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When expressing a set of variables $\{y_i \}$ as a linear combination of another set of variables $\{x_i\}$, the matrix expression for the case $2\times2$ is

$$ \left[\begin{array}{c} y_{1} \\ y_{2} \\ \end{array}\right]= \left[\begin{array}{cccc} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{array}\right] \left[\begin{array}{c} x_{1} \\ x_{2} \\ \end{array}\right] \tag{1}$$

so it is quite straightforward to express the $\{x_i\}$ variables in terms of the $\{y_i\}$ variables by by simply making use of the inverse matrix:

$$ \left[\begin{array}{c} x_{1} \\ x_{2} \\ \end{array}\right]= \left[\begin{array}{cccc} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{array}\right]^{-1} \left[\begin{array}{c} y_{1} \\ y_{2} \\ \end{array}\right] \\ =\frac{1}{a_{11}a_{22}-a_{12}a_{21}} \left[\begin{array}{cccc} a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{array}\right] \left[\begin{array}{c} y_{1} \\ y_{2} \\ \end{array}\right] \tag{2}$$

However, if we choose an arbitrary set of variables, say $\{x_1,y_2\}$, to be expressed in terms of the remaining variables $\{x_2,y_1\}$, would there be a matrix method to find from $(1)$ the corresponding matrix expression of $(3)$?

$${x_1}=\frac{{y_1}-{a_{12}} {x_2}}{{a_{11}}}, \quad {y_2}=\frac{{a_{21}} {y_1}+\left( {a_{11}} {a_{22}}-{a_{12}} {a_{21}}\right) {x_2}}{{a_{11}}} \tag{3}$$

Could that method be scaled for an arbitrary number $n$ of variables $\{y_i\}$ and $\{x_i\}$?

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  • $\begingroup$ Have a look at Schur complement. $\endgroup$
    – Tomek
    Jul 31, 2021 at 11:57

1 Answer 1

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If I have:

$$\begin{bmatrix}Y_1\\Y_2\end{bmatrix}=\begin{bmatrix}A&B\\C&D\end{bmatrix}\begin{bmatrix}X_1\\X_2\end{bmatrix}$$

Where $X_1,X_2$ and $Y_1,Y_2$ are $p\times1,q\times1$ matrices respectively (holding your desired parameters $x_0,x_1,\cdots$), and $A$ is a $p\times p$ matrix, $B$ is a $p\times q$ matrix, $C$ is a $q\times p$ matrix, and $D$ is a $q\times q$ matrix, collectively known as a block matrix. We get as a result: $Y_1=AX_1+BX_2$, $Y_2=CX_1+DX_2$. Suppose your desired parameter $x_i$ was in $X_1$, by way of example (the same process is applied if it were in $X_2$). Then you know that $AX_1=Y_1-BX_2\implies X_1=A^{-1}(Y_1-BX_2)$ and the $i$th element of $X_1$ will hold your parameter, $x_i$. If $A$ is not invertible, you should still be able to solve for the span of possible values of $X_1$ and thus the range of possibilities for $x_i$.

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