<s>[INST] Proving a set of vectors is a basis for the quotient map between two vector spaces I want to see if my work is justifiable. I am tasked with the following: 
I will neglect to prove (a), as the work for this is fairly straight forward. I will center my attention on (b).
$$\text{Implication is} \ \mathscr {B}_{\mathbb R^4 / \ ker \ T}  = 
\left\{
\begin{bmatrix}
\begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
\end{pmatrix}
\end{bmatrix},\begin{bmatrix}
\begin{pmatrix}
0 \\
0 \\
1 \\
0 \\
\end{pmatrix}
\end{bmatrix}
\right\} \text{, where $\mathscr B$ is the basis for $\mathbb R^4 / \ ker \ T$}$$
$$
\implies \text{for some $[v] \in \mathbb R^4 / \ ker \ T$,}$$
$$[v] = a_1 \begin{bmatrix}
\begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
\end{pmatrix}
\end{bmatrix} + a_2 \begin{bmatrix}
\begin{pmatrix}
0 \\
0 \\
1 \\
0 \\
\end{pmatrix}
\end{bmatrix}$$
$$\implies
\begin{bmatrix}
\begin{pmatrix}
x \\
y \\
z \\
t \\
\end{pmatrix}
\end{bmatrix} 
= 
a_1 \begin{bmatrix}
\begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
\end{pmatrix}
\end{bmatrix} + a_2 \begin{bmatrix}
\begin{pmatrix}
0 \\
0 \\
1 \\
0 \\
\end{pmatrix}
\end{bmatrix}$$
$$\implies 
\begin{bmatrix}
\begin{pmatrix}
x \\
y \\
z \\
t \\
\end{pmatrix}
\end{bmatrix} 
=
 \begin{bmatrix}
\begin{pmatrix}
0 \\
a_1 \\
a_2 \\
0 \\
\end{pmatrix}
\end{bmatrix} $$
$$\implies 
\begin{pmatrix}
x \\
y \\
z \\
t \\
\end{pmatrix} - \begin{pmatrix}
0 \\
a_1 \\
a_2 \\
0 \\ 
\end{pmatrix} \in ker \ T$$
$$
\implies 
\begin{pmatrix}
x \\
y-a_1 \\
z-a_2 \\
t \\
\end{pmatrix} = b_1 \begin{pmatrix}
1 \\
1 \\
0 \\
0 \\
\end{pmatrix} + b_2 \begin{pmatrix}
0 \\
0 \\
1 \\
1 \\
\end{pmatrix} \text{, where $b_1, b_2 \in ker \ T$}$$
After some algebra...
$$\implies a_1 = y - x$$
$$\implies a_2 =  z - t$$
Which I think implies $\mathscr B$ is spanning? $x,y,z,t$ are all arbitrary. Now I am left with proving $\mathscr B$ is linearly independent. I may skip a step or two.
$$
\begin{bmatrix}
\begin{pmatrix}
0 \\
a_1 \\
a_2 \\
0 \\ 
\end{pmatrix}
\end{bmatrix} = [0] \iff a_1 = a_2 = 0 $$
$$\begin{bmatrix}
\begin{pmatrix}
0 \\
a_1 \\
a_2 \\
0 \\ 
\end{pmatrix}
\end{bmatrix} = [0] \implies 
\begin{pmatrix}
0 \\
a_1 \\
a_2 \\
0 \\ 
\end{pmatrix} - \begin{pmatrix}
0 \\
0 \\
0 \\
0 \\ 
\end{pmatrix} \in Ker \ T$$
$$\implies
\begin{pmatrix}
0 \\
a_1 \\
a_2 \\
0 \\ 
\end{pmatrix} = \begin{pmatrix}
b1 \\
b1 \\
b2 \\
b2 \\ 
\end{pmatrix} \implies a_1 = a_2 = 0$$
Hence, basis. Has what I done made sense?
 [/INST] It's all correct.
Note that the 4 given vectors form a basis of $\Bbb R^4$, two of them spans the kernel, so the (images of) the other two will be a basis of the quotient space. 
</s>