Expanding a vector to a basis

Question

I'm doing Linear Algebra and I'm asked to calculate the basis of quotient subspaces. Here's my exercise:

Let $F=<(1,0,0,2),(0,1,0,0)>$, and $G=<(-1,1,0,0),(1,0,1,0),(0,0,0,1)>$. Calculate the basis of $F/(F\cap{G})$.

By calculating equations of F and G, i've gotten the basis of $F\cap{G}$={$(1,1,0,2)$}, and my question is about finding the basis of $F/F\cap{G}$. I was told that you could calculate it by expanding the basis of the "quotient" (in this case, $F\cap{G}$) to a basis of the "total space" (in this case, $F$), and then picking the vectors not from $F\cap{G}$ to form the basis.

Since $(1,1,0,2)=1*(1,0,0,2)+1*(0,1,0,0)$, by Steinitz's exchange lemma, both {$(1,1,0,2),(1,0,0,2)$} and {$(1,1,0,2),(0,1,0,0)$} are basis of F, but does this mean that then both {$(1,0,0,2)$} and {$(0,1,0,0)$} are basis of $F/(F\cap{G})$?

It seems a little odd to me since I don't see how $[(1,0,0,2)]=[(0,1,0,0)]$. Thank you very much.

Answer 1

First of all $F\cap G=\langle(-1,1,0,-2)\rangle$ sincee $(-1,1,0,-2)=(−1,1,0,0)+2*(0,0,0,1)\in G$ and $(-1,1,0,-2)=-2*(1,0,0,2)+(0,1,0,0)\in F$

A basis of $F/(F\cap G)$ can be constructed in this way: take a $\{v_1,\dots ,v_n\}$ basis of $(F\cap G)$ and extend it with $\{w_1,\dots, w_k\}$ to a basis of $F$ (i.e. $\{v_1,\dots,v_n,w_1,\dots,w_k\}$ is a basis of $F$). The set $\{[w_1],\dots,[w_k]\}$ is a basis of $F/(F\cap G)$.

In your example, you can take $[(1,0,0,2)]$ (or $[(0,1,0,0)]$ as a basis). In fact you have $(1,0,0,2)= (0,1,0,0) -(-1,1,0,-2) = (0,1,0,0)+F\cap G$ so $[(1,0,0,2)]=[(0,1,0,0)]$

Answer 2

Since it seems like you have a good grasp on the concepts, I'll just note that the issue is with the computation: $F\cap G = \{(-1,1,0,-2)\}$, not $\{(1,1,0,2)\}$. (One way to see that $(1,1,0,2)$ is not in $G$, is the last three coordinates of the vector must be the coefficients in the given basis.)