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Math Help - 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 \mathbb{R}^3.
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  3. #3
    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:

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

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

    \begin{pmatrix}<br />
{\frac{12}{13}}&{\frac{12}{13}}&{0}\\ <br />
{0}&{\frac{5}{13}}&{\frac{12}{13}}\\ <br />
{\frac{5}{13}}&{0}&{\frac{5}{13}}<br />
\end{pmatrix}=\frac{1}{13}\begin{pmatrix}<br />
{12}&{12}&{0}\\ <br />
{0}&{5}&{12}\\ <br />
{5}&{0}&{5}<br />
\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:

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

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

    \begin{pmatrix}<br />
{\frac{12}{13}}&{\frac{12}{13}}&{0}\\ <br />
{0}&{\frac{5}{13}}&{\frac{12}{13}}\\ <br />
{\frac{5}{13}}&{0}&{\frac{5}{13}}<br />
\end{pmatrix}=\frac{1}{13}\begin{pmatrix}<br />
{12}&{12}&{0}\\ <br />
{0}&{5}&{12}\\ <br />
{5}&{0}&{5}<br />
\end{pmatrix}?
    \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 \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 \begin{pmatrix}<br />
{0}\\ <br />
{0}\\<br />
{1}<br />
\end{pmatrix} since those two vectors don't have a k component.

    So ultimately we have:

    \begin{pmatrix}<br />
{0}&{\frac{12}{13}}&{-\frac{5}{13}}\\ <br />
{0}&{\frac{5}{13}}&{\frac{12}{13}}\\ <br />
{1}&{0}&{0}<br />
\end{pmatrix}<br />

    Can you also have:

    \begin{pmatrix}<br />
{0}&{\frac{12}{13}}&{\frac{5}{13}}\\ <br />
{0}&{\frac{5}{13}}&{\frac{12}{13}}\\ <br />
{1}&{0}&{0}<br />
\end{pmatrix}<br />
or \begin{pmatrix}<br />
{0}&{-\frac{12}{13}}&{-\frac{5}{13}}\\ <br />
{0}&{\frac{5}{13}}&{\frac{12}{13}}\\ <br />
{1}&{0}&{0}<br />
\end{pmatrix}<br />
?

    (ie. any situation in which the magnitude of the rows and columns is 1?)
    Last edited by Showcase_22; April 10th 2009 at 04:46 AM. Reason: magnitude of zero? =S
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    Quote Originally Posted by Showcase_22 View Post
    Ah, it's just \begin{pmatrix}<br />
{0}\\ <br />
{0}\\<br />
{1}<br />
\end{pmatrix} since those two vectors don't have a k component.

    So ultimately we have:

    \begin{pmatrix}<br />
{0}&{\frac{12}{13}}&{-\frac{5}{13}}\\ <br />
{0}&{\frac{5}{13}}&{\frac{12}{13}}\\ <br />
{1}&{0}&{0}<br />
\end{pmatrix}<br />
    Yes.

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

    \begin{pmatrix}<br />
{0}&{\frac{12}{13}}&{\frac{5}{13}}\\ <br />
{0}&{\frac{5}{13}}&{\frac{12}{13}}\\ <br />
{1}&{0}&{0}<br />
\end{pmatrix}<br />
or \begin{pmatrix}<br />
{0}&{-\frac{12}{13}}&{-\frac{5}{13}}\\ <br />
{0}&{\frac{5}{13}}&{\frac{12}{13}}\\ <br />
{1}&{0}&{0}<br />
\end{pmatrix}<br />
?

    (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 \bigl[\tfrac{12}{13},\tfrac5{13},0\bigr]^{\mathsf{T}} and \bigl[\tfrac5{13},\tfrac{12}{13},0\bigr]^{\mathsf{T}} is \tfrac{60+60}{169}. You need exactly one coordinate to have a negative sign, so as to make the scalar product \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 \bigl[\tfrac{12}{13},\tfrac5{13},0\bigr]^{\mathsf{T}} and \bigl[\tfrac5{13},\tfrac{12}{13},0\bigr]^{\mathsf{T}} is \tfrac{60+60}{169}. You need exactly one coordinate to have a negative sign, so as to make the scalar product \tfrac{60-60}{169} = 0.
    I get it. So you can also have <br />
\begin{pmatrix}<br />
{0}&{-\frac{12}{13}}&{\frac{5}{13}}\\ <br />
{0}&{\frac{5}{13}}&{\frac{12}{13}}\\ <br />
{1}&{0}&{0}<br />
\end{pmatrix}<br />
since the scalar product of \begin{pmatrix}<br />
{-\frac{12}{13}}\\ <br />
{\frac{5}{13}}\\<br />
{0}<br />
\end{pmatrix} and \begin{pmatrix}<br />
{\frac{5}{13}}\\ <br />
{\frac{12}{13}}\\<br />
{0}<br />
\end{pmatrix} is -\frac{12(5)}{13^2}+\frac{5(12)}{13^2}=0.

    Thanks Opalg.
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