(Example of a Linear Transformation (system))
(Example of a Linear Transformation (system))
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and
 
and
  
<math>\ T(C_2 e2)= \begin{bmatrix}cos(\pi/2) & sin(\pi/2)\\ -sin(\pi/2) & cos(\pi/2)\end{bmatrix} \begin{bmatrix} 0 \\ C_2 \end{bmatrix} = \begin{bmatrix}C_2/ \sqrt(2) \\ C_2/\sqrt(2) \end{bmatrix} </math>
+
<math>\ T(C_2 e_2)= \begin{bmatrix}cos(\pi/2) & sin(\pi/2)\\ -sin(\pi/2) & cos(\pi/2)\end{bmatrix} \begin{bmatrix} 0 \\ C_2 \end{bmatrix} = \begin{bmatrix}C_2/ \sqrt(2) \\ C_2/\sqrt(2) \end{bmatrix} </math>
  
 
Now summing the two vectors:
 
Now summing the two vectors:
  
<math>\ T(C_1 e_1) + T(C_2 e_2) =  
+
<math>\ Y = T(C_1 e_1) + T(C_2 e_2) = \begin{bmatrix}\frac{C_1} {\sqrt(2)} + \frac{C_2}{\sqrt(2)} \\ \frac{-C_1}{\sqrt(2)} + \frac{C_2}{\sqrt(2)} \end{bmatrix}</math>
  
<math> \ X= \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} </math>
+
 
 +
Now summing the two vectors before putting them into the transformation:
 +
 
 +
<math> \ X= C_1 e_1 + C_2 e_2 = \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} </math>
 +
 
 +
<math> \ T(X)=  \begin{bmatrix}cos(\pi/2) & sin(\pi/2) \\ -sin(\pi/2) & cos(\pi/2)\end{bmatrix} \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} </math>
 +
 
 +
<math> \ T(X) = C_1 \begin{bmatrix} cos(\pi/2) \\ -sin(\pi/2) \end{bmatrix} + C_2 \begin{bmatrix}sin(\pi/2) \\ cos(\pi/2) \end{bmatrix} = \begin{bmatrix}\frac{C_1} {\sqrt(2)} + \frac{C_2}{\sqrt(2)} \\ \frac{-C_1}{\sqrt(2)} + \frac{C_2}{\sqrt(2)} \end{bmatrix}= T(C_1 e_1)+T(C_2 e_2)</math>
 +
 
 +
Therefore the transformation is linear.

Revision as of 12:17, 11 September 2008

What Does Linearity Mean?

Linearity describes a special property of a transformation T from Rn to Rm such that any linear combination of inputs yields the respective linear combination of their outputs. A transformation such as this remains closed under the operations of addition and scalar multiplication.

Example of a Linear Transformation (system)

The following linear transformation takes any vector in R2 and maps it to another vector in R2 of same length rotated 45 degrees counter clockwise. Using the standard basis vectors:

$ \ T(X)= \mathbf{A}X $

where $ \mathbf{A} = \begin{bmatrix}cos(\pi/2) & sin(\pi/2)\\ -sin(\pi/2) & cos(\pi/2) \end{bmatrix} $

and $ \ X $ is any vector in R2

Therefore $ \ T(C_1 e1)= \begin{bmatrix}cos(\pi/2) & sin(\pi/2)\\ -sin(\pi/2) & cos(\pi/2)\end{bmatrix} \begin{bmatrix} C_1 \\ 0 \end{bmatrix} = \begin{bmatrix}C_1/ \sqrt(2) \\ -C_1/\sqrt(2) \end{bmatrix} $

and

$ \ T(C_2 e_2)= \begin{bmatrix}cos(\pi/2) & sin(\pi/2)\\ -sin(\pi/2) & cos(\pi/2)\end{bmatrix} \begin{bmatrix} 0 \\ C_2 \end{bmatrix} = \begin{bmatrix}C_2/ \sqrt(2) \\ C_2/\sqrt(2) \end{bmatrix} $

Now summing the two vectors:

$ \ Y = T(C_1 e_1) + T(C_2 e_2) = \begin{bmatrix}\frac{C_1} {\sqrt(2)} + \frac{C_2}{\sqrt(2)} \\ \frac{-C_1}{\sqrt(2)} + \frac{C_2}{\sqrt(2)} \end{bmatrix} $


Now summing the two vectors before putting them into the transformation:

$ \ X= C_1 e_1 + C_2 e_2 = \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} $

$ \ T(X)= \begin{bmatrix}cos(\pi/2) & sin(\pi/2) \\ -sin(\pi/2) & cos(\pi/2)\end{bmatrix} \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} $

$ \ T(X) = C_1 \begin{bmatrix} cos(\pi/2) \\ -sin(\pi/2) \end{bmatrix} + C_2 \begin{bmatrix}sin(\pi/2) \\ cos(\pi/2) \end{bmatrix} = \begin{bmatrix}\frac{C_1} {\sqrt(2)} + \frac{C_2}{\sqrt(2)} \\ \frac{-C_1}{\sqrt(2)} + \frac{C_2}{\sqrt(2)} \end{bmatrix}= T(C_1 e_1)+T(C_2 e_2) $

Therefore the transformation is linear.

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