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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?

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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 $\;\;R=\begin{bmatrix}1&0&\frac{1}{2}\\0&1&-\frac{1}{2}\\0&0&0\end{bmatrix}$.

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$$T(x) = (x+y, z-2y)$$ That's all you need.

How'd I find it? I said that in the vector you gave, $x = y$, so I needed $x - y = 0$. So $x-y$ should be one of the components. Similarly for $y$ and $z$. And there was no need for a third term matching $x$ with $z$, but I could have included it, and written $$ T(x) = (x+y, z-2y, z+2x) $$ for instance.

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