(Example of a Linear Transformation (system))
(Example of a Linear Transformation (system))
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==Example of a Linear Transformation (system)==
 
==Example of a Linear Transformation (system)==
The following linear transformation takes any vector in '''R'''<sup>''2''</sup> and maps it to another vector in '''R'''<sup>''2''</sup> of same length rotated 45 degrees counter clockwise.
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The following linear transformation takes any vector in '''R'''<sup>''2''</sup> and maps it to another vector in '''R'''<sup>''2''</sup> of same length rotated 45 degrees counter clockwise. Using the standard basis vectors:
  
 
<math>\ T(X)= \mathbf{A}X </math>
 
<math>\ T(X)= \mathbf{A}X </math>
  
where <math> \mathbf{A} = \begin{bmatrix}cos(\pi/2) & sin(\pi/2) \end{bmatrix} </math>
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where <math> \mathbf{A} = \begin{bmatrix}cos(\pi/2) & sin(\pi/2)\\ -sin(\pi/2) & cos(\pi/2) \end{bmatrix} </math>
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 +
and <math>\ X </math> is any vector in '''R'''<sup>''2''</sup>
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 +
Therefore <math>\ 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} </math>
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 +
and
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 +
<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>
 +
 
 +
Now summing the two vectors:
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 +
<math>\ T(C_1 e_1) + T(C_2 e_2) =
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<math> \ X= \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} </math>

Revision as of 11:41, 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 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} $

Now summing the two vectors:

$ \ T(C_1 e_1) + T(C_2 e_2) = <math> \ X= \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva