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Thread: Subspaces and bases

  1. #1
    Feb 2013

    Question Subspaces and bases

    S={(1,-1,-4), (3,2,3)}

    T= {(x,y,3y-x) : x,y e R}

    (a) Show that S C T, and write down a a vector in IR3 that does not belong to T

    (b) Show that T is a subspace of IR3.

    (c) Show that S is a basis for T. Write down the dimension of T, and describe T geometrically.

    HELP..... my head is completely messed with this one!!!
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  2. #2
    MHF Contributor

    Mar 2011

    Re: Subspaces and bases

    to show (a) we need to find an x and y such that:

    1) (x,y,3y-x) = (1,-1,-4)

    2) (x,y,3y-x) = (3,2,3)

    it's pretty obvious that for (1) we have to take x = 1, y = -1, so all we have to do is verify that 3(-1) - 1 = -4.

    similarly for (2), we must have x = 3, y = 2, so it boils down to verifying that 3(2) - 3 = 3.

    so to find a vector in R3, let's pick x = 1, y = 1. all we have to do is pick a 3rd coordinate which is NOT 3(1) - 1 = 2. so, for example, (1,1,1) will work just fine.

    (b) we have to verify 3 things:

    1) T is closed under vector addition.

    2) T is closed under scalar multiplication.

    3) the 0-vector is in T.

    suppose u = (x,y,3y-x) and v = (x',y',3y'-x') are in T. we need to show that u+v is in T, as well.

    now u+v = (x+x',y+y',(3y-x)+(3y'-x')). so we need to show the 3rd coordinate of u+v, which is (3y-x)+(3y'-x'), is 3 times the second coordinate of u+v minus the first coordinate of u+v. that is, show:

    (3y-x)+(3y'-x') = 3(y+y')-(x+x')....can you do this?

    next suppose c is ANY real number. we need to show that cu is in T. now cu = c(x,y,3y-x) = (cx,cy,c(3y-x)). so the question now is, does:

    c(3y-x) = 3(cy)-cx ?

    finally, is (0,0,0) in T? that is: does 3(0) - 0 = 0?

    (c) we have to do TWO things to show S is a basis for T:

    1) we need to show S is a linearly independent set. this means that if:

    a(1,-1,-4) + b(3,2,3) = (0,0,0) the only possible choice for a and b is a = b = 0.

    now a(1,-1,-4) + b(3,2,3) = (a,-a,-4a) + (3b,2b,3b) = (a+3b,2b-a,3b-4a). if this equals (0,0,0), this gives us 3 equations in 2 unknowns:

    a+3b = 0
    2b-a = 0
    3b-4a = 0

    if we add equations 1 and 2 together, we get:

    5b = 0, which means b = 0. substituting b = 0 in any of the original 3 equations quickly gives a = 0, so we see that indeed this is the only possibility. so S is linearly independent.

    2) we need to show S SPANS T. that is, given ANY member u of T, we must be able to find a and b such that:

    a(1,-1,-4) + b(3,2,3) = u.

    recall a typical element u of T is u = (x,y,3y-x). so if:

    a(1,-1,-4) + b(3,2,3) = (x,y,3y-x) then:

    (a+3b,2b-a,3b-4a) = (x,y,3y-x). in this equation, "x" and "y" are constants, and we are solving for "a" and "b". again we have 3 equations:

    a+3b = x
    2b-a = y
    3b-4a = 3y-x

    again, if we add the first two equations we obtain: 5b = x+y, thus b = (x+y)/5. substituting this value for b in the first equation gives:

    a + 3(x+y)/5 = x

    a = [5x - 3(x+y)]/5 = (2x-3y)/5

    now, to verify this is a consistent solution, we could check all three equations with these values, but it is simpler just to verify that:

    [(2x-3y)/5](1,-1,-4) + [(x+y)/5](3,2,3) = (1/5)(2x-3y,3y-2x,12y-8x) + (1/5)(3x+3y,2x+2y,3x+3y) = (1/5)(5x,5y,15y-5x) = (x,y,3y-x), as desired. so S spans T.

    the dimension of a subspace is the number of elements in ANY basis. so S is a linearly independent spanning set for T, it IS a basis. since S has 2 elements, the dimension of T is 2. geometrically, a linear subspace of dimension 2 of R3 is called a plane, in this case, the plane determined by the 3 points:

    (1,-1,-4),(3,2,3) and (0,0,0).
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