Krdytkyu

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?

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.