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Thread: Orthogonal Matrix

  1. #1
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    Post Orthogonal Matrix

    Can anyone help me with this question:


    Find a 3 by 3 orthogonal matrix which has 5/13 as its (2,2)-element..
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    Quote Originally Posted by yakuut View Post
    Find a 3 by 3 orthogonal matrix which has 5/13 as its (2,2)-element..
    The columns of the orthogonal matrix have to form an orthonormal basis for the space. So start by finding a unit vector whose middle coordinate is 5/13. Then find two other vectors so that the three vectors form an orthonormal basis for $\displaystyle \mathbb{R}^3$.
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    Super Member Showcase_22's Avatar
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    I know I didn't post this question but I haven't got a clue how to do it either.

    After reading your post, here's what I tried:

    $\displaystyle \left( \frac{5}{13} \right)^2+a^2=1 \Rightarrow \ a=\frac{12}{13}$

    Therefore I just need to arrange $\displaystyle 0, \frac{12}{13}$ and $\displaystyle \frac{5}{13}$ in the matrix so that each vector is linearly independent from the other ones. I got this:

    $\displaystyle \begin{pmatrix}
    {\frac{12}{13}}&{\frac{12}{13}}&{0}\\
    {0}&{\frac{5}{13}}&{\frac{12}{13}}\\
    {\frac{5}{13}}&{0}&{\frac{5}{13}}
    \end{pmatrix}=\frac{1}{13}\begin{pmatrix}
    {12}&{12}&{0}\\
    {0}&{5}&{12}\\
    {5}&{0}&{5}
    \end{pmatrix}$?
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    Quote Originally Posted by Showcase_22 View Post
    I know I didn't post this question but I haven't got a clue how to do it either.

    After reading your post, here's what I tried:

    $\displaystyle \left( \frac{5}{13} \right)^2+a^2=1 \Rightarrow \ a=\frac{12}{13}$

    Therefore I just need to arrange $\displaystyle 0, \frac{12}{13}$ and $\displaystyle \frac{5}{13}$ in the matrix so that each vector is linearly independent from the other ones. I got this:

    $\displaystyle \begin{pmatrix}
    {\frac{12}{13}}&{\frac{12}{13}}&{0}\\
    {0}&{\frac{5}{13}}&{\frac{12}{13}}\\
    {\frac{5}{13}}&{0}&{\frac{5}{13}}
    \end{pmatrix}=\frac{1}{13}\begin{pmatrix}
    {12}&{12}&{0}\\
    {0}&{5}&{12}\\
    {5}&{0}&{5}
    \end{pmatrix}$?
    $\displaystyle \begin{bmatrix}12/13\\5/13\\0\end{bmatrix}$ is fine for the middle column, but the other two columns are not orthogonal to it. You could for example use $\displaystyle \begin{bmatrix}-5/13\\12/13\\0\end{bmatrix}$ for one of them. Can you see a vector (hint: without any fractions in its entries) that is orthogonal to both of these?

    Just to emphasise the point: linear independence is not enough, the columns must be orthogonal to each other.
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  5. #5
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by Opalg View Post
    Can you see a vector (hint: without any fractions in its entries) that is orthogonal to both of these?
    Ah, it's just $\displaystyle \begin{pmatrix}
    {0}\\
    {0}\\
    {1}
    \end{pmatrix}$ since those two vectors don't have a k component.

    So ultimately we have:

    $\displaystyle \begin{pmatrix}
    {0}&{\frac{12}{13}}&{-\frac{5}{13}}\\
    {0}&{\frac{5}{13}}&{\frac{12}{13}}\\
    {1}&{0}&{0}
    \end{pmatrix}
    $

    Can you also have:

    $\displaystyle \begin{pmatrix}
    {0}&{\frac{12}{13}}&{\frac{5}{13}}\\
    {0}&{\frac{5}{13}}&{\frac{12}{13}}\\
    {1}&{0}&{0}
    \end{pmatrix}
    $ or $\displaystyle \begin{pmatrix}
    {0}&{-\frac{12}{13}}&{-\frac{5}{13}}\\
    {0}&{\frac{5}{13}}&{\frac{12}{13}}\\
    {1}&{0}&{0}
    \end{pmatrix}
    $?

    (ie. any situation in which the magnitude of the rows and columns is 1?)
    Last edited by Showcase_22; Apr 10th 2009 at 03:46 AM. Reason: magnitude of zero? =S
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  6. #6
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    Quote Originally Posted by Showcase_22 View Post
    Ah, it's just $\displaystyle \begin{pmatrix}
    {0}\\
    {0}\\
    {1}
    \end{pmatrix}$ since those two vectors don't have a k component.

    So ultimately we have:

    $\displaystyle \begin{pmatrix}
    {0}&{\frac{12}{13}}&{-\frac{5}{13}}\\
    {0}&{\frac{5}{13}}&{\frac{12}{13}}\\
    {1}&{0}&{0}
    \end{pmatrix}
    $
    Yes.

    Quote Originally Posted by Showcase_22 View Post
    Can you also have:

    $\displaystyle \begin{pmatrix}
    {0}&{\frac{12}{13}}&{\frac{5}{13}}\\
    {0}&{\frac{5}{13}}&{\frac{12}{13}}\\
    {1}&{0}&{0}
    \end{pmatrix}
    $ or $\displaystyle \begin{pmatrix}
    {0}&{-\frac{12}{13}}&{-\frac{5}{13}}\\
    {0}&{\frac{5}{13}}&{\frac{12}{13}}\\
    {1}&{0}&{0}
    \end{pmatrix}
    $?

    (ie. any situation in which the magnitude of the rows and columns is 1?)
    No, because those columns are not orthogonal. The scalar product of $\displaystyle \bigl[\tfrac{12}{13},\tfrac5{13},0\bigr]^{\mathsf{T}}$ and $\displaystyle \bigl[\tfrac5{13},\tfrac{12}{13},0\bigr]^{\mathsf{T}}$ is $\displaystyle \tfrac{60+60}{169}$. You need exactly one coordinate to have a negative sign, so as to make the scalar product $\displaystyle \tfrac{60-60}{169} = 0$.
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  7. #7
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by Opalg View Post
    Yes.
    YAY!


    No, because those columns are not orthogonal. The scalar product of $\displaystyle \bigl[\tfrac{12}{13},\tfrac5{13},0\bigr]^{\mathsf{T}}$ and $\displaystyle \bigl[\tfrac5{13},\tfrac{12}{13},0\bigr]^{\mathsf{T}}$ is $\displaystyle \tfrac{60+60}{169}$. You need exactly one coordinate to have a negative sign, so as to make the scalar product $\displaystyle \tfrac{60-60}{169} = 0$.
    I get it. So you can also have $\displaystyle
    \begin{pmatrix}
    {0}&{-\frac{12}{13}}&{\frac{5}{13}}\\
    {0}&{\frac{5}{13}}&{\frac{12}{13}}\\
    {1}&{0}&{0}
    \end{pmatrix}
    $ since the scalar product of $\displaystyle \begin{pmatrix}
    {-\frac{12}{13}}\\
    {\frac{5}{13}}\\
    {0}
    \end{pmatrix}$ and $\displaystyle \begin{pmatrix}
    {\frac{5}{13}}\\
    {\frac{12}{13}}\\
    {0}
    \end{pmatrix}$ is $\displaystyle -\frac{12(5)}{13^2}+\frac{5(12)}{13^2}=0$.

    Thanks Opalg.
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