Matrix rearrangement

Question

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 $C$?

Answer 1

(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 ...

Answer 2

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.