1. ## Linear transformation

I don't understand this definition.

In a linear transformation, $f0,0) \to (0,0)" alt="f0,0) \to (0,0)" />. If $f0,0)" alt="f0,0)" /> (the "to" symbol with the "not" diagonal cross in it) (0,0), the mapping is not a linear transformation.

Now I have to determine if each mapping that moves (x,y) to the given point is a linear transformation. I don't get how it works.

Why is moving (x,y) to (0,x) is a linear transformation and moving it to (x,-1) is not?

Another question I don't get.

Given the linear transformation $\left( \begin{array}{lcr} x' \\y' \end{array} \right) = \left( \begin{array}{lcr} 2&1 \\ 3&4 \end{array} \right)=\left( \begin{array}{lcr} x \\ y \end{array} \right)$, determine the matrix of the linear transformation which moves points (1,0) and (0,1) to points (1,2) and (3,4), respectively.

How do you get the answer $\left( \begin{array}{lcr} 1&3 \\ 2&4 \end{array} \right)$?

2. $f$is linear transformation iff
$f(x+y)=f(x)+f(y)$
$f(\alpha x)=\alpha f(x)$
1: $f((x,y)+(x_1,y_1))=f((x+x_1,y+y_1))=(x+x_1,0)=f((x ,y))+f((x_1+y_1))$
$f(\alpha(x,y))=f((\alpha x,\alpha y))=\alpha x=\alpha f((x,y))$
2: $f(\alpha(x,y))=f((\alpha x,\alpha y))=(\alpha x,-1)$
but $\alpha f(x,y)=(\alpha x,-\alpha)$

$f\left(\begin{array}{cc}x\\y\end{array}\right)=xf\ left(\begin{array}{cc}1\\0\end{array}\right)+yf\le ft(\begin{array}{cc}0\\1\end{array}\right)=x\left( \begin{array}{cc}1\\2\end{array}\right)+y\left(\be gin{array}{cc}3\\4\end{array}\right)=\left(\begin{ array}{cc}1&3\\2&4\end{array}\right)\left(\begin{a rray}{cc}x\\y\end{array}\right)$