System of Linear Equations as Continuous Linear Transformation

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Theorem

Let $x_1, x_2 \in \R$ be real numbers.

Consider the following system of simultaneous linear equations $\paren S$:

\(\ds x_1\) \(=\) \(\ds \frac 1 2 x_1 + \frac 1 3 x_2 + 1\)
\(\ds x_2\) \(=\) \(\ds \frac 1 3 x_1 + \frac 1 4 x_2 + 2\)

Let $\norm {\, \cdot \,}_2$ be the Euclidean norm.

Let $\norm {\, \cdot \,}$ be the supremum operator norm.

Let $I$ be the $2 \times 2$ identity matrix.


Then:

$S$ is expressible as $\paren {I - K} x = y$ where $x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$, $y = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$, $K = \begin{pmatrix} \frac 1 2 & \frac 1 3 \\ \frac 1 3 & \frac 1 4 \end{pmatrix}$;
$\norm K < 1$;
$S$ admits a unique solution.


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Proof

$\paren S$ is expressible as $\paren {I - K} x = y$

We have that:

\(\ds x_1\) \(=\) \(\ds \frac 1 2 x_1 + \frac 1 3 x_2 + 1\)
\(\ds \leadsto \ \ \) \(\ds \frac 1 2 x_1 - \frac 1 3 x_2\) \(=\) \(\ds 1\)

Furthermore:

\(\ds x_1\) \(=\) \(\ds \frac 1 2 x_1 + \frac 1 3 x_2 + 1\)
\(\ds x_2\) \(=\) \(\ds \frac 1 3 x_1 + \frac 1 4 x_2 + 2\)
\(\ds \leadsto \ \ \) \(\ds - \frac 1 3 x_1 + \frac 3 4 x_2\) \(=\) \(\ds 2\)

Hence, $\paren S$ can be rewritten as:

\(\ds \frac 1 2 x_1 - \frac 1 3 x_2\) \(=\) \(\ds 1\)
\(\ds - \frac 1 3 x_1 + \frac 3 4 x_2\) \(=\) \(\ds 2\)


Moreover:

\(\ds I - K\) \(=\) \(\ds \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} \frac 1 2 & \frac 1 3 \\ \frac 1 3 & \frac 1 4 \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \begin{pmatrix} \frac 1 2 & -\frac 1 3 \\ -\frac 1 3 & \frac 3 4 \end{pmatrix}\)

By a simple inspection it is evident that $\paren S$ is expressible as $\paren {I - K} x = y$.

$\Box$

$\norm K < 1$

We have that:

\(\ds \norm {K x}_2^2\) \(=\) \(\ds \paren {\frac 1 2 x_1 + \frac 1 3 x_2}^2 + \paren {\frac 1 3 x_1 + \frac 1 4 x_2}^2\)
\(\ds \) \(\le\) \(\ds \paren {\frac 1 4 + \frac 1 9} \cdot \paren {x_1^2 + x_2^2} + \paren {\frac 1 9 + \frac 1 {16} } \cdot \paren {x_1^2 + x_2^2}\) Cauchy-Schwarz Inequality
\(\ds \) \(=\) \(\ds \paren {\frac 1 4 + \frac 1 9 + \frac 1 9 + \frac 1 {16} } \norm x_2^2\) Definition of Euclidean Norm

Hence:

\(\ds \norm K\) \(=\) \(\ds \sup_{\norm x_2 \mathop \le \mathop 1} \sqrt{ \paren {\frac 1 2 x_1 + \frac 1 3 x_2}^2 + \paren {\frac 1 3 x_1 + \frac 1 4 x_2}^2 }\) Definition of Supremum Operator Norm
\(\ds \) \(\le\) \(\ds \sup_{\norm x_2 \mathop \le \mathop 1} \sqrt{ \paren {\frac 1 4 + \frac 1 9 + \frac 1 9 + \frac 1 {16} } \norm x_2^2 }\)
\(\ds \) \(\le\) \(\ds \sqrt {\frac 1 4 + \frac 1 9 + \frac 1 9 + \frac 1 {16} }\)
\(\ds \) \(\le\) \(\ds \sqrt {\frac {1 + 16 + 16 + 9} {9 \cdot 16} }\)
\(\ds \) \(<\) \(\ds \sqrt {\frac {4 \cdot 16} {9 \cdot 16} }\)
\(\ds \) \(=\) \(\ds \sqrt {\frac 4 9 }\)
\(\ds \) \(=\) \(\ds \frac 2 3\)
\(\ds \) \(<\) \(\ds 1\)

$\Box$

$S$ admits a unique solution

By Neumann series theorem, $\paren {I - K}^{-1}$ exists in $\map {CL} {\R^2}$.

Hence:

\(\ds \forall x \in \R^2: \, \) \(\ds \paren {I - K} x\) \(=\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds \paren {I - K}^{-1} \paren {I - K} x\) \(=\) \(\ds \paren {I - K}^{-1} y\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \paren {I - K}^{-1} y\) Definition of Invertible Continuous Linear Operator

By Inverse of Invertible 2 x 2 Real Square Matrix:

\(\ds \paren {I - K}^{-1}\) \(=\) \(\ds \frac 1 {\frac 1 2 \cdot \frac 3 4 - \paren {- \frac 1 3} \cdot {- \frac 1 3} } \begin{pmatrix} \frac 3 4 & \frac 1 3 \\ \frac 1 3 & \frac 1 2 \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \frac {72}{19} \begin{pmatrix} \frac 3 4 & \frac 1 3 \\ \frac 1 3 & \frac 1 2 \end{pmatrix}\)

Therefore:

\(\ds x\) \(=\) \(\ds \frac {72}{19} \begin{pmatrix} \frac 3 4 & \frac 1 3 \\ \frac 1 3 & \frac 1 2 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \frac {72}{19} \begin{pmatrix} \frac 3 4 \cdot 1 + \frac 1 3 \cdot 2 \\ \frac 1 3 \cdot 1 + \frac 1 2 \cdot 2 \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \frac {72}{19} \begin{pmatrix} \frac {17} {12} \\ \frac 4 3 \end{pmatrix}\)


Sources

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