I have never worked with vectors in $\mathbb{R}^4$. I have the basis
$B=\{e_4=(2,1,0,7), e_5=(-1,2,-2,-5), e_6=(1,-1,5,12)\}$
and
$F=\{e_1=(1,0,-1,1), e_2=(0,1,-1,-1), e_3=(0,0,1,2)\}$
Is the subspace spanned by $B$ the same as $F$? If affirmative, give the change of basis matrix from this basis $B$ to the original basis defined by the vectors $e_1$, $e_2$ and $e_3$.
I think first of all, I have to do this step: $Pb \to F = [[e_4]_F, [e_5]_F, [e_6]_F]$
Can someone either give me a hint, or preferably guide me towards an example of this type of problem?
Here's a hint on the general method for solving this kind of problem, I've left the details to you as you've requested.
First show that $B$ and $F$ are both linearly independent, i.e. if
$$\lambda_1 e_1 + \lambda_2 e_2 + \lambda_3 e_3 = 0$$
then $\lambda_1$ = $\lambda_2$ = $\lambda_3=0$. Similarly for $e_4$, $e_5$, $e_6$.
This tells us that the dimensions of the subspaces of $B$ and $F$ are both $3$.
To show that $\mathrm{span}(B)=\mathrm{span}(F)$, then, we need to show that any vector in $B$ can be written as a linear combination of the basis elements of $F$. In fact, it is sufficient to do that for the basis elements of $B$, since all vectors in $B$ will be a linear combination of the basis elements.
This then specifies the change of basis matrix. Take the basis elements of $B$, and write them as a linear combination of the basis elements of $F$, this then specifies the columns of the matrix $P$ - as you noted, $P = [[e_4]_F, [e_5]_F, [e_6]_F]$.