Results 1 to 6 of 6
Like Tree1Thanks
  • 1 Post By Barioth

Math Help - Inner Product Space question

  1. #1
    Newbie
    Joined
    Dec 2012
    From
    Canada
    Posts
    3

    Inner Product Space question

    Let R4 have the Euclidean inner product. Express w = (-1, 2, 6, 0) in the form w = w1 + w2, where w1 is in the space W spanned by u1 = (-1, 0, 1, 2) and u2 = (0, 1, 0, 1), and w2 is orthogonal to W
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member Barioth's Avatar
    Joined
    Jan 2013
    From
    Canada
    Posts
    54
    Thanks
    16

    Re: Inner Product Space question

    Hi the euclidean inner produit, I guess that is the basic scalar product as <u,v> = u^tv?

    ps. I'm from french school so just wana make sure!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Inner Product Space question

    do you know how to do an orthogonal projection onto W?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2012
    From
    Canada
    Posts
    3

    Re: Inner Product Space question

    I honestly have no idea how to approach this question.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member Barioth's Avatar
    Joined
    Jan 2013
    From
    Canada
    Posts
    54
    Thanks
    16

    Re: Inner Product Space question

    every vector in W are of the form a_1u_1+a_2u_2=(-a_1, 0 , a_1,2a_1)+(0,a_2,0,a_2)=(-a_1,a_2,a_1,[2a_1+a_2])=m

    where a_1 and a_2 are in R^1

    since dimW1+dimW2=4
    we know that dimW2 = 2. There are 2 vector x1 and x2 as <m,x1>=<m,x2> =0, and x1 and x2 are linearly independant.
    let x1=(b,c,d,e)

    <m,x1>=-a_1b+ca_2+da_1+e(2a_1+a_2)=a_1(d-b+2e)+a_2(c+e) so we can take x1=(1,0,1,0)
    doing the same for x2 one that would work is x2=(0,1,2,-1)

    there we've our 2 basic for w2 and w1 we could now try to find out to write the vector w given.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Inner Product Space question

    what we want is a basis for the orthogonal complement of W, W (we're already given the basis for W).

    since every vector in W is orthogonal to all of W, this means:

    (-1,0,1,2).(x,y,z,w) = 0

    (0,1,0,1).(x,y,z,w) = 0

    for any (x,y,z,w) in W.

    from the second equation, we get: y+w = 0.
    from the first, we get: -x+z+2w = 0.

    since dim(W) = 2, we will have "two free parameters". we know that:

    y = -w
    x = z + 2w.

    so if we select z and w, this will determine x and y.

    so let's try z = 1, w = 0 for our first choice. this leads to (1,0,1,0). next, try z = 0, w = 1, this leads to (2,-1,0,1).

    this gives us the basis {(-1,0,1,2),(0,1,0,1),(1,0,1,0),(2,-1,0,1)} for R4.

    since this IS a basis, we can write:

    (-1,2,6,0) = a(-1,0,1,2) + b(0,1,0,1) + c(1,0,1,0) + d(2,-1,0,1), for some UNIQUE real numbers a,b,c,d.

    that is:

    (-1,2,6,0) = (-a+c+2d,b-d,a+c,2a+b+d)

    or:

    -a+c+2d = -1
    b-d = 2
    a+c = 6
    2a+b+d = 0

    we can solve this equation for a,b,c,d like we do any system of linear equations (form an augmented matrix, row-reduce, or any other method). since i'm lazy, and i hate row reduction with a passion, i will do this, instead:

    b = d+2
    a = 6-c

    (c-6)+c+2d = -1
    2(6-c)+d+2+d = 0

    2c+2d = 5
    -2c+2d = -14

    4d = -9, so:
    d = -9/4
    c = 19/4
    b = -1/4
    a = 5/4

    substituting back in, we have:

    (-1,2,6,0) = (5/4)(-1,0,1,2) - (1/4)(0,1,0,1) + (19/4)(1,0,1,0) - (9/4)(2,-1,0,1), which you can verify.

    thus w1 = (5/4)(-1,0,1,2) - (1/4)(0,1,0,1) = (-5/4,-1/4,5/4,9/4) <--this is in W,

    and w2 = (19/4)(1,0,1,0) - (9/4)(2,-1,0,1) = (1/4,9/4,19/4,-9/4),

    furthermore, w2 is orthogonal to W:

    (-1,0,1,2).(1/4,9/4,19/4,-9/4) = -1/4 + 19/4 - 18/4 = 0
    (0,1,0,1).(1/4,9/4,19/4,-9/4) = 9/4 - 9/4 = 0

    since it is orthogonal to the basis vectors of W, and thus any linear combination of them (including w1).
    Last edited by Deveno; January 18th 2013 at 04:21 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inner product space proof question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 18th 2010, 07:39 PM
  2. Inner product space proof question
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 16th 2010, 08:33 PM
  3. inner product space question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 10th 2009, 07:48 PM
  4. inner product space question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 3rd 2009, 08:15 PM
  5. Question about inner product space inequality
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 24th 2009, 08:37 PM

Search Tags


/mathhelpforum @mathhelpforum