Matrix rearrangement I have a matrix A in this form: 
$$
A=
\left[
\begin{array}{cccc}
x_1 & x_2 & 0 & 0 \\
0 & 0& x_1 & x_2 
\end{array} \right]
$$
 Here, $x_1$ and $x_2$ are variables, and A is a $2 \times 4$ matrix. I 
would like to rearrange the matrix product so that
$$
 A^TA=BC
$$
$A^T$ is the transpose of A. $B$ and $C$ are matrices to be found. $B$ is a $4 \times 2$ matrix that only contains constants, and C is a $2 \times 4$ matrix that contains constants and variables ($x_1,x_2$ etc). Is this possible to have such $B$ and $C$?
 A: (Not really an answer but at least some development...)
Let's try it!
The left-hand side is easy to calculate:
$$
A^T A =
\left[
\begin{array}{cccc}
x_1^2 & x_1 x_2 & 0 & 0 \\
x_1 x_2 & x_2^2 & 0 & 0 \\
0 & 0 &x_1^2 & x_1 x_2  \\
0 & 0 &x_1 x_2 & x_2^2  \\
\end{array}
\right]
$$
The right-hand side should have this form also. Let's see what happens if we set
$$
B_0 = 
\left[
\begin{array}{cc}
b_{00} & b_{01} \\
b_{10} & b_{11} \\
\end{array}
\right]
\qquad
B_1 = 
\left[
\begin{array}{cc}
b_{20} & b_{21} \\
b_{30} & b_{31} \\
\end{array}
\right] \qquad
\Rightarrow \qquad
B_1 = 
\left[
\begin{array}{c}
B_0 \\ B_1
\end{array}
\right]
$$
and 
$$
C_0=
\left[
\begin{array}{cc}
c_{00} & c_{01} & \\
c_{10} & c_{11} & \\
\end{array}
\right]\qquad 
C_1=
\left[
\begin{array}{cc}
c_{02} & c_{03} \\
 c_{12} & c_{13} \\
\end{array}
\right]\qquad
\Rightarrow \qquad
C=
\left[
\begin{array}{cc} C_0 & C_1 \\
\end{array}
\right]
$$
Now their product is
$$
BC = \left[
\begin{array}{cc}
B_0 C_0 & B_0 C_1 \\
B_1 C_0 & B_1 C_1
\end{array}
\right]
$$
Now we get the equations
$$
\begin{array}{cl}
B_0 C_1 = B_1 C_0 &= 
\left[
\begin{array}{cc}
0& 0 \\
0 & 0 \\
\end{array}
\right] \\
B_0 C_0 = B_1C_1
&= 
\left[
\begin{array}{cc}
x_1^2 & x_1 x_ 2 \\
x_1 x_2 & x_2^2 \\
\end{array}
\right] \\
\end{array}
$$
Now, some insight would be needed to continue ...
A: We have $$A^T A = M = 
\left[
\begin{array}{cccc}
x_1^2 & x_1 x_2 & 0 & 0 \\
x_1 x_2 & x_2^2 & 0 & 0 \\
0 & 0 &x_1^2 & x_1 x_2  \\
0 & 0 &x_1 x_2 & x_2^2  \\
\end{array}
\right]$$
We will try to write this as $M = CB$, where $B$ is a $2*4$ matrix of constants, and $C$ is a $4*2$ matrix of constants and variables. I applied a transpose to the original problem in order to switch $B$ and $C$, because it's easier to have the right matrix be the matrix of constants. (ie. you have $ M = M^{t} = B^{t}C^{t}$, where $B^{t}$ would be your original $B$ . However, I work with its transpose).
Now, because $B$ is a $2*4$ matrix of constants, it has a constant kernel of dimension at least $2$. There exists at least $2$ constant linearly independent vectors $v_{1}$, $v_{2}$, such that $B{v_1} = 0$ and $B{v_2} = 0$ . Eg. if $B=
\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1& 0 & 0 
\end{array} \right]$, an example of such vectors would be $v_1 = 
\left[
\begin{array}{c}
0 \\
0 \\
1 \\
0 \\ 
\end{array} \right]$, $v_2 = 
\left[
\begin{array}{c}
0 \\
0 \\
0 \\
1 \\ 
\end{array} \right]$. Regardless of $B$, we can always find such vectors. Now, if $M= CB$, that means $M$ "inherits" the kernel-spanning constant vectors from $B$: there must exist constant linear independent vectors $v_1,v_2$ such that $Mv_1 = 0$ and $Mv_2 = 0$. The kernel of $M$ must always contain this constant subspace of dimension $2$ spanned by $v_1, v_2$.  For $x_1 = 1 $, $x_2 = 0$, then $M$ is
$$ M = 
\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0  & 0 & 0 & 0 \\
0 & 0 &1 & 0  \\
0 & 0 &0 & 0  \\
\end{array}
\right]$$
and its kernel is spanned by $\left[
\begin{array}{c}
0 \\
0 \\
0 \\
1 \\ 
\end{array} \right], 
\left[
\begin{array}{c}
0 \\
1 \\
0 \\
0 \\ 
\end{array} \right]$ . On the other hand, for $x_1 = 0$, $x_2 = 1$ gives
$$ M = 
\left[
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0  & 1 & 0 & 0 \\
0 & 0 &0 & 0  \\
0 & 0 &0 & 1  \\
\end{array}
\right]$$ and its kernel is  spanned by  $\left[ \begin{array}{c}
0 \\
0 \\
1 \\
0 \\ 
\end{array} \right], 
\left[
\begin{array}{c}
1 \\
0 \\
0 \\
0 \\ 
\end{array} \right]$. However, these two subspaces are disjoint (except the null space). They cannot both contain linearly independent constant vectors $v_1,v_2$ that span the kernel of $B$ in $M = CB$. Therefore, no decomposition of the form $M=CB$ with $B$ constant $2*4$ matrix exists. The kernel of $M$ "moves too much" for that to happen.  
