LMorphisms MA265F11Walther - Rhea

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 R^m

$ 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.