Results 1 to 2 of 2

Math Help - Orthogonally diagonalizable

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    2

    Orthogonally diagonalizable

    given the basis:

    x1= (1,-2,2)
    how did they get the orthonormal basis
    of (1/3,-2/3,2/3)

    x2= (2,1,0)
    how did the get the orthonomal basis
    1/sqr(5)*x2

    x3=(-2,0,1)
    how did the get the orthonomal basis
    1/3sqr(5)(-2,4,5)

    please help with steps
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,567
    Thanks
    1409
    Quote Originally Posted by mathisdullnboring View Post
    given the basis:

    x1= (1,-2,2)
    how did they get the orthonormal basis
    of (1/3,-2/3,2/3)

    x2= (2,1,0)
    how did the get the orthonomal basis
    1/sqr(5)*x2

    x3=(-2,0,1)
    how did the get the orthonomal basis
    1/3sqr(5)(-2,4,5)

    please help with steps
    I think you need help with the definitions. A single vector is NOT a "basis", othonormal or not, unless you have a vector space of dimension 1 which I do not believe is the case here. I think you are asking how to go from the basis {(1,-2,2), (2,1,0), (-2,0,1)} of R^3 to the orthonormal basis {(1/3,-2/3,2/3), 1/sqr(5)(2,1,0),1/3sqr(5)(-2,4,5)}.

    I wrote out an entire long, complicated, explanation of the "Gram-Schmidt" orthogonalization process, to create a basis of vectors "orthogonal" (perpendicular) to one another from a general basis and then, applying it to this example, started getting "0"s! The first two vectors are already orthogonal to the each other, and the first to the second but not the second to the third. We can go ahead and "normalize" the first two. (1,-2,2) has length \sqrt{1^2+ (-2)^2+ (2)^2}= \sqrt{9}= 3. To get a vector of length 1 in the same direction as (1,-2,2), divide by 3: (1/3, -2/3, 2/3). That is the first vector in your new "orthonormal" basis. To get a vector of length 1 in the same direction as (2,1,0), divide by its length: \sqrt{2^2+ 1^2+ 0^2}= \sqrt{5}. (1/\sqrt{5})(2,1,0) is the "normalized" vector in the same direction as (2,1,0).

    The third is the hardest because (-2,0,1) is not already perpendicular to (2,1,0) and so not to (2/\sqrt{5}, 1/\sqrt{5}, 0). To construct a vector that is, we need to divide (-2,0,1) into two "components", one parallel to (2/\sqrt{5},1/\sqrt{5},0), the other perpendicular. And, since (2/\sqrt{5},1/\sqrt{5},0) is in the same direction as (2,1,0) we can get simplify the calculations by using (2,1,0). To get the one parallel to (2, 1,0), "project" (-2,0,1) onto (2,1,0). I'm sure you already have a formula for that. The projection of vector \vec{u} on vector \vec{v} is \frac{\vec{u}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}. Here, [tex](-2,0,1)\cdot(2,1,0)= -4[tex] and we have already seen that |(2,1,0)|= 5 so the projection of (-2,0,1) onto (2,1,0) is (-4/5)(2,1,0)= (-8/5, -4/5, 0). That is parallel to (2,1,0) and so (-2,0,1)- (-8/5, -4/5, 0)= (-2/5, 4/5, 1) is orthogonal to (2,1,0)- you can check the dot product: (-2/5, 4/5, 1)\cdot(2,1,0)= -4/5+ 4/5+ 0= 0. We might also check that (-2/5, 4/5, 1)\cdot(1,-2,2)= -2/5- 8/5+ 2= -2+2= 0 so (-2/5, 4/5, 1) is still perpendicular to (1,-2,2).

    Now, we need only normalize (-2/5, 4/5, 1). It's length is \sqrt{(-2/5)^2+ (4/5)^2+ 1}= \sqrt{45/25}= (3/5)\sqrt{5}. Dividing (-2/5, 4/5, 1) by (3/5)\sqrt{5} gives (\frac{1}{3\sqrt{5}}(-2, 4, 5), the third vector in your orthonormal basis.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. diagonalizable
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: April 15th 2011, 07:53 AM
  2. How do I orthogonally diagonalize a matrix?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: July 23rd 2009, 10:33 AM
  3. Orthogonally diagonalize the matrix
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: April 22nd 2009, 02:39 AM
  4. diagonalizable
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: July 23rd 2008, 11:10 AM
  5. Orthogonally diagonalize the symmetric matrix
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 15th 2008, 07:26 AM

Search Tags


/mathhelpforum @mathhelpforum