Proving the 3-dimensional representation of S3 is reducible The 3-dimensional representation of the group S3 can be constructed by introducing a vector $(a,b,c)$ and permute its component by matrix multiplication.
For example, the representation for the operation $(23):(a,b,c)\rightarrow(a,c,b)$ is
$
D(23)=\left(\begin{matrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \end{matrix}\right)
$
and so forth. 
The exercise is to prove this representation is reducible. The hint tells me to find a common eigenvector for all 6 matrices which is just $(1,1,1)$. How do I proceed from here? Any help is appreciated.
 A: Here's another way to prove it's reducible, although it may depend on stuff you haven't learned yet. The order of the group is the sum of the squares of the degrees of the irreducible representations. So a group of order 6 can't have an irreducible representation of degree 3; $3^2\gt6$. 
A: You asked: The exercise is to prove this representation is
reducible. The hint tells me to find a common eigenvector for all
6 matrices which is just $(1,1,1)$. How do I proceed from here?
Any help is appreciated.
Representation is
irreducible
if and only if the corresponding $FG$-module has no non-trivial
$FG$-submodule. If you have managed to find a common eigenvector
$v$, then you have $vg=\lambda_g v$ for each $g\in G$; which
implies that $\operatorname{span}(v)$ is an $FG$-submodule.
In short: Looking for 1-dimensional submodules is the same thing
as looking for common eigenvectors. If your the whole $FG$-module
has dimension 3, it suffices to find out whether it has a
1-dimensional submodule, in order to decide whether it is
irreducible or not.
It is perhaps worth mentioning that the same approach would work for any $S_n$ and that this representation is called permutation representation of $S_n$.
Another interesting fact is that the permutation representation can be decomposed into this trivial representation and an irreducible representation of degree $n-1$. We have a question about this on this site; link to this MO thread is given there in comments.
Note: My answer is more-or-less the same as Benjalim's answer (which is deleted at the moment, so it is visible only for 10k+ users), with the exception that my answer uses modules and his answer avoids modules and uses only representations. (Both approaches, $FG$-modules and representations, are equivalent in the sense that we can get module from a representation and vice-versa. Hence we can describe properties of representation using the properties of the corresponding $FG$-module.)
A: I offer an elementary proof available to a beginner.
The group $S_3$ comprises the identity element $\mathbb I$, the transpositions $(12)$, $(23)$, $(31)$, the clockwise permutation $(123)$ and the counterclockwise permutation $(132)$.
It is implied here that a number is replaced by the preceding number in parentheses.
In 3 dimensions, the elements of $S_3$ are represented by
$$
 \mathbb{I}\;=\;
 \begin{bmatrix}
   1 & 0 & 0\\
   0 & 1 & 0\\
   0 & 0 & 1
 \end{bmatrix}~~,~~~~
  {(123)}\;=\;
 \begin{bmatrix}
   0 & 0 & 1\\
   1 & 0 & 0\\
   0 & 1 & 0
 \end{bmatrix}~~,~~~~
 {(132)}\;=\;
 \begin{bmatrix}
   0 & 1 & 0\\
   0 & 0 & 1\\
   1 & 0 & 0
 \end{bmatrix}~~,
 $$
$$
   {(23)}\;=\;
 \begin{bmatrix}
   1 & 0 & 0\\
   0 & 0 & 1\\
   0 & 1 & 0
 \end{bmatrix}~~,~~~~~
 {(13)}\;=\;
 \begin{bmatrix}
   0 & 0 & 1\\
   0 & 1 & 0\\
   1 & 0 & 0
 \end{bmatrix}~~,~~~~~
  {(12)}\;=\;
 \begin{bmatrix}
   0 & 1 & 0\\
   1 & 0 & 0\\
   0 & 0 & 1
 \end{bmatrix}~~.~~~~
 $$
Following Dirac (Principles of Quantum Mechnaics, 1958), introduce the operators
$$
 X_1\,=\,\mathbb{I}~~,
 $$
$$
 X_2\,=\,\frac{(12)\,+\,(23)\,+\,(31) }{3}~~,
 $$
$$
 X_3\,=\,\frac{(123)\,+\,(132) }{2}~~.
 $$
It can be shown by inspection that each $X_i$ commutes with all the elements of $S_3$.
Inserting the matrix representations of the group elements into the expressions for $X_i\,$, $\,$we get:
$$
 X_1\,=\;
 \begin{bmatrix}
   1 & 0 & 0\\
   0 & 1 & 0\\
   0 & 0 & 1
 \end{bmatrix}~~~~,~~~~
 X_2\,=\;
   \begin{bmatrix}
   1 & 1 & 1\\
   1 & 1 & 1\\
   1 & 1 & 1
 \end{bmatrix}~~~~,~~~~
  X_3\,=\;
   \begin{bmatrix}
   0 & 1 & 1\\
   1 & 0 & 1\\
   1 & 1 & 0
 \end{bmatrix}~~~.
 $$
Since in this representation we have $X_3 = X_2-X_1$, and since $X_2$ and $X_1$ are linearly independent, we see that in this representation the space of commutants is two-dimensional.
A corollary of the Schur lemma says that a non-zero square matrix commuting with all of the matrices of an irreducible representation is a constant multiple of the unit matrix. For completely reducible representations, this works also in the reciprocal direction: if the unit operator is the only one (up to multiplication by a number)
that commutes with all operators of a completely reducible representation, then the representation is irreducible. [A propos, the said corollary works only in finite or countable dimensions, while the reciprocal statement is valid in any dimensions.]
Since in our situation we have a commutant $X_2$ different from $\mathbb I$, the representation must be reducible. Indeed, the vector $(1,\,1,\,1)$ happens to be a common eigenvector of all 3-dim representations of the group elements. This fact can be either checked directly or derived more elegantly from the fact that it is also an eigenvector of $X_2$ and that $X_2$ is a commutant.
To conclude, the line defined by the vector $(1,\,1,\,1)$ is an invariant subspace and is home to a subrepresentation of $S_3$.
