Compacting a Matrix Suppose you have a matrix that is zero-valued everywhere except the diagonal.
As an example, take the identity matrix, $I$.  For this example, let's say you are using the $4\times 4$ version.
Is there an operation that can produce a $2\times 2$ matrix consisting of only the non-zero values?  If not, is there a notation to express this intent?
Example:
$$    I = \pmatrix{
    i_{11} &0    &0    &0\\
    0    &i_{22} &0    &0\\
    0    &0    &i_{33} &0\\
    0    &0    &0    &i_{44}}
$$
$$    I_2 =\pmatrix{ 
    i_{11} &i_{22}\\
    i_{33} &i_{44}}$$
 A: Sometimes one writes $\mbox{diag}(D)$ to represent the diagonal elements of $D$. This is also a Matlab command; to get your $I_2$ from $I$ in Matlab, you'd use the following code:
I2 = reshape(diag(I),2,2)';

The transpose operator ' is necessary to get the elements arranged in the order you specify. (Why you chose that order isn't clear to me, but that's not relevant.)
In math notation I'd write this:
For $I \in \mathbb R ^{4 \times 4}$, define  $I_2 \in \mathbb R ^{2 \times 2}$ by
$$
    I_2[i,j] = I[2(i-1)+j,2(i-1)+j] \mbox{ for } i,j \in \{1,2\}
$$
A: Here is a solution that you can find by solving a linear system. First we will obtain a $4\times 4$ matrix that looks like your target matrix ($I_2$) but padded. We let
$$A = \begin{bmatrix}a_1 & 0 & 0 & 0\\0 & a_2 & 0 & 0\\0 & 0 & a_3 & 0\\0 & 0 & 0 & a_4\end{bmatrix} \quad{}\text{and}\quad X = \begin{bmatrix}
 x_{1,1} & x_{1,2} & x_{1,3} & x_{1,4} \\
 x_{2,1} & x_{2,2} & x_{2,3} & x_{2,4} \\
 x_{3,1} & x_{3,2} & x_{3,3} & x_{3,4} \\
 x_{4,1} & x_{4,2} & x_{4,3} & x_{4,4}
\end{bmatrix}$$
When you multiply these two two matrices you obtain
$$AX = 

\begin{bmatrix}
 a_1 x_{1,1} & a_1 x_{1,2} & a_1 x_{1,3} & a_1 x_{1,4} \\
 a_2 x_{2,1} & a_2 x_{2,2} & a_2 x_{2,3} & a_2 x_{2,4} \\
 a_3 x_{3,1} & a_3 x_{3,2} & a_3 x_{3,3} & a_3 x_{3,4} \\
 a_4 x_{4,1} & a_4 x_{4,2} & a_4 x_{4,3} & a_4 x_{4,4}
\end{bmatrix}

$$
You want all the entries of this matrix to be $0$ except the four entries in the upper left corner (which we want to be $a_1,a_2,a_3$ and $a_4$). Solving the equations formed leads to 
$$ X =
\begin{bmatrix}
 1 & a_2/a_1 & 0 & 0 \\
 a_3/a_2 & a_4/a_2 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0
\end{bmatrix}
$$
and when you multiply $A$ and $X$ together you obtain
$$AX = \begin{bmatrix}
 a_1 & a_2 & 0 & 0 \\
 a_3 & a_4 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0
\end{bmatrix}$$
Note that the solution assumes that $a_1 \neq 0$ and $a_2 \neq 0$. If one or more of these is zero, we cannot solve the system. In your case it is okay because you say that the entries on the diagonal are non-zero. As a final step we need to remove two of the columns and two of the rows. Let $B$ be given by
$$B = 
\begin{bmatrix}
 1 & 0 \\
 0 & 1 \\
 0 & 0 \\
 0 & 0
\end{bmatrix}.$$
Then $B^T A X B = I_2$. Note that the matrix $X$ depends on the entries of matrix $A$. I am not sure if there is a matrix that will work on all such diagonal matrices.
A: An easy solution can be found if you can "see" that adding columns or rows in the diagonal matrix won't affect the diagonal elements but can "copy" values as needed.  For example, with these two transform matrices $A= \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$ and $B= \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{bmatrix}$, $A I B^T$ will give you the desired compressed form $I_2.$
Note for notation improvement: usually $I$ is "reserved" for identity matrix, most people'd prefer if you use $D$ instead for representing arbitrary diagonal matrices.
