Let $f : \mathbb{R}^{2} \to \mathbb{R}^{2}$ be the linear transformation defined by $$f(\vec{x}) = \left[\begin{array}{cc} 1 &-3\cr 0 &-3 \end{array}\right] \vec{x}.$$ Let $$\begin{array}{lcl} \mathcal{B} & = & \lbrace \left<1,2\right>, \left<2,5\right> \rbrace, \\ \mathcal{C} & = & \lbrace \left<-1,-1\right>, \left<2,3\right> \rbrace, \end{array}$$ be two different bases for $\mathbb{R}^{2}$.

1. Find the matrix $\lbrack f \rbrack_{\mathcal{B}}^{\mathcal{B}}$ for $f$ relative to the basis $\mathcal{B}$.
   
   $\lbrack f \rbrack_{\mathcal{B}}^{\mathcal{B}} =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$
   
   
2. Find the matrix $\lbrack f \rbrack_{\mathcal{C}}^{\mathcal{C}}$ for $f$ relative to the basis $\mathcal{C}$.
   
   $\lbrack f \rbrack_{\mathcal{C}}^{\mathcal{C}} =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$
   
   
3. Find the transition matrix $\lbrack I \rbrack_{\mathcal{C}}^{\mathcal{B}}$ such that $[v]_{\mathcal{B}} = \lbrack I \rbrack_{\mathcal{C}}^{\mathcal{B}} [v]_{\mathcal{C}}$.
   
   $\lbrack I \rbrack_{\mathcal{C}}^{\mathcal{B}} =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$
   
   
4. Find the transition matrix $\lbrack I \rbrack_{\mathcal{B}}^{\mathcal{C}}$ such that $[v]_{\mathcal{C}} = \lbrack I \rbrack_{\mathcal{B}}^{\mathcal{C}} [v]_{\mathcal{B}}$.
   Reminder: $\lbrack I \rbrack_{\mathcal{B}}^{\mathcal{C}} = \left( \lbrack I \rbrack_{\mathcal{C}}^{\mathcal{B}} \right)^{-1}$.
   
   $\lbrack I \rbrack_{\mathcal{B}}^{\mathcal{C}} =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$
   
   
5. On paper, check that $\lbrack I \rbrack_{\mathcal{B}}^{\mathcal{C}} \lbrack f \rbrack_{\mathcal{B}}^{\mathcal{B}} \lbrack I \rbrack_{\mathcal{C}}^{\mathcal{B}} = \lbrack f \rbrack_{\mathcal{C}}^{\mathcal{C}}$.