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

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

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<x,y\right>\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<1,-1\right>` and
   `\vec{v_2}= \left<2,-3\right>`, 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<x,y\right>\in\mathbb{R}^2` can be written as
`\\ \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}.` Therefore, the
`\mathcal{C}` -- coordinates of
`\vec{x}` are
`\\ \lbrack \vec{x} \rbrack_{\mathcal{C}}=\left<-x+2y,-x+y\right>.`

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<-1+2,-1+1\right>=\left<1,0\right>`

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

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]`