Find linear map depending on parameter t I'm currently sitting on the following problem:
For what $t \in \mathbb{R}$ exists a linear map $\varphi_t: \mathbb{R}^{1x3}\rightarrow \mathbb{R}^{2}$
$\varphi_t:\left\{ 
\begin{array}{lcl} 
\left( 0, -1, 0 \right) & \mapsto & 
\left( \begin{array}{c} -1 \\ 1  \end{array}\right), \\ 
\left( 0, -1, 1 \right) & \mapsto & 
\left( \begin{array}{c} t \\ 1 \end{array} \right), \\
\left( t-1, -1, 1 \right) & \mapsto & 
\left( \begin{array}{c} t \\ t \end{array} \right)
\end{array} \right.$ 
Give a linear mapping for every $t$ by specifying $Image$ $\varphi(\left( a, b, c \right))$ $a,b,c \in \mathbb{R}$.
For what $t$ is this mapping is unique?
This looks like it should just be a calculation, however I don't even know where to start on this one. Could someone enlighten me please or give a hint?
 A: If $t = 1$, the last two domain vectors $(0,-1,1)$ and $(t-1,-1,1)$  are equal, while the first two vectors are linearly independent, so the three domain vectors span a plane in $\mathbb{R}^3$. So there is no unique linear map, since to define a linear map we can choose any third vector not in the span of those two vectors and send this to whatever point in $\mathbb{R}^2$ we like.
If $t \neq 1$, the three domain vectors are linearly independent, hence they form a basis for $\mathbb{R}^3$. Then the given definition of $\phi_t$ defines a unique linear map, because a linear map is determined by its values on a basis.
To find the image, we can proceed as follows. First, we can find out how to express any vector $(x,y,z)$ as a linear combination of the given basis $\{ (0, -1, 0), (0, -1, 1), (t - 1, -1, 1) \}$ by solving the augmented matrix: $$\left[ \begin{array}{ccc | c} 0 & 0 & t - 1 & x\\ -1 & -1 & -1 & y\\ 0 & 1 & 1 & z \end{array} \right].$$ If we reduce this, the coefficients are $\displaystyle -y-z, z - \frac{x}{t-1}$ and $\displaystyle \frac{x}{t-1}$.
Therefore, for any vector $(x,y,z)$, we have $$ \displaystyle \phi_t(x,y,z) = (-y-z) \phi_t(0,-1,0) + (z - \frac{x}{t-1})\phi_t (0, -1, 1) + \frac{x}{t-1}\phi_t (t-1, -1, 1) = (-y-z)(-1,1) + (z-\frac{x}{t-1})(t, 1) + \frac{x}{t-1}(t, t) = … = (y+z+zt, -y) = y(1, -1) + z(1+t, 0).$$
Therefore, the image can be described as $$\text{image of $\phi_t$} = \{ \phi_t(x, y, z): (x, y, z) \in \mathbb{R}^3 \} = \{ y(1, -1) + z(1 + t, 0): y, z \in \mathbb{R} \} = \text{span$\{ (1,-1),(1+t, 0)\}$},$$
which is just all of $\mathbb{R}^2$.

Remark 1: Actually, you don't need to do any calculations if you are only interested in finding what the image of $\phi_t$ is. We can see that the given codomain vectors $(-1, 1), (t, 1)$ and $(t, t)$ will span all of $\mathbb{R}^2$ no matter what $t$ is, so the image is always all of $\mathbb{R}^2$.
Remark 2: Instead of finding out how $(x,y,z)$ can be written as a linear combination of those basis vectors, which is what I did above, you could also just find out what $\phi_t$ does to each standard basis vector $e_i$. Then $\phi_t(x,y,z) = x\phi_t(e_1) + y\phi_t(e_2) + z\phi_t(e_3)$.
A: To define a unique map, you need to know the value of $\varphi_t$ on each of the basis vectors, and show that these values are unique.
Let us try to find the value on $e_1$
Note $$\varphi_t((t-1,-1,1)-(0,-1,1)) = \varphi_t((t-1,-1,1)) - \varphi_t((0,-1,1)) = {t \choose t} - {t \choose 1} = {0 \choose t-1} $$
But we also have 
$$\varphi_t((t-1,-1,1)-(0,-1,1)) = \varphi_t((t-1,0,0))$$
So we have $$(t-1)\varphi_t((1,0,0)) = (t-1){0 \choose 1}$$ 
Note that this uniquely defines $\varphi_t((e_1)) = {0 \choose 1}$ if and only if $t \neq 1$.
Now, for the value on $e_2$, we know that 
$\varphi_t((0,-1,0)) = {-1 \choose 1}$
This means that $$\varphi_t(e_2) = {1 \choose -1}$$
Lastly, for $e_3$
$$\varphi_t(e_3) = \varphi_t((0,-1,1)-(0,-1,0)) = {t \choose 1} - {-1 \choose 1} = {t+1 \choose 0}$$
So as long as $t \neq 1$, we have a unique map.
The image is given by $span \Big( {0 \choose 1}, {1 \choose -1} \Big) = \mathbb{R}^2$
Note: it is not at all necessary to find the image on the specific basis $\{ e_1, e_2, e_3\}$. Any basis would have worked. The setup of this problem allowed for some quick algebra to do this. However, the quickest way to solve this problem in general is to take the determinant of the $3 \times 3$ matrix whose rows are the $3$ vectors in $\mathbb{R}^3$. You'll find that the determinant is $0$ when $t=1$, and non-zero otherwise.
