Results 1 to 3 of 3

Math Help - basis of the subspace of R4 consisting of all vectors perpendicular to v.

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    29

    basis of the subspace of R4 consisting of all vectors perpendicular to v.

    v=[3 -5 7 9]^T

    Find a basis of the subspace of R4 consisting of all vectors perpendicular to v.

    I have no clue how to do this...

    please help?

    Thank you,
    Diggidy
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Diggidy View Post
    v=[3 -5 7 9]^T

    Find a basis of the subspace of R4 consisting of all vectors perpendicular to v.

    I have no clue how to do this...

    please help?

    Thank you,
    Diggidy
    Here is a hint Find the "plane" with v as a normal vector that passes though the origin. It is

    3x_1-5x_2+7x_3+9x_4=0

    Now solve this under-determined system to find three vectors perpendicular to v
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,306
    Thanks
    1282
    Same thing, really: a vector <w, x , y, z> is perpendicular to <3, -5, 7, 9> if and only if there dot product, 3w- 5x+ 7y+ 9z= 0.

    You can solve for any one of those variables in terms of the other 3, then write <w, x, y, z> in terms of those three.

    Rather than do that problem for you consider the space of all vector perpendicular to <-2, 4, 6, 5>. That would consist of all vectors <w, x, y, z> such that -2w+ 4x+ 6y+ 5z= 0. Solving for w, w= 2x+ 3y+ (5/2)z. Then, <w, x, y, z>= <2x+ 3y+ (5/2)z, x, y, z>= <2x, x, 0, 0> + <3y, 0, y, 0> + <(5/2)z, 0, 0, z>= x<2, 1, 0, 0>+ y<3, 0, 1, 0> + z<5/2, 0, 0, 1>.

    And now, a basis is obvious.

    If you don't like fractions, that last one can be written (z/2)<5, 0, 0, 2>
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Basis for a perpendicular vector
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 4th 2011, 05:48 PM
  2. Replies: 1
    Last Post: April 5th 2011, 05:21 PM
  3. Replies: 4
    Last Post: November 17th 2010, 11:12 AM
  4. Basis for subspace from a set of vectors.
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: June 18th 2010, 07:44 AM
  5. Basis perpendicular to a subspace spanned by 2 vectors
    Posted in the Advanced Algebra Forum
    Replies: 17
    Last Post: April 19th 2010, 06:06 PM

Search Tags


/mathhelpforum @mathhelpforum