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Math Help - Matrix equation

  1. #1
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    Matrix equation

    Let M = { [a -b] , a,b E R }
    ..............[b a]

    Solve X^4+X^2+I=0, where X E M

    I attempted this question a while ago. I remember finding the eigenvalues and eigenvectors (X^4+X^2=-I and treat X^4+X^2 as a matrix itself) but it led to nowhere. I then tried to use a linear map but no luck either.

    I would love to hear ideas on how to solve this one.

    Thanks.
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  2. #2
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    You would solve for the eigenvalues by solving the numeric equation x^4+ x^2- 1= 0. If we let y= x^2, then we have y^2+ y- 1= 0 which, by the quadratic formula, has roots \frac{-1\pm i\sqrt{3}}{2}. The eigenvalues of the original equation are the roots of those and so will be very messy!
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  3. #3
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    M = \begin{bmatrix}a & -b \\ b & a\end{bmatrix} = \gamma\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}

    Where \gamma = \sqrt{a^2 + b^2} and \theta = \tan^{-1}(b/a).

    So M is a rotation by \theta with a scaling by \gamma.

    Immediately using a geometric interpretation, we can see that a rotation by 60 degrees is a solution for X. Also a rotation by 120 degrees is another solution.
    There are more solutions, but hopefully this helps you get started finding roots.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    An alternative: we can use the natural field isomorphism:

     \varphi: \mathcal{M}\rightarrow{\mathbb{C}},\quad  \varphi\begin{bmatrix}{a}&{-b}\\{b}&{\;\;a}\end{bmatrix}=a+bi

    so, solving

    X^4+X^2+I=0 in \mathcal{M}

    it is equivalent to solve

    z^4+z^2+1=0 in \mathbb{C}.

    But,

    z^4+z^2+1=(z^2+1)^2-z^2=(z^2+z+1)(z^2-z+1)=0

    z=\dfrac{-1\pm\sqrt{3}i}{2},\;z=\dfrac{1\pm\sqrt{3}i}{2}

    As a consequence the roots of the given equation are the elements of the set:

    \mathcal{S}=\left\{{\dfrac{1}{2}\begin{bmatrix}{-1}&{-\sqrt{3}}\\{\sqrt{3}}&{-1}\end{bmatrix},\;\dfrac{1}{2}\begin{bmatrix}{-1}&{\sqrt{3}}\\{-\sqrt{3}}&{-1}\end{bmatrix},  \;\dfrac{1}{2}\begin{bmatrix}{1}&{-\sqrt{3}}\\{\sqrt{3}}&{1}\end{bmatrix},\;\dfrac{1}  {2}\begin{bmatrix}{1}&{-\sqrt{3}}\\{\sqrt{3}}&{1}\end{bmatrix}}\right\}


    Fernando Revilla
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  5. #5
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    Thanks for your help guys.

    FernandoRevilla, I haven't learn anything about natural field isomorphism so I don't understand what you did there...Is that some kind of a linear map? Is there any link about it so I can have a read? I tried google but nothing really came up.

    EDIT: you can actually prove that its a linear map. =) Thanks that's a very pretty solution!

    snowtea, could you please explain how you got 60 as a solution using geometric interpretation? I didn't quite see that =S

    I get everything up to "So M is a rotation by with a scaling by gamma." X^4 and X^2 mean rotations by 4*theta and 2*theta respectively(with scalings too). But what about X^4+X^2+I? How did you see that 60 is a solution so quickly?
    Last edited by sakodo; January 8th 2011 at 04:08 PM.
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  6. #6
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    Let X be a rotation by 60 degrees (no scaling).

    Draw a vector v.
    Multiplying by I keeps it the same.
    Multiplying it by X^2 rotates it by 120 degrees.
    Multiplying it by X^4 rotates it by 240 degrees.

    (X^4+X^2+I) v is the sum of all the above vectors.
    You can show this is always 0 using some simple trigonometry.
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  7. #7
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    Quote Originally Posted by snowtea View Post
    Let X be a rotation by 60 degrees (no scaling).

    Draw a vector v.
    Multiplying by I keeps it the same.
    Multiplying it by X^2 rotates it by 120 degrees.
    Multiplying it by X^4 rotates it by 240 degrees.

    (X^4+X^2+I) v is the sum of all the above vectors.
    You can show this is always 0 using some simple trigonometry.
    Thanks that makes sense now.
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