# Conjugation of matrices.

• Sep 30th 2010, 06:01 AM
bram kierkels
Conjugation of matrices.
def.: $SO(3)$ is the collection of lineair orthogonal transformations of $\mathbb{R}^3$, the elements are $3\times 3$ matrices.

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

(Hint: Maby you want to use: if $A \in SO(3)\ \mbox{then}\ \exists p \in \mathbb{R}^3,\ p\neq 0\ \mbox{and}\ Ap=p$)
• Sep 30th 2010, 07:14 AM
HallsofIvy
Okay, what is the definition of "conjugate matrices"?
• Sep 30th 2010, 08:22 AM
Swlabr
Quote:

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!
• Oct 2nd 2010, 03:47 AM
bram kierkels
Quote:

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:

$A= CBC^{-1}$
• Oct 2nd 2010, 06:23 AM
tonio
Quote:

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

$A= CBC^{-1}$

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

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
• Oct 2nd 2010, 11:25 AM
Bruno J.
Using the hint : by an appropriate conjugation, you can put any $A \in SO(3)$ in the form $\left(\begin{matrix}1 &0 & 0 \\ 0 & a & b \\ 0&c & d \end{matrix}\right)$ by an appropriate base change... and in fact we can also suppose $\left(\begin{matrix} a &b \\ c& d\end{matrix}\right)=\left(\begin{matrix} \cos \theta &-\sin \theta \\ \sin \theta& \cos \theta\end{matrix}\right)$. (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 $SO(3) \cong SU(2)/\{\pm \mbox{id.}\}$, and the fact that two Möbius transformations are conjugate if and only if they have the same multiplier. But that's more complicated.