Mapping from $\Bbb{R}^2$ onto $\Bbb{R}^3$

Question

For the linear transformation:

P: $\Bbb{R}^3$ → $\Bbb{R}^3$, P($x,y,z$) = $(4z-x,4y-z,4x-y)$

I have calculated the corresponding matrix as

\begin{matrix} -1 & 0 & 4 \\ 0 & 4 & -1 \\ 4 & -1 & 0 \\ \end{matrix}

If this working is correct and I want to calculate the linear transformation for a mapping of $\Bbb{R}^2$ → $\Bbb{R}^3$ do I just add a $z$ value of 0 on the end?

For example the matrix for,

Q: $\Bbb{R}^2$ → $\Bbb{R}^3$, Q$(x,y)$ = $(x − y,3y,y − 2x)$

would be

\begin{matrix} 1 & 0 & -2 \\ -1 & 3 & 1 \\ 0 & 0 & 0 \\ \end{matrix}

Answer 1

Your first matrix is correct. How did you calculate it? Because you don't seem to have calculated the second matrix in the same way at all.

Plug the values $\begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} 1\\0 \end{pmatrix}$ and $\begin{pmatrix} 0\\1 \end{pmatrix}$ into your function $Q$, and see what you get. Put them next to each other. That's the resulting matrix. I get $\begin{pmatrix} 1&-1\\ 0&3\\ -2&1\end{pmatrix}.$

Answer 2

The first matrix is correct.

For a linear transformation $\Bbb R^2\to\Bbb R^3$ your matrix will simply have only two columns instead of three but the process is the same.

---

From what you have tried to do I can see that the first matrix was correct only because it was symmetric. The image of the first basis vector is the first column of the matrix, not the first row.