1. ## Linear transformations

Consider the basis $\displaystyle S=\{v1,v2\}$ where $\displaystyle v1=(1,1)$ and $\displaystyle v2=(1,0)$ and is $\displaystyle T:R^2=>R^2$ the linear operator such that $\displaystyle T(v1)=(1,-2)$ and $\displaystyle T(v2)=(-4,1)$. Find a formula for $\displaystyle T(x,y)$ and use this formula to get $\displaystyle T(5,-3)$

2. So we use linearity properties to know the range by $\displaystyle T$ of the canonic basis of $\displaystyle \mathbb{R}^{2}$ : $\displaystyle (1,0)$ and $\displaystyle (0,1)$.

$\displaystyle T((1,0))=(-4,1)$
$\displaystyle T((0,1))=T((1,1)-(1,0))=T((1,1))-T((1,0))=(1,-2)-(-4,1)=(5,-3)$

So, using again the linearity of $\displaystyle T$, we can conclude :

$\displaystyle \forall (x,y) \in \mathbb{R}^{2},\ T((x,y))=(-4x+5y,x-3y)$

Therefore $\displaystyle T((5,-3))=(-35,14)$

3. Thank you. How do you find: $\displaystyle T(x,y)=(-4x+5y,x-3y)$

4. Just write $\displaystyle (x,y)$ in the basis $\displaystyle \{(1,0),(0,1)\}$:

$\displaystyle T((x,y))=T((x,0)+(0,y))=T(x(1,0)+y(0,1))=xT((1,0)) +yT((0,1))$

But we know that $\displaystyle T((1,0))=(-4,1)$ and $\displaystyle T((0,1))=(5,-3)$

So we get:

$\displaystyle T((x,y))=xT((1,0))+yT((0,1))=x(-4,1)+y(5,-3)=(-4x,x)+(5y,-3y)$$\displaystyle =(-4x+5y,x-3y)$

5. ok thank you