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Given the matrix $A = 1/7\begin{bmatrix} 6 & 3 & a_{13} & \\ -2 & 6 & a_{23} \\3 & a_{32} & a_{33} \end{bmatrix}$

How do replace the $a_{ij}$'s with real entries such that the matrix becomes a rotation matrix?

I know it should have a determinant of - 1 and all eigenvalues should be either 1 or -1, but it turns out to be a four variables equation, how do I find those values?

Edit: I forgot the scalar in front of the matrix.

Leroy
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1 Answers1

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Each column has to have length $1$, so $$3^3+6^2+a_{32}^2=7^2.$$ This determines $a_{32}$ up to sign. But the first two columns must be orthogonal, and that will determine the sign of $a_{32}$.

Also the rows have to have length $1$. Therefore $$6^3+3^2+a_{13}^2=7^2$$ etc. So one gets the entries in the last column up to sign. Orthogonality of rows means that one sign determines the others. Finally a rotation matrix has determinant $1$, so this determines the final sign.

Angina Seng
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  • You can also get the last column as the cross product of the other two. – Jyrki Lahtonen Aug 30 '17 at 05:17
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    @JyrkiLahtonen Gordon Pall (1940) showed that all 3 by 3 rational orthogonal matrices come from one type of quaternion calculation, the same material as we get with "Pythagorean quadruples." The matrix is formula (10) http://zakuski.math.utsa.edu/~kap/Pall_Automorphs_1940.pdf – Will Jagy Aug 30 '17 at 16:43
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    Thanks for the reference @WillJagy! Because I played with reflection groups in my youth I have been in the habit of getting rational orthogonal matrices as products of two reflections. Like here. I'm not sure I get all those matrices from that recipe. That we get them all from quaternions (probably rational) does not necessarily surprise me - even though I didn't know about it. – Jyrki Lahtonen Aug 30 '17 at 17:51
  • @JyrkiLahtonen right, it may not be clear but you just use integers. So $7 = 2^2 + 1^2 + 1^2 + 1^2$ is the only correct sum up to order and $\pm$ signs, so matrix (10) always gives different versions of the 2,3,6 entries. My favorite paper is Jones and Pall (1939), this is in the same area http://zakuski.math.utsa.edu/~kap/Jones_Pall_1939.pdf – Will Jagy Aug 30 '17 at 18:10