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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- Apr 9th 2009, 07:46 PMyakuutOrthogonal Matrix
Can anyone help me with this question:

Find a 3 by 3 orthogonal matrix which has 5/13 as its (2,2)-element.. - Apr 10th 2009, 12:31 AMOpalg
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 $\displaystyle \mathbb{R}^3$.

- Apr 10th 2009, 02:39 AMShowcase_22
I know I didn't post this question but I haven't got a clue how to do it either. (Crying)

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}$? - Apr 10th 2009, 03:21 AMOpalg
$\displaystyle \begin{bmatrix}12/13\\5/13\\0\end{bmatrix}$ 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. - Apr 10th 2009, 03:45 AMShowcase_22
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?) - Apr 10th 2009, 04:04 AMOpalg
Yes. (Clapping)

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 $\displaystyle \tfrac{60-60}{169} = 0$. - Apr 10th 2009, 04:26 AMShowcase_22
YAY! (Cool)

Quote:

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 $\displaystyle \tfrac{60-60}{169} = 0$.

\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 $\displaystyle -\frac{12(5)}{13^2}+\frac{5(12)}{13^2}=0$. (Clapping)

Thanks Opalg.