I have the $(x,y)$-plane $$\left\{(x,y,z)\in \mathbb{R}^3 | x,y\in \mathbb{R}, z = 0 \right\}.$$ I need a transformation matrix to transform this to the plane
$$ \left\{ (x,y,z) \in \mathbb{R}^3 | x+y+z=0 \right\}.$$
I found the Matrix:
$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & -1 & 0 \end{pmatrix}.$$
Is this correct?
On
there are many choices of this type of transformation: for exempla tack an as transform a base of the first plane in the base of the second plane and complete this transformation s.t transform a base of the first (beginnin) space to a base of the second (arrival) space that is: $\{(1,0,0),(0,1,0)\}$ this is a base for the first plane; $\{(1,0,-1),(0,1,-1)\}$ this is a base for the second plane, so my transformation could transform the first system in the second system above we complete this two system in a base respectively of $\Bbb{R}^3$ (beginning and arrival spaces) so we choice (simply) respectively the vectors $(0,0,1)$ and $(0,0,1)$ and the transformation that choice has matrix in the canonical bases of $\Bbb{R}^3$ the matrix $P= \left( \begin{array}{ccc} 1 & 0&0 \\ 0 & 1&0\\ -1&-1&1 \end{array} \right)$
This matrix does send the first plane bijectively to the second one, but it is not invertible. For example, you could take $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ as a third column.