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Sergeu [11.5K]
3 years ago
6

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.
Mathematics
1 answer:
alexira [117]3 years ago
5 0

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]

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