Lie algebras homeomorphism problem

In my homework problem I have to prove that

$F: (\Bbb{R}^3,\times) \to (so(3),[,]),\ F(v)=\begin{pmatrix}0&-v_3&v_2 \\ v_3& 0&-v_1\\ -v_2&v_1&0 \end{pmatrix}=\hat{v}$ is a homeomorphism of Lie Algebras. Furthermore, for $w \in \Bbb{R}^3$ we have $\hat{v}w=v \times w$.

I have proved the above facts. The last part of the problem sounds like this:

Prove that for $A \in O(3)$ we have $Ad_A \hat{v}=\widehat{Av}$, i.e. $A\hat v A^{-1}=\widehat{Av}$. (where taking "$\text{^}$" means applying F to the corresponding vector)

Is there an easy way to prove this last part, without brute force computations?

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Are the words "The last part of the problem..." a quote from the homework assignent? – Henning Makholm Jan 22 '12 at 13:27
You can try to prove the corresponding relation for the Lie algebra $[a,\hat{\nu}]= \widehat{a \nu}$ with $a\in so(3)$ and then think what happens if $\det A = -1$. – Fabian Jan 22 '12 at 13:48
Hm, since $\mathtt{A}^{-1}=\mathtt{A}^\top$, you simply have to multiply $\mathtt{A}\hat{v}\mathtt{A}^\top$ out. Do you mean this by brute force computation? It is just two matrix multiplications, or do I overlook something?? – Hauke Strasdat Jan 22 '12 at 14:39
@HaukeStrasdat: Yes, that is what I mean by brute force. I've done that, and the results are not pleasing to watch and write. There are two matrix multiplications, but we don't know either of the matrices, so there are a lot of unknowns. The resulting matrix cannot be written on the width of an A4 paper. I was thinking that there is a trick, a theorem, something related to the given homeomorphism that can simplify things a bit. – Beni Bogosel Jan 22 '12 at 14:58

$O(3)$ is generated by matrices of the form $\exp(\hat w)$, $\hat w \in so(3)$ and $-I$. It is clear for $-I$ and for $A = \exp(\hat w)$ we have $$Ad_A \hat v = \exp(ad_{\hat w} v) = \hat v + [\hat w, \hat v] + \frac{1}{2} [\hat w, [\hat w, \hat v]] + \cdots = \hat v + F(w \times v) +\frac{1}{2} F(w \times (w \times v)) + \cdots$$ $$= F(v) + F(\hat w v) + \frac{1}{2}F(\hat w^2 v) + \cdots = F(v + \hat w v + \frac{1}{2}\hat w^2 v + \cdots ) = F(\exp(\hat w) v) = \widehat{Av}.$$
EDIT: err...it isn't true for $A = -I$. But it looks to be true for $A \in SO(3)$.