Linear transformation and its matrix

[lukaszh]

Hello everybody,
I have a problem. There is a linear trasformation \xi:\mathbb{R}^2\mapsto\mathbb{R}^2 and:
\xi\begin{pmatrix}3\\1\end{pmatrix}=\begin{pmatrix}2\\-4\end{pmatrix}
\xi\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}0\\2\end{pmatrix}
How to find a matrix for this linear transformation for basis \mathcal{B}=\left\{\begin{pmatrix}1\\2\end{pmatrix},\begin{pmatrix}2\\1\end{pmatrix}\right\}

Thank you so much :-)

 

[jojo12345]

There are a number of ways to do this. What have you tried so far? In general if you have a finite dimensional vector space V with a basis \{ e_{1},e_{2},...,e_{n}\} the components of a linear transformation T:V\rightarrow V are uniquely determined by Te_{i}=T^{j}_{i}e_{j} where the sum over j from 1 to n is implied. Furthermore, if you have a second basis \{f_1,f_2,...,f_n\} then the new and old basis vectors are related through e_{i}=A^{j}_{i}f_{j}.

 

[HallsofIvy]

Hello everybody,
I have a problem. There is a linear trasformation \xi:\mathbb{R}^2\mapsto\mathbb{R}^2 and:
\xi\begin{pmatrix}3\\1\end{pmatrix}=\begin{pmatrix}2\\-4\end{pmatrix}
\xi\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}0\\2\end{pmatrix}
How to find a matrix for this linear transformation for basis \mathcal{B}=\left\{\begin{pmatrix}1\\2\end{pmatrix},\begin{pmatrix}2\\1\end{pmatrix}\right\}

Thank you so much :-)

Write each of the vectors in B in terms of the first basis, <3, 1>, and <1, 1>, so that you can apply the linear transformation to them. For example, <1, 2>= a<3, 1>+ b<1, 1>= <3a+ b, a+ b>. 3a+ b= 1, a+ b= 2. Subtracting the second equation from the first, 2a= -1 so a= -1/2. Then b= 5/2: <1, 2>= (-1/2)<3,1>+ (5/2)<1, 1>. Applying the \xi to that, \xi<1, 2>= (-1/2)\xi<3,1>+ (5/2)\xi<1,1>= (-1/2)<2, -4>+ (5/2)<0, 2>= <-1, 1>. Now write that in terms of <1, 2> and <2, 1>: <-1, 1>= a<1, 2>+ b<2, 1>= <a+ 2b, 2a+ b>. a+ 2b= -1 and 2a+ b= 1. Subtracting twice the first equation from the second, -3b= 3 so b= -1. Then a= 1. \xi<1, 2>= (1)<1, 2>+ (-1)<2, 1> so the first column of the matrix representing \xi in this basis is [1 -1].

Do the same with <2, 1> to find the second column.

 

[lukaszh]

Thank you both very much :-)