1. Basis (theoretical question)

Imagine I have $V=span \{(a,b,c,d),(e,f,g,h)\}$.
They give me a vector $(i,j,k,l)$ and they ask me to find another vector $(m,n,o,p)$ such that $span ((i,j,k,l),(m,n,o,p))=V$.
My attempt : I realized that $(i,j,k,l)$ can be written as a linear combination of $(a,b,c,d)$ and $(e,f,g,h)$. (It's not always true I know, but it is in the particular example I got assigned)
So I must find a linear independent vector from $(i,j,k,l)$ but that spans $V$ with $(i,j,k,l)$.
I don't know how to do it. (I'd be glad if there's a general method to find it)
Another question : say I found a vector and want to test if it spans $V$ with $(i,j,k,l)$. If I see that I can write $(a,b,c,d)$ and $(e,f,g,h)$ as a combination linear of $(i,j,k,l)$ and the vector I found, does this implies that the latter 2 vectors span $V$?
P.S.: I've all the vectors if you want them... but I wanted to know the general case, hence my choice of using letters instead of numbers.

2. I reformulate and detail just in case : $(a,b,c,d)=(1,1,0,1)=\alpha$ and $(e,f,g,h)=(0,-1,1,1)=\beta$.
$(i,j,k,l)=(2,1,1,3)=\gamma$. I noticed that $\gamma = \alpha + 2 \beta$.
I must find a vector $\zeta$ such that $span \{ \gamma, \zeta \} = span \{ \alpha, \beta \}$.
My main question is : say I found a vector $\zeta$ and I want to check if it does span $V$ along with $\gamma$.
Is it enough to check that both $\alpha$ and $\beta$ are linear combination of $\gamma$ and $\zeta$? My intuition says yes because if $\alpha$ and $\beta$ can be written as a comb. linear of $\gamma$ and $\zeta$, so does any vector spanned by $\alpha$ and $\beta$. And I also must check that $\zeta$ is a comb. linear of $\alpha$ and $\beta$. If it is then I can conclude that $span \{ \gamma, \zeta \} = span \{ \alpha, \beta \}$.
Is there a way to find $\zeta$, other than having a mathematical eye?

3. Originally Posted by arbolis
I reformulate and detail just in case : $(a,b,c,d)=(1,1,0,1)=\alpha$ and $(e,f,g,h)=(0,-1,1,1)=\beta$.
$(i,j,k,l)=(2,1,1,3)=\gamma$. I noticed that $\gamma = \alpha + 2 \beta$.
I must find a vector $\zeta$ such that $span \{ \gamma, \zeta \} = span \{ \alpha, \beta \}$.
My main question is : say I found a vector $\zeta$ and I want to check if it does span $V$ along with $\gamma$.
Is it enough to check that both $\alpha$ and $\beta$ are linear combination of $\gamma$ and $\zeta$? My intuition says yes because if $\alpha$ and $\beta$ can be written as a comb. linear of $\gamma$ and $\zeta$, so does any vector spanned by $\alpha$ and $\beta$. And I also must check that $\zeta$ is a comb. linear of $\alpha$ and $\beta$. If it is then I can conclude that $span \{ \gamma, \zeta \} = span \{ \alpha, \beta \}$.
Is there a way to find $\zeta$, other than having a mathematical eye?

Okay, so you want to find two vectors that span your vector space and are linearly independent, and you are given two that already do. Firstly, note that any spanning set of n vectors is automatically linearly independent for a vector space of dimension n. So here, if you can find two vectors that span your set you are done.

The "trick" to doing this sort of question is showing that you can construct your original basis vectors with your new vectors, as then this linear combination of your new vectors will span the space and your vectors are linearly independent by my first paragraph.

Hint: there is nothing that says you cannot choose one of the original basis vectors for your new basis...

4. Originally Posted by arbolis
My attempt : I realized that $(i,j,k,l)$ can be written as a linear combination of $(a,b,c,d)$ and $(e,f,g,h)$. (It's not always true I know, but it is in the particular example I got assigned)

Every vector in your space is a linear combination of your basis vectors. It would be silly to ask a question such as this for vectors outwith the vector space. "Find two vectors not in V that span V" would be a strange question...

5. Originally Posted by arbolis
I reformulate and detail just in case : $(a,b,c,d)=(1,1,0,1)=\alpha$ and $(e,f,g,h)=(0,-1,1,1)=\beta$.
$(i,j,k,l)=(2,1,1,3)=\gamma$. I noticed that $\gamma = \alpha + 2 \beta$.
I must find a vector $\zeta$ such that $span \{ \gamma, \zeta \} = span \{ \alpha, \beta \}$.
My main question is : say I found a vector $\zeta$ and I want to check if it does span $V$ along with $\gamma$.
Is it enough to check that both $\alpha$ and $\beta$ are linear combination of $\gamma$ and $\zeta$? My intuition says yes because if $\alpha$ and $\beta$ can be written as a comb. linear of $\gamma$ and $\zeta$, so does any vector spanned by $\alpha$ and $\beta$. And I also must check that $\zeta$ is a comb. linear of $\alpha$ and $\beta$. If it is then I can conclude that $span \{ \gamma, \zeta \} = span \{ \alpha, \beta \}$.
Is there a way to find $\zeta$, other than having a mathematical eye?
one simple but important point: two non-zero vectors are linearly independent if and only if one is not a scalar multiple of the other. so, in your problem, $\alpha, \beta$ are linearly independent, and

since they span V, we have $\dim V=2.$ now they give you another non-zero vector, say $\gamma.$ if i choose any non-zero vector in V, say $\zeta,$ which is not a scalar multiple of $\gamma,$ then, by the point

i mentioned, $\beta, \zeta$ will be linearly independent. but since $\dim V = 2,$ any set of linearly independent vectors has at most 2 elements. that means $span \{\gamma, \zeta \}=V.$ for example, in your problem,

neither $\alpha$ nor $\beta$ is a scalar multiple of $\gamma.$ so you can simply choose $\zeta=\alpha$ or $\zeta=\beta.$

6. Originally Posted by Swlabr
Every vector in your space is a linear combination of your basis vectors. It would be silly to ask a question such as this for vectors outwith the vector space. "Find two vectors not in V that span V" would be a strange question...
Yeah I realized this, hehe.

Originally Posted by Swlabr
Hint: there is nothing that says you cannot choose one of the original basis vectors for your new basis...
,
Originally Posted by NonCommAlg
neither nor is a scalar multiple of so you can simply choose or
Wow. I thought about it, but when I checked if I could write $\alpha$ as a linear combination of $\beta$ and $\gamma$ it seems I made an error of arithmetic. I've redone it and it works now... Thanks a lot. Silly me, I now realize why it works.