Finding linear transformation such that $\operatorname{im} \phi = \ker \phi = \operatorname{span}(\alpha_1, \alpha_2)$ Here i am completely lost.
I have to find a formula for linear transformation $\phi : \mathbb{R}^4 \rightarrow \mathbb{R}^4$ such that $\operatorname{im} \phi=\ker \phi = \operatorname{span} (\alpha_1, \alpha_2)$, where $\alpha_1 =[1,1,2,1]$ and $\alpha_2=[1,2,1,1]$. I have no idea how to approach this question and what should i do with it.
 A: Hint 
Complete the set $\{\alpha_1,\alpha_2\}$ on a basis $\mathcal B$ of $\Bbb R^4$: we choose for example $\mathcal B=(\alpha_1,\alpha_2,e_3,e_4)$. It suffices to define $\phi$ on the vectors of the basis $\mathcal B$ by
$$\phi(\alpha_1)=\phi(\alpha_2)=(0,0,0,0)\quad\text{and}\quad \phi(e_3)=\alpha_1\quad;\quad\phi(e_4)=\alpha_2$$
Remark We can express $\phi$ relative to the standard basis $\mathcal B_c$ using the change matrix from $\mathcal B$ to $\mathcal B_c$.
A: A geometric object like a linear subspace (line, plane, etc.) can be represented in two ways among others: as an image of a transformation or as the inverse image of $0$. 
One way the object in this case can be written as:
$$\mathbf{P}:=\{(x_1,x_2,x_3,x_4): (x_1,x_2,x_3,x_4) = c_1(1,1,2,1)+c_2(1,2,1,1), c_i \in \mathbb R\}$$
This is just another way to write the image of the following function:
$$f:\mathbb{R}^2 \longrightarrow \mathbb{R}^4 ~~\text{as}~~ f(c_1,c_2):= c_1(1,1,2,1)+c_2(1,2,1,1)$$
Notice, this is an injective function.
Also, we want to submerse $\mathbb{R}^4$ into $\mathbb{R}^2$:
$$\sigma:\mathbb{R}^4 \longrightarrow \mathbb{R}^2$$
such that we have:
$$\mathbf{P} = \text{Im} (f\circ \sigma) = (f\circ \sigma)^{-1}((0,0,0,0))$$
Now, another way to see $\mathbf{P}$ is an object orthogonal to two other vectors in $\mathbb{R}^4$, say, $(a_1,a_2,a_3,a_4)$ and $(b_1,b_2,b_3,b_4)$ (computed the usual way) i.e.:
$$\mathbf{P}:=\left\{[x_1,x_2,x_3,x_4]^t: 
\begin{bmatrix}
a_1 & a_2 & a_3 & a_4 \\ 
b_1 & b_2 & b_3 & b_4
\end{bmatrix}
\begin{bmatrix}
x_1\\ x_2 \\ x_3 \\ x_4
\end{bmatrix}
=
\begin{bmatrix}
0\\ 0
\end{bmatrix}
\right\}
$$
Let the transformation wrt to the matrix 
$\begin{bmatrix}
a_1 & a_2 & a_3 & a_4 \\ 
b_1 & b_2 & b_3 & b_4
\end{bmatrix}$
be the submersion $s$.
That is:
$$ \mathbb{R}^2 \overset{f}{\longrightarrow} \mathbb{R}^4 \overset{s}{\longrightarrow} \mathbb{R}^2 \qquad \text{and}~~\text{Im}(f) = \text{ker}(s) = \mathbf{P}$$
Notice that $s(P\oplus P')=s(P')$ and rank of $s = 2$.
In fact, the transformation $s$ is our required submersion $\sigma$.
$$ \mathbb{R}^4 \overset{\sigma}{\longrightarrow} \mathbb{R}^2 \overset{f}{\longrightarrow} \mathbb{R}^4 \overset{\sigma}{\longrightarrow} \mathbb{R}^2 \overset{f}{\longrightarrow} \mathbb{R}^4$$
In short, the idea is our required transformation on $\mathbb{R}^4 = \mathbf{P} \oplus \mathbf{P'}$ is one that pinches the plane $\mathbf P$ to $0$ and flattens the remaining space $\mathbf P'$ to it.
