The natural representation of $SO(n)$ is irreducible for $n\ge 3$ The natural representation $(\pi,\mathbb C^n)$ of $SO(n)$ is the one for which 
$$\pi (g)z = g^{-1}z$$
for $g\in SO(n)$ and $z \in \mathbb C^n$ (the product $g^{-1}z$ is just the usual matrix product).
(Or maybe the representation is $gz$, not $g^{-1}z$. The exercise isn't clear what it meant. I don't think this changes much?)
I'm supposed to show this representation is irreducible if $n\ge 3$.
As a first part of the exercise, I proved that given two pairs $v_1, v_2$ and $u_1,u_2$ of orthonormal vectors in $\mathbb R^n$ then there exists a $x\in SO(n)$ for which
$$ xv_1 = u_1\\ xv_2=u_2 $$
Also, I proved that for all elements $c$ in the centralizer of $SO(n)$, the following holds:
$$\langle cv_1,v_2 \rangle = \langle cu_1,u_2 \rangle $$
But I have no idea what these preliminary steps have to do with the original problem.
 A: We want to show that the centralizer only consists of elements $\lambda\cdot I_n$ for some $\lambda \in \mathbb{C}$. Schur's lemma then immediately yields that the representation is irreducible. 
First, we note that we can express an arbitrary vector $x$ as $\sum_{i=1}^n \langle x,e_i\rangle e_i$, because $e_1,\ldots,e_n$ form an orthonormal basis. We want to use this expression to find out how $T$ works on the basis vectors. In other words, we want to compute $\langle Te_i,e_j\rangle$ for all $i\neq j$. However, we note that by the previous exercise we have $$\langle Te_i,e_j\rangle=\langle T(-e_i),-e_j\rangle=-\langle Te_i,e_j\rangle $$Hence,$\langle Te_i,e_j\rangle=0$ for all $i\neq j$, so we already know that $T$ is a diagonal matrix. It remains to show that all the diagonal elements must equal each other. Consider the matrix $P_{i,j}$, which is the identity matrix but with rows (or columns) $i$ and $j$ swapped. Using the fact that $P_{i,j}$, or $-P_{i,j}$, is in $SO(n)$, and that $TP_{i,j}=P_{i,j}T$, we get that $T_{i,i}=T_{j,j}$. We can repeated this argument and conclude that every diagonal element has to be the same. 
