Decompose $P$ into the direct sum of irreducible representations. Note: I need help with part (c).
Consider the representation $P: S_3 \rightarrow GL_3$ where $P_{\sigma}$ is the permutation matrix associated to $\sigma$.
a) Determine the character $\chi_P : S_3 \rightarrow \mathbb{C}$
b) Find all the irreducible representations of $S_3$.
c) Decompose $P$ into the direct sum of irreducible representations.  That is, find a single matrix $Q$ so that $Q^{-1}P_{\sigma}Q$ is block diagonal where the blocks along the diagonal are either $T_{\sigma}$, $\Sigma_{\sigma}$ or $A_{\sigma}$
My Attempt
Here is my overall progress for the problem:
I let $e$ to be the identity permutation, $x = (1 \ 2 \ 3)$ and $y = (1 \ 2)$
Then, I let $P_x = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$, $P_y = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $P_e = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
The conjugacy classes are as followed:


*

*$\{e\}$

*$\{y, xy, yx\}$

*$\{x, x^2\}$


The character table shows that:
$$\chi (\{e\}) = 3$$
$$\chi (\{x, x^2\}) = 0$$
$$\chi (\{y, xy, yx\}) = 1$$
By the theorem I have applied, I have found three representations, which is congruent to the number of conjugacy classes.  They are:


*

*trivial representation $T$

*sign representation $\Sigma$

*two-dimensional representation $A$, presented as the symmetries of equilateral triangle


The sum of their dimensions corresponds to the theorem I applied:
$$d_1^2 + d_2^2 + d_3^2 = |S_3| = 6$$
The only possibility for equality to hold is $d_1 = d_2 = 1$ and $d_3 = 2$.
Another character table shows that:
$\chi_T (\{e\}) = 1$
$\chi_T (\{x, x^2\}) = 1$
$\chi_T (\{y, yx, xy\}) = 1$
$\chi_A (\{e\}) = 2$
$\chi_A (\{x, x^2\}) = -1$
$\chi_A (\{y, xy, yx\}) = 0$
$\chi_{\Sigma} (\{e\}) = 1$
$\chi_{\Sigma} (\{x, x^2\}) = 1$
$\chi_{\Sigma} (\{y, yx, xy\}) = -1$
Now, I am stuck in determining what is the matrix $Q$ for $Q^{-1}P_{\sigma}Q$.
I know that I need to do "change of basis" and work out the vectors and stuff like this, but I can't seem to find the thorough approach.
EDIT:
Here is what I currently have:
For the trivial representation, I have the vector $(1 , 1 , 1)$ spanning the invariant subspace.
For the two-dimensional representation, I need to find two vectors $v$ and $w$ such that:
$P_x v = -v/2 + \sqrt{3}w/2$
$P_x w = -v/2 - \sqrt{3}w/2$
$P_y v = v$
$P_y w = -w$
I found the $Q$ matrix, which is:
$$\begin{bmatrix} 1 & 1 & -\frac{(1 + \sqrt{3})}{2} \\ 1 & 1 & \frac{(1 + \sqrt{3})}{2} \\ 1 & -2 & 0 \end{bmatrix}$$
But it is wrong.
Any advices or comments you have?
 A: The permutation representation of $S_3$ on $\mathbb{C}^3$ (though everything works over the rationals in exactly the same way) decomposes as the direct sum of the trivial representation and the two dimensional reflection representation, which you have called $A$. 
More explicitly, the subspace spanned by the vector $v=(1,1,1)^t$ is a copy of the trivial representation, with $S_3$-stable complement spanned by the vectors $\alpha_1=(1,-1,0)^t$ and $\alpha_2=(0,1,-1)^t$, isomorphic to $A$. The matrix $Q$ can therefore be taken to have these three vectors as its columns.
In case these vectors $\alpha_1$ and $\alpha_2$ seemed to come out of the blue, here is where they actually came from: the usual Hermitian inner product on $\mathbb{C}^3$ is $S_3$-invariant, so the orthogonal complement of $v=(1,1,1)^t$ is an $S_3$-stable complement to its span. But this orthogonal complement consists exactly of the vectors whose coordinates sum to $0$, and the $\alpha$'s are just what I consider the most obvious basis of this space.
