# Linear transformation and its matrix

1. Jan 13, 2009

### 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 :-)

2. Jan 13, 2009

### 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 $$e_{i}=A^{j}_{i}f_{j}$$.

3. Jan 14, 2009

### HallsofIvy

Staff Emeritus
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.

4. Jan 14, 2009

### lukaszh

Thank you both very much :-)