How would you express a 45 degree rotation matrix as a 3 x 3?

I'm looking for an example of a real 3 x 3 matrix such that A^8=1, but A^4 does not equal 1. Am I correct that if C were such a 2 x2 rotation matrix, then matrix C would be the following:


(sqrt2)/2..... (-sqrt2)/2


(sqrt2)/2..... (sqrt2)/2





Can I express this as a 3 x 3 by adding a 1 in upper right corner and some zeros to make it 3 x 3 like this:





1 .................. 0 ................ 0


0 ......... (sqrt2)/2 ......... -sqrt2/2


0 ........ (sqrt2)/2 ......... sqrt2/2





I'm reasoning the determinants would be the same and the characteristic polynomial for the 3 x 3 would be (t-1)det(C).





I'm interested in any other 3 x 3 real matrices where A^8=1 but A^4 does not =1 as well. Thank you.

How would you express a 45 degree rotation matrix as a 3 x 3?
Everything you have said above is correct.


Also you could use a rotation in the opposite direction:


(1/√2 1/√2)


(-1/√2 1/√2)





Also to change it to 3 by 3 you could put the one (and extra zeros) in any position along the main diagonal.





Additionally you could rotate by 3π/4 or 5π/4 or 7π/4 (note that 7π/8 is same as -π/8 and that 5π/4 is -3π/4).





So all up you have 4 different rotations possible and can place the one in 3 different places so a total of 12 solutions based on rotation matrices derived from two by two matrices.








Note that putting the 1 in the first row makes a matrix which rotates in the yz plane. Putting the 1 in the middles makes a matrix which rotates in the xz plane and putting 1 bottom right makes a matrix which rotates in the xy plane.


So there are also an infinite number of solutions which perform rotations in arbitary other planes.


You can see how you can generate one rotated in any plane on the wikipedia link I have provided below.


http://en.wikipedia.org/wiki/Rotation_ma...