Let f : \mathbb{R}^{2} \to \mathbb{R}^{2} be the linear transformation defined by f(\vec{x}) = \left[\begin{array}{cc} 4 &3\cr 2 &1 \end{array}\right] \vec{x}. Let \begin{array}{lcl} \mathcal{B} & = & \lbrace \left<1,-1\right>, \left<2,-3\right> \rbrace, \\ \mathcal{C} & = & \lbrace \left<1,1\right>, \left<-2,-1\right> \rbrace, \end{array} be two different bases for \mathbb{R}^{2}. Find the matrix \lbrack f \rbrack_{\mathcal{B}}^{\mathcal{C}} for f relative to the basis \mathcal{B} in the domain and \mathcal{C} in the codomain.

Remember that [tex]\lbrack f \rbrack_{\mathcal{B}}^{\mathcal{C}}[/tex] is the matrix whose columns are the images under f of the vectors of the basis [tex]\mathcal{B}[/tex] written in the coordinates of the basis [tex]\mathcal{C}[/tex]. Then we have to do the following:

Find the [tex]\mathcal{C}[/tex] -- coordinates of any vector [tex]\vec{x}=\left<x,y\right>\in\mathbb{R}^2[/tex], that is, [tex]\lbrack \vec{x} \rbrack_{\mathcal{C}}[/tex].
Calculate the images under f of the vectors [tex]\vec{v_1}=\left<1,-1\right>[/tex] and [tex]\vec{v_2}= \left<2,-3\right>[/tex], that is, [tex]f(\vec{v_1})[/tex] and [tex]f(\vec{v_2})[/tex].
Find the [tex]\mathcal{C}[/tex] -- coordinates of [tex]f(v_1)[/tex] and [tex]f(v_2)[/tex], that is, [tex]\lbrack f(v_1) \rbrack_{\mathcal{C}}[/tex] and [tex]\lbrack f(v_2) \rbrack_{\mathcal{C}}[/tex].

For (1), note that any vector [tex]\vec{x}=\left<x,y\right>\in\mathbb{R}^2[/tex] can be written as [tex]\\ \vec{x}=\left<x,y\right>=(-x+2y)\left<1,1\right>+(-x+y)\left<-2,-1\right>=(-x+2y)\vec{v_1}+(-x+y)\vec{v_2}.[/tex] Therefore, the [tex]\mathcal{C}[/tex] -- coordinates of [tex]\vec{x}[/tex] are [tex]\\ \lbrack \vec{x} \rbrack_{\mathcal{C}}=\left<-x+2y,-x+y\right>.[/tex]

For (2), we calculate:

[tex]f(\vec{v_1}) = \left[\begin{array}{cc} 4 &3\cr 2 &1 \end{array}\right] \left[\begin{array}{c}\phantom{-}1 \cr -1 \end{array}\right]=\left[\begin{array}{c}1 \cr 1 \end{array}\right][/tex]

[tex]f(\vec{v_1}) = \left[\begin{array}{cc} 4 &3\cr 2 &1 \end{array}\right] \left[\begin{array}{c}\phantom{-}2 \cr -3 \end{array}\right]=\left[\begin{array}{c}-1 \cr \phantom{-}1 \end{array}\right][/tex]

Now we use the results obtained in steps (1) and (2) for finding [tex]\lbrack f(v_1) \rbrack_{\mathcal{C}}[/tex] and [tex]\lbrack f(v_2) \rbrack_{\mathcal{C}}[/tex] as requested in (3):

[tex]\lbrack f(v_1) \rbrack_{\mathcal{C}}=\left<-1+2,-1+1\right>=\left<1,0\right>[/tex]

[tex]\lbrack f(v_2) \rbrack_{\mathcal{C}}\left<1+2,1+1\right>=\left<3,2\right>.[/tex]

Therefore, the matrix [tex]\lbrack f \rbrack_{\mathcal{B}}^{\mathcal{C}}[/tex] for f relative to the basis [tex]\mathcal{B}[/tex] in the domain and [tex]\mathcal{C}[/tex] in the codomain is given by

[tex]\lbrack f \rbrack_{\mathcal{B}}^{\mathcal{C}}=\left[\begin{array}{cc} 1 &3\cr 0 &2 \end{array}\right][/tex]