1. ## Linear transformations

Determine the linear transformation:

$T:IR^2=>IR^3$

$T(-1,1)=(3,2,1)$

$T(0,1)=(1,1,0)$

2. Sorry for being late. I don't think this problem has a unique solution.
I've found one handily.
$T(x,y)=(y-2x,y-x,-x)$.

3. $x \in \mathbb{R}^2 \ ,\ T:\mathbb{R}^2\rightarrow \mathbb{R}^3 \ , \ A\in \mathbb{R}^{3\times2}\ , \ T(\vec{x})=A\vec{x}$

$T(\begin{bmatrix} -1\\1\end{bmatrix})=A(-\vec{e}_1+\vec{e}_2)=\begin{bmatrix} 3\\2\\1 \end{bmatrix}=\vec{v}_1$

$T(\begin{bmatrix} 0\\1\end{bmatrix})=A\vec{e}_2=\begin{bmatrix} 1\\1\\0 \end{bmatrix}=\vec{v}_2$

$A\vec{e}_1 = A\vec{e}_2-A(-\vec{e}_1+\vec{e}_2)=\vec{v}_2-\vec{v}_1$
$A=[A\vec{e}_1 \ \ A\vec{e}_2]=[\vec{v}_2-\vec{v}_1 \ \ \ \vec{v}_2]=\begin{bmatrix} -2&1\\-1&1\\-1&0 \end{bmatrix}$

4. It really depends on what you mean by "determine". math2009 found a matrix form for it, in the standard basis, which certainly determines it!

Another way to do this, almost the same thing, would be to say that the standard bases, (1, 0) and (0, 1) can be written in terms of (-1, 1) and (0,1) as (1, 0)= A(-1, 1)+ B(0, 1)= (-A, A+ B) so -A= 1, A+ B= 0 so A= -1, B= 1: (1, 0)= -(-1,1)+ (0,1) and so T(1,0)= -T(-1, 1)+ T(0,1)= -(3,2,1)+ (1,1,0)= (-2, -1, -1). We already know that T(0,1)= (1,1,0)
That is basically what math2009 did.

Now, for any (x,y), T(x,y)= xT(1,0)+ yT(0,1)= x(-2, -1, -1)+ y(1, 1, 0)= (-2x+y, -x+y, -x).

That is T(x,y)= (-2x+ y, -x+ y, -x) which "determines" T by showing what it does to any (x, y).