| (8 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
| − | + | [[Category:bonus point project]] | |
| + | [[Category:linear algebra]] | ||
| + | [[Category:MA265]] | ||
| + | <center><font size= 4> | ||
| + | '''Linear Transformations and Isomorphisms''' | ||
| + | </font size> | ||
| + | |||
| + | Student project for [[MA265]] | ||
| + | </center> | ||
| + | ---- | ||
---- | ---- | ||
| Line 9: | Line 18: | ||
A <u>vector transformation</u> can transform a vector from R<sup>n</sup> to R<sup>m</sup> | A <u>vector transformation</u> can transform a vector from R<sup>n</sup> to R<sup>m</sup> | ||
| − | <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> | + | <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> |
| − | + | ||
| − | <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> | + | |
<br> | <br> | ||
| Line 31: | Line 32: | ||
<math>X=\left(\begin{array}{c}-1\\-2\end{array}\right)</math><br> | <math>X=\left(\begin{array}{c}-1\\-2\end{array}\right)</math><br> | ||
| − | <math>f(\left(\begin{array}{c}-1\\-2\end{array}\right | + | <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> |
<br> | <br> | ||
| Line 67: | Line 68: | ||
<br> <u>Example 1:</u> | <br> <u>Example 1:</u> | ||
| − | <br> | + | <br> We must check conditions (1) and (2) |
| − | <br><math> | + | <br>(1): |
| + | |||
| + | <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> | ||
<br><math>V=\left(\begin{array}{c}v_1\\v_2\end{array}\right)=\left(\begin{array}{c}2\\5\end{array}\right)</math> | <br><math>V=\left(\begin{array}{c}v_1\\v_2\end{array}\right)=\left(\begin{array}{c}2\\5\end{array}\right)</math> | ||
| Line 77: | Line 80: | ||
<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> | <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> | ||
| − | <br><math> | + | <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> |
| + | |||
| + | <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> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <br> and, | ||
| + | |||
| + | <math>\left(\begin{array}{c}5\\0\end{array}\right)NOT=\left(\begin{array}{c}1\\0\end{array}\right)</math> | ||
| + | |||
| + | <br> | ||
| + | |||
| + | 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> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <br> is NOT a linear transform, therefore we don't need to check (2). | ||
| + | |||
| + | <u>Example 2:</u><br> | ||
<br> We must check conditions (1) and (2) | <br> We must check conditions (1) and (2) | ||
| − | <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> |
| + | |||
| + | <br><math>X=\left(\begin{array}{c}x_1\\x_2\end{array}\right)</math> | ||
| + | |||
| + | <br><math>Y=\left(\begin{array}{c}x_1\\x_2\end{array}\right)</math> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <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> | ||
<br> | <br> | ||
| + | |||
| + | Therefore, condition (1) passes. | ||
| + | |||
| + | (2): | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | |||
| + | |||
| + | 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.
