Rotating coordinates of a matrix Say I have a 2D image represented by a matrix $ A_{ij} $ with $i,j\in[0,..,N-1]$, so that to each entry $(i,j)$ corresponds its value $A(i,j)$. I have a new coordinates system with the indices being $u=i+j$ and $v=i-j$, and I want to get the matrix in this coordinate system $A_{u,v}$ by making:
\begin{eqnarray}
A_{uv}=\sum_{ij}R_{ui}^{\dagger}.A_{ij}.R_{jv}
\end{eqnarray}
I want to get the rotation matrices that follow the $u,v$ transformation of coordinates above. How is this achieved? I understand this corresponds to a change of basis from the canonical for $i,j$ to the one given by $u,v$, my problem is on how to get the entries of $A(u,v)$.
I am not sure I quite understand why you desire the form you mentioned, but we can compute a transformation matrix that given you the desired result. So, you are interested in finding a transformation as follows:
$$ \begin{bmatrix} u' \\ v'\end{bmatrix} = \mathbf{T} \begin{bmatrix} u \\ v\end{bmatrix} $$
where $u$ and $v$ are the original image coordinates and $u'$ and $v'$ are the transformed image coordinates.
You want a matrix so that $u' = u + v$ and $v' = u - v$. We do some algebra to find this matrix.
$$
\begin{align}
\begin{bmatrix} u+v \\ u-v \end{bmatrix} = \begin{bmatrix} \mathbf{T}_{00} &amp; \mathbf{T}_{10} \\ \mathbf{T}_{01} &amp; \mathbf{T}_{11} \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} 
\end{align}
$$
Then, multiplying through we have the following system of equations:
$$
\begin{align}
u + v &amp;= \mathbf{T}_{00} u + \mathbf{T}_{10} v \\
u - v &amp;= \mathbf{T}_{01} u + \mathbf{T}_{11} v
\end{align}
$$
Solving, you have the following:
$$
\begin{align}
\mathbf{T} = \begin{bmatrix} 1 &amp; 1 \\ 1 &amp; -1 \end{bmatrix}
\end{align}
$$
Unfortunately, this is not a proper rotation matrix because $\det[\mathbf{T}] \neq 1$, but the transformation matrix exists.
I am not sure I quite understand why you desire the form you mentioned, but we can compute a transformation matrix that given you the desired result. So, you are interested in finding a transformation as follows:
$$ \begin{bmatrix} u' \\ v'\end{bmatrix} = \mathbf{T} \begin{bmatrix} u \\ v\end{bmatrix} $$
where $u$ and $v$ are the original image coordinates and $u'$ and $v'$ are the transformed image coordinates.
You want a matrix so that $u' = u + v$ and $v' = u - v$. We do some algebra to find this matrix.
$$
\begin{align}
\begin{bmatrix} u+v \\ u-v \end{bmatrix} = \begin{bmatrix} \mathbf{T}_{00} &amp; \mathbf{T}_{10} \\ \mathbf{T}_{01} &amp; \mathbf{T}_{11} \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} 
\end{align}
$$
Then, multiplying through we have the following system of equations:
$$
\begin{align}
u + v &amp;= \mathbf{T}_{00} u + \mathbf{T}_{10} v \\
u - v &amp;= \mathbf{T}_{01} u + \mathbf{T}_{11} v
\end{align}
$$
Solving, you have the following:
$$
\begin{align}
\mathbf{T} = \begin{bmatrix} 1 &amp; 1 \\ 1 &amp; -1 \end{bmatrix}
\end{align}
$$
Unfortunately, this is not a proper rotation matrix because $\det[\mathbf{T}] \neq 1$, but the transformation matrix exists.