrix. What you need to look at is whether the coordinates are transformed in the right way.
With the triangle spanning over the vertices (0,0),(1,0),(0,1), you need to check that (0,0) is transformed to (0,0), (1,0) is transformed to (1,1) and finally that (0,1) is transformed to (-1,1).
This is actually pretty simple in this case, because (1,0) and (0,1) are the standard coordinates of the coordinate system. Therefore the columns of the transformation matrix display the resulting coordinates!
Therefore it's true.
To come back to the lemma on uniform distributions: Since all matrix multiplications are liner, the first part of the lemma is a given. The second part is what we checked above!
A: Yes, the coordinate change is
\begin{gather*}
X = U - V \\ Y = U + V.
\end{gather*}
What is happening is the triangle with vertices at (0,0), (1,0) and (0,1) is first rotated around the origin by $\pi/4$ (or $45^\circ$). This corresponds to the transformation
\begin{gather*} X' = U\cos(\pi/4) - V\sin(\pi/4) \\
Y' = U\sin(\pi/4) + V\cos(\pi/4).
\end{gather*}
The triangle is then stretched by a factor of $\sqrt{2}$ in each direction. (This is why the determinant of $\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$ is $2$.) This stretching corresponds to the transformation
\begin{gather*}
X = \sqrt{2}X' \\ 
Y = \sqrt{2}Y'.
\end{gather*}
Notice that $$\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} \sqrt{2} & 0 \\ 0 & \sqrt{2} \end{pmatrix}\begin{pmatrix} \cos(\pi/4) & -\sin(\pi/4) \\ \sin(\pi/4) & \cos(\pi/4)\end{pmatrix}.$$
This is just the composition of the two transformations (rotation and dilation). You then have to use the result you mentioned about linear transformations to conclude that the resulting distribution is uniform. 
Note also that this is only true for linear (or affine) transformations. For instance, a common mistake is that if $\theta \sim U(0,2\pi)$ and $R \sim U(0,1)$, then $X = R\cos(\theta), Y = R\sin(\theta)$ is uniformly distributed on the unit disk, which is not true. 
