Rotation matrix always has eigenvalue 1

I've been trying to disprove the statement:

"If A is a 2x2 real matrix without real eigenvalues then A is a rotation matrix."

However I have read in numerous places that rotation matrices always have 1 as an eigenvalue, so the above statement would not hold, because if A does not have any real eigenvalues, then it can't have 1 as an eigenvalue and hence can't be a rotation matrix, however I am struggling to prove this in the 2x2 case.

My (poor) attempt:

The 2x2 rotation matrix is always of the form:

\begin{pmatrix} cos{\theta} & -sin{\theta} \\ sin{\theta} & cos{\theta} \end{pmatrix}

So it has characteristic equation

(cos{\theta} - x)^2 + sin^2{\theta} = 0

where x is our eigenvalue.

which reduces down to

x^2 - 2xcos{\theta} + 1 = 0

which when solving with the quadratic formula we get

x = cos{\theta} \pm \sqrt{cos^2{\theta} - 1}

however I can't see how x = 1 is always a solution to this? In fact, I'm fairly certain it's not, any advice at all would be great.

(edited 11 years ago)

Reply 1

ghostwalker

Original post by TheIrrational

...

I refer you to wiki

1 is an eigenvalue of a 3x3 rotation matrix, but not a 2x2 rotation matrix.

Reply 2

TheIrrational

Original post by ghostwalker

I refer you to wiki

1 is an eigenvalue of a 3x3 rotation matrix, but not a 2x2 rotation matrix.

Ah god, I overlooked matrix size when I was reading through these. Thanks, that would explain why it doesn't work for me.

Edit: Was easy enough to disprove by counter example anyway.