Let $\;A\in \mathbb M^{3\times 3}\;$ as $\;A=\begin{pmatrix} A_{11} \; A_{12} \;A_{13}\\ A_{21} \; A_{22} \;A_{23}\\ A_{31} \; A_{32} \;A_{33}\\ \end{pmatrix}\;$ and consider $\;x=\begin{pmatrix} x_1\\ x_2\\ x_3\\ \end{pmatrix}\in\mathbb R^3\;$. I need to find a transformation matrix $\;B\;$ such that: $\;ABx=-x_3 A_1 + A_2 + x_1 A_3\;$ where $\;A_i=\begin{pmatrix} A_{1i}\\ A_{2i}\\ A_{3i}\\ \end{pmatrix}\;\;i=1,2,3\;$
My attempt:
If $\;B=\begin{pmatrix} 0\;\;\;0\;-1 \\ 0\;\;\;\frac{1}{x_2}\;\;0 \\ 1\;\;\;0\;\;\;0\\ \end{pmatrix}\;$ then $\;ABx=\begin{pmatrix} A_{11} \; A_{12} \;A_{13}\\ A_{21} \; A_{22} \;A_{23}\\ A_{31} \; A_{32} \;A_{33}\\ \end{pmatrix}\begin{pmatrix} 0\;\;\;0\;-1 \\ 0\;\;\;\frac{1}{x_2}\;\;0 \\ 1\;\;\;0\;\;\;0\\ \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ x_3\\ \end{pmatrix}\;=\begin{pmatrix} A_{11} \; A_{12} \;A_{13}\\ A_{21} \; A_{22} \;A_{23}\\ A_{31} \; A_{32} \;A_{33}\\ \end{pmatrix} \begin{pmatrix} -x_3\\ 1\\ x_1\\ \end{pmatrix}\;=-x_3 \begin{pmatrix} A_{11}\\ A_{21}\\ A_{31}\\ \end{pmatrix} + \begin{pmatrix} A_{12}\\ A_{22}\\ A_{32}\\ \end{pmatrix} + x_1 \begin{pmatrix} A_{13}\\ A_{23}\\ A_{33}\\ \end{pmatrix}\;$
Questions:
Although I concluded to what I wanted, I'm a bit unsure if $\;B\;$ is indeed a transformation matrix.
- If such a transformation exists, then what's its geometric meaning?
- Instead of this $\;B\;$, could it be a rotation matrix ?
Any help would be valuable. Thanks in advance!
The problem is unsolvable for most matrices $A$. It's possible for matrices $A$ whose second column is all zeros.
We have a particular matrix $A$ and would like to find a matrix $B$ such that for all $x\in\mathbb R^3$:
$$ABx = A\tag{1}\begin{pmatrix}-x_3\\1\\x_1\\\end{pmatrix}$$
This is an equality between the functions $x\to ABx$ and the function of $x$ defined on the right hand side. The problem is that the function on the right hand side is non-linear (basically because of the $1$ in the second coordinate), so it can't be represented by a matrix $AB$. The problem with your solution is that $B$ depends on $x$.
If the second column of $A$ is not zero, then this is impossible. To see this, take $x=(0,\lambda,0)$. On the right you get $A_2$, and on the left you get $\lambda(AB)_2$, that is, $\lambda$ times the second column of $AB$. Thus we get $\forall \lambda,\ A_2=\lambda(AB)_2$. So, a constant on the left equals something non-constant on the right. Not possible - unless both sides are always equal to zero, so $A_2$ is the zero vector.