Can anyone help me with this question:
Find a 3 by 3 orthogonal matrix which has 5/13 as its (2,2)-element..
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.
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?)
Yes.
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 $\displaystyle \tfrac{60-60}{169} = 0$.
YAY!
I get it. So you can also have $\displaystyleNo, 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 $\displaystyle -\frac{12(5)}{13^2}+\frac{5(12)}{13^2}=0$.
Thanks Opalg.