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Math Help - finding basis vecto space subspace

  1. #1
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    finding basis vecto space subspace

    (9,0,8)
    Last edited by jecowap; December 2nd 2009 at 11:44 AM. Reason: question wrong
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  2. #2
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    Quote Originally Posted by jecowap View Post
    assuming that S and T are subspaces of V3(C) generated by
    <(1,1,0),(i,1+i,1),(1+i,1+i,0)> and <(1,0,1),(i,-i,0),(0,i,i)> respectively
    determine C-bases for S, T, SnT, S+T
    Finding bases for S and T is easy: if the given sets are independent, then they are bases. If not, then you can solve for one (or two) of the vectors in terms of the others.

    For example, to see if <(1,0,1), (i,,-i,0), (0,i,i)> is an independent set, look at a(1,0,1)+ b(i,-i,0)+ c(0,i,i)= (0,0,0). That is the same as (a+ ib, -ib+ ic, a+ ic)= (0,0,0) so we have the three equations a+ ib= 0, -ib+ ic= 0, a+ ic= 0. The middle equation is the same as ib= ic or b= c. The first and second equations, both reduce to a= -ib. If, for example, we take b= 1, then a= -i and c= 1 so -i(1,0,1)+ (i, -i, 0)+ (0,i,i)= (-i+i, -i+i, -i+i)= (0,0,0). That set is not independent and we can now see that (i,-i,0)= i(1,0,1)- (0,i,i) so we don't need the vector (i,-i,0)- it can be written using the other two. So <1,0,1),(0,i,i)> is enough for a basis. (Actually, any two of the three vectors forms a basis.)
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Finding bases for S and T is easy: if the given sets are independent, then they are bases. If not, then you can solve for one (or two) of the vectors in terms of the others.

    For example, to see if <(1,0,1), (i,,-i,0), (0,i,i)> is an independent set, look at a(1,0,1)+ b(i,-i,0)+ c(0,i,i)= (0,0,0). That is the same as (a+ ib, -ib+ ic, a+ ic)= (0,0,0) so we have the three equations a+ ib= 0, -ib+ ic= 0, a+ ic= 0. The middle equation is the same as ib= ic or b= c. The first and second equations, both reduce to a= -ib. If, for example, we take b= 1, then a= -i and c= 1 so -i(1,0,1)+ (i, -i, 0)+ (0,i,i)= (-i+i, -i+i, -i+i)= (0,0,0). That set is not independent and we can now see that (i,-i,0)= i(1,0,1)- (0,i,i) so we don't need the vector (i,-i,0)- it can be written using the other two. So <1,0,1),(0,i,i)> is enough for a basis. (Actually, any two of the three vectors forms a basis.)

    thanks so much
    but for the other set i get to the abc point and the first and second are equal (a+c+ci=0) and b=0
    could you possibly help me further with where to go from this point?
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