# Orthogonal Matrix

• Apr 9th 2009, 07:46 PM
yakuut
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..
• Apr 10th 2009, 12:31 AM
Opalg
Quote:

Originally Posted by yakuut
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 $\displaystyle \mathbb{R}^3$.
• Apr 10th 2009, 02:39 AM
Showcase_22
I know I didn't post this question but I haven't got a clue how to do it either. (Crying)

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

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 AM
Opalg
Quote:

Originally Posted by Showcase_22
I know I didn't post this question but I haven't got a clue how to do it either. (Crying)

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

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.
• Apr 10th 2009, 03:45 AM
Showcase_22
Quote:

Originally Posted by Opalg
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?)
• Apr 10th 2009, 04:04 AM
Opalg
Quote:

Originally Posted by Showcase_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:

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

Yes. (Clapping)

Quote:

Originally Posted by Showcase_22
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 $\displaystyle \tfrac{60-60}{169} = 0$.
• Apr 10th 2009, 04:26 AM
Showcase_22
Quote:

Originally Posted by Opalg
Yes. (Clapping)

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 $\displaystyle \tfrac{60-60}{169} = 0$.
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 $\displaystyle -\frac{12(5)}{13^2}+\frac{5(12)}{13^2}=0$. (Clapping)

Thanks Opalg.