1. ## Conjugation of matrices.

def.: $\displaystyle SO(3)$ is the collection of lineair orthogonal transformations of $\displaystyle \mathbb{R}^3$, the elements are $\displaystyle 3\times 3$ matrices.

I want to show that two element of $\displaystyle SO(3)$ are conjugated if and only if they have the same trace.

(Hint: Maby you want to use: if $\displaystyle A \in SO(3)\ \mbox{then}\ \exists p \in \mathbb{R}^3,\ p\neq 0\ \mbox{and}\ Ap=p$)

2. Okay, what is the definition of "conjugate matrices"?

3. Originally Posted by HallsofIvy
Okay, what is the definition of "conjugate matrices"?
C'mon, this is a tough question. At least assume the OP has some knowledge!

4. Originally Posted by HallsofIvy
Okay, what is the definition of "conjugate matrices"?
Two matrices A and B are conjugated if there exists an orthogonal transformation, such that:

$\displaystyle A= CBC^{-1}$

5. Originally Posted by bram kierkels
Two matrices A and B are conjugated if there exists an orthogonal transformation, such that:

$\displaystyle A= CBC^{-1}$

No. Two square matrices $\displaystyle A,B$ are conjugate if there exists SOME invertible matrix $\displaystyle P$ s.t. $\displaystyle A=PBP^{-1}$

One direction of your problem is almost trivial: ANY two conjugate matrices, whether orthogonal or not, have the same trace (and the same determinant, and the same eigenvalues...).
The other direction looks highly suspicious to me, but I cannot prove it isn't true.

Tonio

6. Using the hint : by an appropriate conjugation, you can put any $\displaystyle A \in SO(3)$ in the form $\displaystyle \left(\begin{matrix}1 &0 & 0 \\ 0 & a & b \\ 0&c & d \end{matrix}\right)$ by an appropriate base change... and in fact we can also suppose $\displaystyle \left(\begin{matrix} a &b \\ c& d\end{matrix}\right)=\left(\begin{matrix} \cos \theta &-\sin \theta \\ \sin \theta& \cos \theta\end{matrix}\right)$. (What this means is essentially that a rotation of three-dimensional Euclidean space is always a rotation around an axis; the axis is fixed, and its orthogonal complement is rotated like a plane. So two rotations are conjugate only if the angles of rotation are equal up to a sign.)

Can you take it from there?

Another way might be to use the isomorphism $\displaystyle SO(3) \cong SU(2)/\{\pm \mbox{id.}\}$, and the fact that two Möbius transformations are conjugate if and only if they have the same multiplier. But that's more complicated.