Question

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.

Answer 1

Answer:

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

Step-by-step explanation:

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

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

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

For (2), we calculate:

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

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

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

$\lbrack f(v_1) \rbrack_{\mathcal{C}}=\left=\left$

$\lbrack f(v_2) \rbrack_{\mathcal{C}}\left=\left.$

Therefore, 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 is given by

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