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.

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 $S_3$.
