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Using Banach fixed-point theorem to show existence and uniqueness of the following system of linear equations

0

$begingroup$

Consider the following system of linear equations :
$$ begin{cases}2x_1 + x_2 = 5, \ x_1 - 5x_2 = -3.end{cases}$$

Using basic linear algebra, it is easy to show that $x_1 = 2$ and $x_2 = 1$ is the unique solution.

But I noticed that
$$T(x_1,x_2) := begin{pmatrix}frac{1}{2}(5-x_2)\ frac{1}{5}(3+x_1) end{pmatrix}$$
is a contraction :
begin{equation}begin{aligned}d(T(x_1,x_2),T(y_1,y_2)) &=sqrt{left(left(frac{1}{2}(5-x_2)right)- left(frac{1}{2}(5-y_2)right)right)^2 + left(left(frac{1}{5}(3+x_1)right) - left(frac{1}{5}(3+y_1)right)right)^2} \&=sqrt{frac{1}{4}(x_2-y_2)^2 + frac{1}{25}(x_1-y_1)^2} \&leq frac{1}{2}sqrt{(x_1-y_1)^2 + (x_2-y_2)^2} \&= frac{1}{2}d((x_1,x_2),(y_1,y_2))end{aligned}end{equation}

Hence, by the Banach fixed-point theorem it has a unique fixed point.

What puzzles me is that depending how we proceed to isolate $x_1$ and $x_2$, this approach may (surprisingly) fail. For instance, the following function is not a contraction :

$$S(x_1,x_2) := begin{pmatrix}-3+5x_2\ 5-2x_1 end{pmatrix}.$$

Am I missing something here?

linear-algebra contraction-operator

share|cite|improve this question

asked Mar 26 at 0:50

user420974user420974

294

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's not a contraction but it's an expansion, and its inverse is a contraction.
  $endgroup$
  – Ertxiem
  Mar 26 at 1:04
  

  
  
  
  

add a comment | 

0

$begingroup$

Consider the following system of linear equations :
$$ begin{cases}2x_1 + x_2 = 5, \ x_1 - 5x_2 = -3.end{cases}$$

Using basic linear algebra, it is easy to show that $x_1 = 2$ and $x_2 = 1$ is the unique solution.

But I noticed that
$$T(x_1,x_2) := begin{pmatrix}frac{1}{2}(5-x_2)\ frac{1}{5}(3+x_1) end{pmatrix}$$
is a contraction :
begin{equation}begin{aligned}d(T(x_1,x_2),T(y_1,y_2)) &=sqrt{left(left(frac{1}{2}(5-x_2)right)- left(frac{1}{2}(5-y_2)right)right)^2 + left(left(frac{1}{5}(3+x_1)right) - left(frac{1}{5}(3+y_1)right)right)^2} \&=sqrt{frac{1}{4}(x_2-y_2)^2 + frac{1}{25}(x_1-y_1)^2} \&leq frac{1}{2}sqrt{(x_1-y_1)^2 + (x_2-y_2)^2} \&= frac{1}{2}d((x_1,x_2),(y_1,y_2))end{aligned}end{equation}

Hence, by the Banach fixed-point theorem it has a unique fixed point.

What puzzles me is that depending how we proceed to isolate $x_1$ and $x_2$, this approach may (surprisingly) fail. For instance, the following function is not a contraction :

$$S(x_1,x_2) := begin{pmatrix}-3+5x_2\ 5-2x_1 end{pmatrix}.$$

Am I missing something here?

linear-algebra contraction-operator

share|cite|improve this question

asked Mar 26 at 0:50

user420974user420974

294

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's not a contraction but it's an expansion, and its inverse is a contraction.
  $endgroup$
  – Ertxiem
  Mar 26 at 1:04
  

  
  
  
  

add a comment | 

$begingroup$

Consider the following system of linear equations :
$$ begin{cases}2x_1 + x_2 = 5, \ x_1 - 5x_2 = -3.end{cases}$$

Using basic linear algebra, it is easy to show that $x_1 = 2$ and $x_2 = 1$ is the unique solution.

But I noticed that
$$T(x_1,x_2) := begin{pmatrix}frac{1}{2}(5-x_2)\ frac{1}{5}(3+x_1) end{pmatrix}$$
is a contraction :
begin{equation}begin{aligned}d(T(x_1,x_2),T(y_1,y_2)) &=sqrt{left(left(frac{1}{2}(5-x_2)right)- left(frac{1}{2}(5-y_2)right)right)^2 + left(left(frac{1}{5}(3+x_1)right) - left(frac{1}{5}(3+y_1)right)right)^2} \&=sqrt{frac{1}{4}(x_2-y_2)^2 + frac{1}{25}(x_1-y_1)^2} \&leq frac{1}{2}sqrt{(x_1-y_1)^2 + (x_2-y_2)^2} \&= frac{1}{2}d((x_1,x_2),(y_1,y_2))end{aligned}end{equation}

Hence, by the Banach fixed-point theorem it has a unique fixed point.

What puzzles me is that depending how we proceed to isolate $x_1$ and $x_2$, this approach may (surprisingly) fail. For instance, the following function is not a contraction :

$$S(x_1,x_2) := begin{pmatrix}-3+5x_2\ 5-2x_1 end{pmatrix}.$$

Am I missing something here?

linear-algebra contraction-operator

share|cite|improve this question

asked Mar 26 at 0:50

user420974user420974

294

$endgroup$

share|cite|improve this question

asked Mar 26 at 0:50

user420974user420974

294

share|cite|improve this question

asked Mar 26 at 0:50

user420974user420974

294

asked Mar 26 at 0:50

user420974user420974

294

asked Mar 26 at 0:50

user420974user420974

294