(Example of a Non-Linear Transformation)
(Example of a Linear Transformation (system))
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and <math>\ X </math> is any vector in '''R'''<sup>''2''</sup>
 
and <math>\ X </math> is any vector in '''R'''<sup>''2''</sup>
  
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>
+
Therefore <math>\ T(C_1 e_1)= \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>
  
 
and
 
and

Revision as of 15:33, 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 linear combination of their respective 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 e_1)= \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) = Y $

Therefore the transformation is linear.

Example of a Non-Linear Transformation

Let a physical system be modeled by the following non-linear transformation:

$ \ T(x,y)= x\ln(y) $ Proof that it is not linear:

$ \ T(x_1, y_1) + T(x_2,y_2) = x_1 \ln(y_1)+ x_2 \ln(y_2) = U $

Now Summing the input signals and then putting it into the transformation yields:

$ \ T(x_1+x_2,y_1+y_2)= (x_1+x_2)\ln(y_1+y_2)=x_1 \ln(y_1+y_2) + x_2 \ln(y_1+y_2) = W $

$ \ U \ne W $ Therefore, transformation is not linear.

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