Problem
(1 point) Let $f: R^{2} \rightarrow R^{2}$ be the linear transformation defined by \[ f(x)=\left[\begin{array}{cc} 2 & -4 \\ -5 & 0 \end{array}\right] \mathbf{x} \] Let \[ \begin{array}{l} \mathcal{B}=\{\langle-1,1\rangle,\langle 2,-1\rangle\} \\ \mathcal{C}=\{\langle 1,-1\rangle,\langle 1,-2\rangle\} \end{array} \] be two different bases for $R^{2}$. a. Find the matrix $[f]_{\mathcal{B}}^{\mathcal{B}}$ for $f$ relative to the basis $\mathcal{B}$. \[ [f]_{\mathcal{B}}^{\mathcal{B}}=[ \] b. Find the matrix $[f]_{\mathcal{C}}^{\mathcal{C}}$ for $f$ relative to the basis $\mathcal{C}$. c. Find the transition matrix $[I]_{\mathcal{C}}^{\mathcal{B}}$ from $\mathcal{C}$ to $\mathcal{B}$ \[ \left[I_{\mathcal{C}}^{\mathcal{B}}=\left[\begin{array}{lll} \square & \\ \square & & \\ \square \end{array}\right]\right. \] d. Find the transition matrix $[I]_{\mathcal{B}}^{\mathcal{C}}$ from $\mathcal{B}$ to $\mathcal{C}$. (Note: $[I]_{\mathcal{B}}^{\mathcal{C}}=\left([I]_{\mathcal{C}}^{\mathcal{B}}\right)^{-1}$,) \[ [I]_{\mathcal{B}}^{\mathcal{C}}=\left[\begin{array}{lll} \square & & \\ \square & & \\ \square & & \\ & & \end{array}\right] \] e. On paper, check that $[I]_{\mathcal{B}}^{\mathcal{C}}[f]_{\mathcal{B}}^{\mathcal{B}}[I]_{\mathcal{C}}^{\mathcal{B}}=[f]_{\mathcal{C}}^{\mathcal{C}}$
Solution
Step 1 ：First, we apply the transformation \(f\) to each vector in the basis \(\mathcal{B}\).
Step 2 ：We then express the result as a linear combination of the basis vectors in \(\mathcal{B}\).
Step 3 ：The coefficients of these linear combinations will form the columns of the matrix representation.
Step 4 ：By performing these steps, we find that the matrix \([f]_{\mathcal{B}}^{\mathcal{B}}\) for \(f\) relative to the basis \(\mathcal{B}\) is \(\left[\begin{array}{cc} 4 & -12 \\ -1 & -2 \end{array}\right]\).
Step 5 ：\(\boxed{[f]_{\mathcal{B}}^{\mathcal{B}} = \left[\begin{array}{cc} 4 & -12 \\ -1 & -2 \end{array}\right]}\)
