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[[Category:bonus point project]]
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[[Category:linear algebra]]
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[[Category:MA265]]
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<center><font size= 4>
 
'''Linear Transformations and Isomorphisms'''
 
'''Linear Transformations and Isomorphisms'''
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</font size>
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Student project for [[MA265]]
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</center>
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----
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----
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<u>Vector Transformations:</u>
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A <u>vector transformation </u>is a function that is performed on a vector. (i.e. f:X-&gt;Y)
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A <u>vector transformation</u> can transform a vector from R<sup>n</sup> to R<sup>m</sup>
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<math>f:\left(\begin{array}{c}x_1\\x_2\\.\\.\\a_n\end{array}\right)-> \left(\begin{array}{c}y_1\\y_2\\.\\.\\y_m\end{array}\right)</math><br>Where<br><math>X = \left(\begin{array}{c}x_1\\x_2\\.\\.\\x_n\end{array}\right)</math><br> and<br><math>Y = \left(\begin{array}{c}y_1\\y_2\\.\\.\\y_m\end{array}\right)</math>
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<br>
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<u>Example 1:</u>
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<u></u>
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<br>
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<math>f(\left(\begin{array}{c}x_1\\x_2\end{array}\right))= \left(\begin{array}{c}x_1^2\\0\end{array}\right)</math><br>
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<math>X=\left(\begin{array}{c}-1\\-2\end{array}\right)</math><br>
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<math>f(\left(\begin{array}{c}-1\\-2\end{array}\right)= \left(\begin{array}{c}-1^2\\0\end{array}\right)= \left(\begin{array}{c}1\\0\end{array}\right)</math>
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<br>
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<u>Example 2:</u>
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<br>
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<math>f(\left(\begin{array}{c}x_1\\x_2\end{array}\right))= \left(\begin{array}{c}-x_1\\x_1 - x_2\\x_1\end{array}\right)</math><br>
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<math>X=\left(\begin{array}{c}-1\\4\end{array}\right)</math><br>
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<math>f(\left(\begin{array}{c}-1\\4\end{array}\right))= \left(\begin{array}{c}-(-1)\\-1 - 4\\-1\end{array}\right)= \left(\begin{array}{c}1\\- 5\\-1\end{array}\right)</math>
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<br>
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<u>Linear Transformations</u>'':''
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A function L:V-&gt;W is a <u>linear transformation </u>of V to W if the following are true:
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<br>
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(1) L(u+v) = L(u) + L(v)
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(2) L(c*u) = c*L(u)
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<br>
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In other words, a <u>linear transformation </u>is a <u>vector transformation </u>that also meets (1) and (2) denoted from now on as L:V -&gt;W
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<br>
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Let's return to examples 1 and 2 to see if they are <u>linear transformations</u>.<br>
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<br> <u>Example 1:</u>
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<br> We must check conditions (1) and (2)
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<br>(1):
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<br><math>L(\left(\begin{array}{c}u_1\\u_2\end{array}\right))= \left(\begin{array}{c}u_1^2\\0\end{array}\right)</math>
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<br><math>V=\left(\begin{array}{c}v_1\\v_2\end{array}\right)=\left(\begin{array}{c}2\\5\end{array}\right)</math>
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<br><math>L(\left(\begin{array}{c}u_1 + v_1\\u_2 + v_2\end{array}\right))= \left(\begin{array}{c}(u_1 + v_1)^2\\0\end{array}\right)</math>
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<br><math>L(\left(\begin{array}{c}-1 + 2\\-2 + 5\end{array}\right))= \left(\begin{array}{c}(-1 + 2)^2\\0\end{array}\right)= \left(\begin{array}{c}1\\0\end{array}\right)</math>
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<br><math>L(\left(\begin{array}{c}-1\\-2\end{array}\right))= \left(\begin{array}{c}-1^2\\0\end{array}\right)= \left(\begin{array}{c}1\\0\end{array}\right)</math>
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<br><math>L(\left(\begin{array}{c}-1\\-2\end{array}\right)= \left(\begin{array}{c}(-1)^2\\0\end{array}\right)+L(\left(\begin{array}{c}2\\5\end{array}\right))= \left(\begin{array}{c}(2)^2\\0\end{array}\right)</math>
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<br><math>\left(\begin{array}{c}1\\0\end{array}\right)+\left(\begin{array}{c}4\\0\end{array}\right)=\left(\begin{array}{c}5\\0\end{array}\right)</math>
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<br> and,
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<math>\left(\begin{array}{c}5\\0\end{array}\right)NOT=\left(\begin{array}{c}1\\0\end{array}\right)</math>
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<br>
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Therefore since, <br><math>L(\left(\begin{array}{c}u_1 + v_1\\u_2 + v_2\end{array}\right))NOT= L(\left(\begin{array}{c}u_1\\u_2\end{array}\right))+ L(\left(\begin{array}{c}v_1\\v_2\end{array}\right))</math>
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<br><math>L(\left(\begin{array}{c}u_1\\u_2\end{array}\right))= \left(\begin{array}{c}u_1^2\\0\end{array}\right)
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</math>
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<br> is NOT a linear transform, therefore we don't need to check (2).
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<u>Example 2:</u><br>
  
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''Vector Transformations:''
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<br> We must check conditions (1) and (2)
  
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A ''vector transformation'' is a function that is performed on a vector. (i.e. f:X->Y)
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<math>L(\left(\begin{array}{c}u_1\\u_2\end{array}\right))= \left(\begin{array}{c}-u_1\\u_1 - u_2\\u_1\end{array}\right)</math>
  
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<math>f:\left(\begin{array}{c}x_1\\x_2\\.\\.\\a_n\end{array}\right)-> \left(\begin{array}{c}y_1\\y_2\\.\\.\\y_n\end{array}\right)</math>
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<br><math>X=\left(\begin{array}{c}x_1\\x_2\end{array}\right)</math>  
  
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Where
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<br><math>Y=\left(\begin{array}{c}x_1\\x_2\end{array}\right)</math>  
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<math>X = \left(\begin{array}{c}x_1\\x_2\\.\\.\\x_n\end{array}\right)</math>
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and
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<math>Y = \left(\begin{array}{c}y_1\\y_2\\.\\.\\y_n\end{array}\right)</math>
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<br><math>L(\left(\begin{array}{c}X + Y\end{array}\right))=L(\left(\begin{array}{c}x_1 + y_1\\x_2 + y_2\end{array}\right))= \left(\begin{array}{c}-(x_1 + y_1)\\(x_1 + y_1) - (x_2 + y_2)\\(x_1 + y_1)\end{array}\right)</math>
  
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<br><math>L(\left(\begin{array}{c}X\end{array}\right))=L(\left(\begin{array}{c}x_1\\x_2\end{array}\right))= \left(\begin{array}{c}-(x_1)\\(x_1) - (x_2)\\(x_1)\end{array}\right)</math>
  
−
''Linear Transformations:''
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<br><math>L(\left(\begin{array}{c}Y\end{array}\right))=L(\left(\begin{array}{c}y_1\\y_2\end{array}\right))= \left(\begin{array}{c}-(y_1)\\(y_1) - (y_2)\\(y_1)\end{array}\right)</math>
  
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A function L:V->W is a ''linear transformation'' of V to W if the following are true:
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<br><math>\left(\begin{array}{c}-(x_1)\\(x_1) - (x_2)\\(x_1)\end{array}\right)+ \left(\begin{array}{c}-(y_1)\\(y_1) - (y_2)\\(y_1)\end{array}\right)= \left(\begin{array}{c}-(x_1 + y_1)\\(x_1 + y_1) - (x_2 + y_2)\\(x_1 + y_1)\end{array}\right)</math>  
  
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(1) L(u+v) = L(u) + L(v)
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<br>
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(2) L(c*u) = c*L(u)
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In other words, a ''linear transformation'' is a ''vector transformation'' that also meets (1) and (2).
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Therefore, condition (1) passes.  
  
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(2):
  
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<br><math>L(\left(\begin{array}{c}u_1\\u_2\end{array}\right))= \left(\begin{array}{c}-u_1\\u_1 - u_2\\u_1\end{array}\right)</math>
  
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<br><math>L(\left(\begin{array}{c}c*u_1\\c*u_2\end{array}\right))= \left(\begin{array}{c}-c*u_1\\c*u_1 - c*u_2\\c*u_1\end{array}\right)= \left(\begin{array}{c}-c*u_1\\c*(u_1 - u_2)\\c*u_1\end{array}\right)</math>
  
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<br><math>c*L(\left(\begin{array}{c}u_1\\u_2\end{array}\right))= c*\left(\begin{array}{c}-u_1\\u_1 - u_2\\u_1\end{array}\right)= \left(\begin{array}{c}-c*u_1\\c*(u_1 - u_2)\\c*u_1\end{array}\right)</math>
  
  
  
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<math>\left(\begin{array}{cccc}1&2&3&4\\5&6&7&8\end{array}\right)</math>
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Therefore, condition (2) passes, and we find that it is a <u>linear transformation.</u>
  
 
[[Category:MA265Fall2011Walther]]
 
[[Category:MA265Fall2011Walther]]

Latest revision as of 08:10, 11 April 2013


Linear Transformations and Isomorphisms

Student project for MA265

Vector Transformations:

A vector transformation is a function that is performed on a vector. (i.e. f:X->Y)

A vector transformation can transform a vector from Rn to Rm

$ f:\left(\begin{array}{c}x_1\\x_2\\.\\.\\a_n\end{array}\right)-> \left(\begin{array}{c}y_1\\y_2\\.\\.\\y_m\end{array}\right) $
Where
$ X = \left(\begin{array}{c}x_1\\x_2\\.\\.\\x_n\end{array}\right) $
and
$ Y = \left(\begin{array}{c}y_1\\y_2\\.\\.\\y_m\end{array}\right) $


Example 1:


$ f(\left(\begin{array}{c}x_1\\x_2\end{array}\right))= \left(\begin{array}{c}x_1^2\\0\end{array}\right) $

$ X=\left(\begin{array}{c}-1\\-2\end{array}\right) $

$ f(\left(\begin{array}{c}-1\\-2\end{array}\right)= \left(\begin{array}{c}-1^2\\0\end{array}\right)= \left(\begin{array}{c}1\\0\end{array}\right) $


Example 2:


$ f(\left(\begin{array}{c}x_1\\x_2\end{array}\right))= \left(\begin{array}{c}-x_1\\x_1 - x_2\\x_1\end{array}\right) $

$ X=\left(\begin{array}{c}-1\\4\end{array}\right) $

$ f(\left(\begin{array}{c}-1\\4\end{array}\right))= \left(\begin{array}{c}-(-1)\\-1 - 4\\-1\end{array}\right)= \left(\begin{array}{c}1\\- 5\\-1\end{array}\right) $


Linear Transformations:

A function L:V->W is a linear transformation of V to W if the following are true:


(1) L(u+v) = L(u) + L(v)

(2) L(c*u) = c*L(u)


In other words, a linear transformation is a vector transformation that also meets (1) and (2) denoted from now on as L:V ->W


Let's return to examples 1 and 2 to see if they are linear transformations.


Example 1:


We must check conditions (1) and (2)


(1):


$ L(\left(\begin{array}{c}u_1\\u_2\end{array}\right))= \left(\begin{array}{c}u_1^2\\0\end{array}\right) $


$ V=\left(\begin{array}{c}v_1\\v_2\end{array}\right)=\left(\begin{array}{c}2\\5\end{array}\right) $


$ L(\left(\begin{array}{c}u_1 + v_1\\u_2 + v_2\end{array}\right))= \left(\begin{array}{c}(u_1 + v_1)^2\\0\end{array}\right) $


$ L(\left(\begin{array}{c}-1 + 2\\-2 + 5\end{array}\right))= \left(\begin{array}{c}(-1 + 2)^2\\0\end{array}\right)= \left(\begin{array}{c}1\\0\end{array}\right) $


$ L(\left(\begin{array}{c}-1\\-2\end{array}\right))= \left(\begin{array}{c}-1^2\\0\end{array}\right)= \left(\begin{array}{c}1\\0\end{array}\right) $


$ L(\left(\begin{array}{c}-1\\-2\end{array}\right)= \left(\begin{array}{c}(-1)^2\\0\end{array}\right)+L(\left(\begin{array}{c}2\\5\end{array}\right))= \left(\begin{array}{c}(2)^2\\0\end{array}\right) $


$ \left(\begin{array}{c}1\\0\end{array}\right)+\left(\begin{array}{c}4\\0\end{array}\right)=\left(\begin{array}{c}5\\0\end{array}\right) $


and,

$ \left(\begin{array}{c}5\\0\end{array}\right)NOT=\left(\begin{array}{c}1\\0\end{array}\right) $


Therefore since,
$ L(\left(\begin{array}{c}u_1 + v_1\\u_2 + v_2\end{array}\right))NOT= L(\left(\begin{array}{c}u_1\\u_2\end{array}\right))+ L(\left(\begin{array}{c}v_1\\v_2\end{array}\right)) $


$ L(\left(\begin{array}{c}u_1\\u_2\end{array}\right))= \left(\begin{array}{c}u_1^2\\0\end{array}\right) $


is NOT a linear transform, therefore we don't need to check (2).

Example 2:


We must check conditions (1) and (2)

$ L(\left(\begin{array}{c}u_1\\u_2\end{array}\right))= \left(\begin{array}{c}-u_1\\u_1 - u_2\\u_1\end{array}\right) $


$ X=\left(\begin{array}{c}x_1\\x_2\end{array}\right) $


$ Y=\left(\begin{array}{c}x_1\\x_2\end{array}\right) $


$ L(\left(\begin{array}{c}X + Y\end{array}\right))=L(\left(\begin{array}{c}x_1 + y_1\\x_2 + y_2\end{array}\right))= \left(\begin{array}{c}-(x_1 + y_1)\\(x_1 + y_1) - (x_2 + y_2)\\(x_1 + y_1)\end{array}\right) $


$ L(\left(\begin{array}{c}X\end{array}\right))=L(\left(\begin{array}{c}x_1\\x_2\end{array}\right))= \left(\begin{array}{c}-(x_1)\\(x_1) - (x_2)\\(x_1)\end{array}\right) $


$ L(\left(\begin{array}{c}Y\end{array}\right))=L(\left(\begin{array}{c}y_1\\y_2\end{array}\right))= \left(\begin{array}{c}-(y_1)\\(y_1) - (y_2)\\(y_1)\end{array}\right) $


$ \left(\begin{array}{c}-(x_1)\\(x_1) - (x_2)\\(x_1)\end{array}\right)+ \left(\begin{array}{c}-(y_1)\\(y_1) - (y_2)\\(y_1)\end{array}\right)= \left(\begin{array}{c}-(x_1 + y_1)\\(x_1 + y_1) - (x_2 + y_2)\\(x_1 + y_1)\end{array}\right) $


Therefore, condition (1) passes.

(2):


$ L(\left(\begin{array}{c}u_1\\u_2\end{array}\right))= \left(\begin{array}{c}-u_1\\u_1 - u_2\\u_1\end{array}\right) $


$ L(\left(\begin{array}{c}c*u_1\\c*u_2\end{array}\right))= \left(\begin{array}{c}-c*u_1\\c*u_1 - c*u_2\\c*u_1\end{array}\right)= \left(\begin{array}{c}-c*u_1\\c*(u_1 - u_2)\\c*u_1\end{array}\right) $


$ c*L(\left(\begin{array}{c}u_1\\u_2\end{array}\right))= c*\left(\begin{array}{c}-u_1\\u_1 - u_2\\u_1\end{array}\right)= \left(\begin{array}{c}-c*u_1\\c*(u_1 - u_2)\\c*u_1\end{array}\right) $


Therefore, condition (2) passes, and we find that it is a linear transformation.

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