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Math Help - Basis (theoretical question)

  1. #1
    MHF Contributor arbolis's Avatar
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    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.
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  2. #2
    MHF Contributor arbolis's Avatar
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    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?
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by arbolis View Post
    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...
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by arbolis View Post
    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...
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    Quote Originally Posted by arbolis View Post
    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.
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  6. #6
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Swlabr View Post
    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.

    Quote Originally Posted by Swlabr View Post
    Hint: there is nothing that says you cannot choose one of the original basis vectors for your new basis...
    ,
    Quote 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.
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