Find a Linear Transformation from $\mathbb{R^3}$ to $\mathbb{R^2}$ . Assuming I have a linear transformation with it's effect on $2$ vectors given:
$$
T(1,1,1) = (1,0), T(1,-1,1) = (0,1)
$$
And I'm supposed to find a linear transformation which follows these conditions, and determine whether it is unique.
I'm pretty stuck with it, since I know that both vectors (arguments) do not form a basis to $ \mathbb{R}^3 $, so I'm pretty much confused about what I should do next.
What is the approach for these kind of problems?
 A: The general transformation from $\mathbb R^3$ to $\mathbb R^2$ is 
$$T(x,y,z) = (ax+by+cz, dx+ey+fz)$$
From your conditions we have 
$$\begin{align*}
a+b+c &= 1, \\
d+e+f & = 0, \\
a-b+c &=0, \\
d-e+f &= 1.
\end{align*}
$$
Solve, for $a$, $b$, $c$, $d$, $e$ and $f$ to get all such transformations $T$. There are infinitely many solutions.
A: That it's not unique can be seen as we can consider a normal $n$ to both $(1,1,1)$ and $(1,-1,1)$ (by using cross product for example, let's say we chose $n=(1,0,-1)$). And then use it to form an addition to the map that will not make any impact on these two vectors $T(u)+(u\cdot n)e$ (where $e$ is an arbitrary vector and $u$ is the vector to be transformed) would have no impact.
To find a mapping we could find a orthogonal basis for the plane spanned $(1,1,1)$ and $(1,-1,1)$ and see where the base vectors should be mapped. We can take the first vectors as one of the base vectors and then find another. To do this we see that $(0,2,0)$ (the difference of the vectors) is also in the plane so $(1,-1-2t,1)$ is also in the plane and the dot product with the first is $(1,1,1)\cdot(1,-1-2t,1) = 1+-1-2t+1$ and requiring it to be zero gives $-2t-1=0$ which makes $t = -1/2$. So the vectors are $(1,1,1)$ and $(1,-2,1) = {3\over2}(1,-1,1) - {1\over2}(1,1,1)$.
So we have $T(1,1,1) = (1,0)$ and $T(1,-2,1) = {3\over2}(0,1) + {1\over2}(1,0) = (1/2, 3/2)$ and since the vectors are orthogonal we can find a transformation:
$$T(x,y,z) = (1,0){(1,1,1)(x,y,z)\over (1,1,1)(1,1,1)} + (1/2, 3/2){(1,-2,1)(x,y,z)\over(1,-2,1)(1,-2,1)}\\
= (1,0){x+y+z\over3} + (1/2,3/2){x-2y+z\over6} \\
= (1,0){4x+4y+4z\over12} + (1,3){2x-4y+2z\over12} \\
= ({x+z\over2}, {x-2y+z\over4})$$
