I tried to reason about this problem using the $RREF$ of the matrix of the linear transformation. Because I know that if I can get the $RREF$ to show homogeneous equations:
$x_1 = -x_2$
$x_2: free$
$x_3 = 2x_2$
Then we can write $\vec{x} = x_2\left( \begin{array}{ccc} -1 \\ 1 \\ 2 \end{array} \right)$ and just use the $RREF$ we found as the matrix of the linear transformation. However, I can't find a way to construct the $RREF$ such that $x_3 = 2x_2$ when we're in $\mathbb{R^3}$. Is this even possible?
Best Answer
If you let $x_3$ be your free variable instead of $x_2$,
you can get a linear transformation $T:\mathbb{R^3}\rightarrow\mathbb{R^2}$ by taking
$T(x,y,z)=A\begin{bmatrix}x\\y\\z\end{bmatrix}$ where A is any matrix with RREF $\;\;R=\begin{bmatrix}1&0&\frac{1}{2}\\0&1&-\frac{1}{2}\end{bmatrix}$.
Similarly, you can find a linear transformation $T:\mathbb{R^3}\rightarrow\mathbb{R^3}$ by taking
$T(x,y,z)=A\begin{bmatrix}x\\y\\z\end{bmatrix}$ where A is any matrix with RREF $\;\;R=\begin{bmatrix}1&0&\frac{1}{2}\\0&1&-\frac{1}{2}\\0&0&0\end{bmatrix}$.