I'm not exactly sure how to find the image of an arbitrary vector
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2 Answers
$\begingroup$$v_1$ and $v_2$ are linearly independent and form a basis. So you only need to write an arbitrary vector as linear combination of them.
Let $$\begin{bmatrix} x \\ y \end{bmatrix}=a\begin{bmatrix} 1 \\ 1 \end{bmatrix}+b\begin{bmatrix}2 \\ 3\end{bmatrix}$$ or $$\begin{bmatrix} 1 & 2 \\ 1 & 3\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}$$ $$\begin{bmatrix} a \\ b \end{bmatrix}=\begin{bmatrix}1&2\\1&3\end{bmatrix}^{-1}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3&-2\\-1 & 1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix} 3x-2y\\-x+y \end{bmatrix}$$ So $$T\begin{bmatrix} x\\y\end{bmatrix}=T\left( (3x-2y)v_1+(-x+y)v_2\right)$$ $$=(3x-2y)T(v_1)+(-x+y)T(v_2)$$ $$=(3x-2y)\begin{bmatrix}-11\\4\end{bmatrix}+(-x+y)\begin{bmatrix}-28\\13\end{bmatrix}$$ $$=\begin{bmatrix}-5x-6y\\-x+5y\end{bmatrix}$$
$\endgroup$ $\begingroup$Let $e_1=\left( \begin{array}{c} 1 \\ 0 \end{array} \right) $ and $e_2=\left( \begin{array}{c} 0 \\ 1 \end{array} \right) $ then $T\left( \begin{array}{c} x \\ y \end{array} \right) =xT(e_1)+yT(e_2)$ or $v_1=e_1+e_2$ and $v_2=2e_1+3e_2$ so $e_1=3v_1-v_2$ and $e_2=v_2-2v-1$, then $T\left( \begin{array}{c} x \\ y \end{array} \right) = = (3x-2y)T(v_1)+(y-x)T(v_2)=\left( \begin{array}{c} 5x-6y \\ -x+5y \end{array} \right)$
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