Jtdylktuy

I'm wondering if I have a valid answer to this.

It is exactly (e) of the following:

I first state that the two vector spaces are isomorphic because they have equal dimension. I then define a linear transform $S: mathbb R^4 / Ker T to mathbb R^2$ by

$$S begin{pmatrix}
0 \
a \
b \
0 \
end{pmatrix} to
begin{pmatrix}
a \
b \
end{pmatrix}: a,b in mathbb R$$

Where $begin{pmatrix}
0 \
a \
b \
0 \
end{pmatrix}$ is an arbitrary element in the departure space.

If this mapping translates basis to basis, this is an isomorphism.

$$
S begin{pmatrix}
0 \
a \
b \
0 \
end{pmatrix} to
begin{pmatrix}
a \
b \
end{pmatrix} in mathbb R^2 implies Sleft(a begin{pmatrix}
0 \
1 \
0 \
0 \
end{pmatrix} +
bbegin{pmatrix}
0 \
0 \
1 \
0 \
end{pmatrix}right) = a vec e_1 + b vec e_2 $$
$$
text{ as
$Sbegin{pmatrix}
0 \
1 \
0 \
0 \
end{pmatrix} =
begin{pmatrix}
1 \
0 \
end{pmatrix}$ and
$
Sbegin{pmatrix}
0 \
0 \
1 \
0 \
end{pmatrix} =
begin{pmatrix}
0 \
1 \
end{pmatrix}$}$$

Is this a valid answer?

Your answer is not valid, since you don't even mention $F^4/ker T$ in it. An explicit isomorphism would be$$begin{array}{rccc}Psicolon&F^4/ker T&longrightarrow&F^2\&begin{pmatrix}a\b\c\dend{pmatrix}+ker T&mapsto&Tbegin{pmatrix}a\b\c\dend{pmatrix}.end{array}$$