Linear transformation : Linear transformation is a linear mapping of elements from one vector space to another vector space .
\[\left( \begin{array}{l}x\\y\end{array} \right) = a\left( \begin{array}{l}1\\2\end{array}\right)+b\left( \begin{array}{l}2\\1\end{array} \right) =\left( \begin{array}{l}a+2b\\2a+b\end{array} \right)\]
\[ \Rightarrow x = a+2b , y = 2a+b\]
\[\Rightarrow a = \frac{-x+2y}{3}, b =\frac{2x-y}{3} \]
\[\left( \begin{array}{l}x\\y\end{array} \right) =\frac{-x+2y}{3}\left( \begin{array}{l}1\\2\end{array}\right)+\frac{2x-y}{3}\left( \begin{array}{l}2\\1\end{array} \right)\]
\[\Rightarrow T\left( \begin{array}{l}x\\y\end{array} \right)=\frac{-x+2y}{3}T\left( \begin{array}{l}1\\2\end{array}\right)+\frac{2x-y}{3}T\left( \begin{array}{l}2\\1\end{array} \right) \]
\[=\frac{-x+2y}{3}\left( \begin{array}{l}3\\-3\end{array} \right)+\frac{2x-y}{3}\left( \begin{array}{l}3\\0\end{array} \right) \]
\[ = \left( \begin{array}{l}-x+2y\\x-2y\end{array} \right)+\left( \begin{array}{l}2x-y\\0\end{array} \right)\]
\[=a\left( \begin{array}{l}3a+3b\\-3\end{array} \right)+b\left( \begin{array}{l}3\\0\end{array} \right)\]
Matrix Representation
<span class=”has-inline-color has-vivid-red-color”>Question (28):</span> Let \(C =\{\left( \begin{array}{l}1\\2\end{array}\right), \left( \begin{array}{l}2\\1\end{array} \right)\}\) be a basis of \(R^2\) and \(T:R^2\rightarrow R^2\) be defined by \(T\left( \begin{array}{l}x\\y\end{array} \right)=\left( \begin{array}{l}x + y\\x – 2y\end{array} \right)\) . If \(T[C]\) represents the basis of \(T\) with respect to \(C\) then which of the following is true?\[\]
(A) \(T[C]=\left( {\begin{array}{*{20}{c}}{ – 3}&{ – 2}\\3&1\end{array}} \right)\) (B) \(T[C]=\left( {\begin{array}{*{20}{c}}{ 3}&{ – 2}\\-3&1\end{array}} \right)\)(C)\(T[C]=\left( {\begin{array}{*{20}{c}}{ – 3}&{ – 1}\\3&2\end{array}} \right)\) (D)\(T[C]=\left( {\begin{array}{*{20}{c}}{ 3}&{ – 1}\\-3&2\end{array}} \right)\)
Solution:
\[T\left( \begin{array}{l}1\\2\end{array}\right) =\left( \begin{array}{l}3\\-3\end{array} \right) = -3\left( \begin{array}{l}1\\2\end{array}\right)+3\left( \begin{array}{l}2\\1\end{array} \right)\]
\[T\left( \begin{array}{l}2\\1\end{array} \right) =\left( \begin{array}{l}3\\0\end{array} \right) =-1\left( \begin{array}{l}1\\2\end{array}\right)+2\left( \begin{array}{l}2\\1\end{array} \right) \]
\[\Rightarrow T[C]=\left( {\begin{array}{*{20}{c}}{ – 3}&{ – 1}\\3&2\end{array}} \right)\]
Therefore option \((C)T[C]=\left( {\begin{array}{*{20}{c}}{ – 3}&{ – 1}\\3&2\end{array}} \right)\) is correct