(→What does Linearity Mean?) |
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==What Does Linearity Mean?== | ==What Does Linearity Mean?== | ||
| − | Linearity describes | + | Linearity describes the special property of a transformation '''T''' from '''R'''<sup>''n''</sup> to '''R'''<sup>''m''</sup> such that any linear combination of its inputs yields a 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 '''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> | ||
| + | |||
| + | where <math> \mathbf{A} = \begin{bmatrix}cos(\pi/2) & sin(\pi/2)\\ -sin(\pi/2) & cos(\pi/2) \end{bmatrix} </math> | ||
| + | |||
| + | and <math>\ X </math> is any vector in '''R'''<sup>''2''</sup> | ||
| + | |||
| + | 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 | ||
| + | |||
| + | <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: | ||
| + | |||
| + | <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> | ||
| + | |||
| + | |||
| + | 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) = Y</math> | ||
| + | |||
| + | Therefore the transformation is linear. | ||
| + | |||
| + | ==Example of a Non-Linear Transformation== | ||
| + | Let a physical system be modeled by the following non-linear transformation: | ||
| + | |||
| + | <math>\ T(x,y)= x\ln(y) </math> | ||
| + | Proof that it is not linear: | ||
| + | |||
| + | <math>\ T(x_1, y_1) + T(x_2,y_2) = x_1 \ln(y_1)+ x_2 \ln(y_2) = U</math> | ||
| + | |||
| + | Now Summing the input signals and then putting it into the transformation yields: | ||
| + | |||
| + | <math>\ 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</math> | ||
| + | |||
| + | <math>\ U \ne W</math> | ||
| + | Therefore, transformation is not linear. | ||
Latest revision as of 12:50, 12 September 2008
What Does Linearity Mean?
Linearity describes the special property of a transformation T from Rn to Rm such that any linear combination of its inputs yields a 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.
