Results 1 to 4 of 4

Math Help - Orthonormality

  1. #1
    Member
    Joined
    Sep 2009
    From
    Minnesota
    Posts
    80
    Thanks
    8

    Orthonormality

    Given the vectors <1/2, 1/2, 1/2, 1/2>,<1/2, 1/2, -1/2, -1/2>,<1/2, -1/2, 1/2, -1/2>, I'm asked to find a vector such that all four will be orthonormal. On such vector is <1/2, -1/2, -1/2, 1/2>. I'm also asked to find how many such vectors there are (given that the vector has to be a unit vector). Intuition from looking at the first two dimensions tells me that the answer is likely two; however I don't know how to prove this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Lord Voldemort View Post
    Given the vectors <1/2, 1/2, 1/2, 1/2>,<1/2, 1/2, -1/2, -1/2>,<1/2, -1/2, 1/2, -1/2>, I'm asked to find a vector such that all four will be orthonormal. On such vector is <1/2, -1/2, -1/2, 1/2>. I'm also asked to find how many such vectors there are (given that the vector has to be a unit vector). Intuition from looking at the first two dimensions tells me that the answer is likely two; however I don't know how to prove this.
    I don't quite understand your question. You can use Gram-Schmidt to normalize the vectors if that is what your intentions are for this problem.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,444
    Thanks
    1863
    dwsmkth, the three given vectors are already orthogonal so "Gram-Schmidt" is not necessary.

    Any vector, <a, b, c, d> that is normal to the three given must satisfy (1/2)a+ (1/2)b+ (1/2)c+ (1/2)d= 0, (1/2)a+ (1/2)b-(1/2)c-(1/2)d= 0, and (1/2)a-(1/2)b+ (1/2)c- (1/2)d= 0. An obvious first step is to multiply all equations by 2 to get a+ b+ c+ d= 0, a+ b- c- d= 0, and a- b+ c- d= 0. Adding the first two equations gives 2a+ 2b= 0 so b= -a. Adding the last two gives 2a-2d= 0 so d= a. Putting b= -a and d= a into the third equation, a- a- c- a= -c-a= 0 so c= -a. That is, any vector perpendicular to all of the three given vectors is of the form <a, -a, -a, a>. The length of that vector is \sqrt{a^2+ (-a)^2+ (-a)^2+ a^2}= \sqrt{4a^2}= 2|a|. Note the "absolute value"- \sqrt{a^2}= |a| since a itself can be either positive or negative while the square root is always positive. There are exactly two values of a that make 2|a|= 1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2009
    From
    Minnesota
    Posts
    80
    Thanks
    8
    Oh duh... I got the three equations but then thought it would result in an infinite number of solutions in haste. Thanks so much though.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum