Rearranging a nested matrix expression into standard form

Question

I have four equations and four unknowns, which appear in the following two expressions: $$A\left[ \begin{array}{cc} x \\ y \end{array} \right] = \left[ \begin{array}{cc} b_1^T\left[ \begin{array}{cc} w \\ z \end{array} \right] \\ b_2^T\left[ \begin{array}{cc} w \\ z \end{array} \right] \end{array} \right]$$ and $$C\left[ \begin{array}{cc} x \\ y \end{array} \right] = \left[ \begin{array}{cc} d_1^T\left[ \begin{array}{cc} w \\ z \end{array} \right] \\ d_2^T\left[ \begin{array}{cc} w \\ z \end{array} \right] \end{array} \right]$$ Ideally, I want to solve for $(x,y,w,z)$. One approach is to find $E$ and $f$ such that the above equations can be written as $Ez=f$, where $z=(x,y,w,z)$, then invert $E$. We can assume $A\in\mathbb{R}^{2\times 2}$ and $C\in\mathbb{R}^{2\times 2}$ are invertible.

Question: How does one write this system in the form $Ez=f$ where $z=(x,y,w,z)$? Is this even necessary to solve for $(x,y,w,z)$ or is there a more direct approach?

Answer 1

$b_1^T\begin{bmatrix}w\\z\end{bmatrix}$ is just the dot product $b_{11}w+b_{12}z$ and similarly for the other terms on the right-hand side, so you can write the two equations as $$A\begin{bmatrix}x\\y\end{bmatrix}-B^T\begin{bmatrix}w\\z\end{bmatrix} = 0 \\ C\begin{bmatrix}x\\y\end{bmatrix}-D^T\begin{bmatrix}w\\z\end{bmatrix} = 0,$$ where the $b_i$ and $d_i$ are the columns of $B$ and $D$, respectively. You can consolidate this into $$\begin{bmatrix}A&-B^T\\C&-D^T\end{bmatrix}\begin{bmatrix}x\\y\\w\\z\end{bmatrix} = 0.$$

Answer 2

If a 2x2-matrix $A$ has the rows $a_1^T$ and $a_2^T$, then, setting $u = \binom xy$, we have $Au = \binom{a_1^Tu}{a_2^Tu}$. Hence you have, setting $v = \binom wz$, $Au = Bv$ and $Cu = Dv$, i.e., $Au-Bv = 0$ and $Cu-Dv = 0$. Hence $$ \begin{pmatrix}A&-B\\C&-D\end{pmatrix}\begin{pmatrix}x\\y\\w\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\\0\end{pmatrix}. $$