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Math Help - Orthogonally diagonalise a 3x3 matrix

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    Orthogonally diagonalise a 3x3 matrix

    I've been at this for a while and I'm now stumped.

    The matrix, A is

    5 -4 0
    -4 3 -4
    0 - 4 1


    Here's what I've got so far.


    eigenvalues are λ = 9, λ = -3, λ = 3


    corresponding eigen vectors are (2k, -2k, k), (k/2, k, k), (-k, k/2, k)


    {(2, -2, 1)} is a basis for s(9)
    {(1,2,2)} is a basis for s(-3)
    {(-2,-1,2)} is a basis for s(3)


    So that gives the set

    E = {(2,-2,1),(1,2,2),(-2,-1,2)} which is an eigenvector basis of A (not sure if this is even needed as I think I need an orthonormal basis)

    I think I need to find an orthonormal eigenvector basis next, but I can't understand how to do that (if that is actually next). As that should give me the transition matrix.

    So looking for someone to hold my hand to the end of this process please
    Last edited by marky82; February 9th 2013 at 09:22 AM.
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    Super Member ILikeSerena's Avatar
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    Re: Orthogonally diagonalise a 3x3 matrix

    Hi marky82!

    An orthonormal basis requires 2 things: orthogonal and normal.
    Orthogonal means that he vectors are orthogonal.
    Normal means that they have length 1.

    Are your current vectors orthogonal?
    And what is their length?
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    Re: Orthogonally diagonalise a 3x3 matrix

    Quote Originally Posted by ILikeSerena View Post
    Hi marky82!

    An orthonormal basis requires 2 things: orthogonal and normal.
    Orthogonal means that he vectors are orthogonal.
    Normal means that they have length 1.


    Are your current vectors orthogonal?
    And what is their length?

    Hello again

    the eigenvectors are othogonal as they correspond to the distinct eigenvalues of the symmetric matrix above.

    I have no idea on how to find the length so not sure what is next :P
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    Super Member ILikeSerena's Avatar
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    Re: Orthogonally diagonalise a 3x3 matrix

    Two vectors are orthogonal if they are perpendicular.
    This is equivalent to their dot product being zero.

    What are the dot products of each pair of vectors in the basis?

    The (regular) length of a vector is \sqrt{x^2+y^2+z^2}
    What are the lengths of the vectors in your basis?

    To normalize a vector, you divide each of its coordinates by its length...
    Thanks from marky82
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    Re: Orthogonally diagonalise a 3x3 matrix

    Quote Originally Posted by ILikeSerena View Post
    Two vectors are orthogonal if they are perpendicular.
    This is equivalent to their dot product being zero.

    What are the dot products of each pair of vectors in the basis?

    The (regular) length of a vector is \sqrt{x^2+y^2+z^2}
    What are the lengths of the vectors in your basis?

    To normalize a vector, you divide each of its coordinates by its length...
    Sorry for the late reply, I decided to give my brain a break and have another go at it today, to see if I can finish it
    Last edited by marky82; February 10th 2013 at 02:23 AM.
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