If $\rho$ is the standard representation of $S_3$ and $W$ is a subspace, show that $(\rho,W)$ has no invariant one dimensional invariant subspace. We define the standard representation of $S_3$ as $\rho:S_3\to\text{GL}(V)$ on the standard basis $\rho_{\sigma}(e_i)=e_{\sigma(i)}.$
Let $W=\{(a,b,c)\mid a+b+c=0\}.$
It is claimed that $(\rho,W)$ has no invariant one dimensional subspace. However, I'm not quite sure how to show this. 
Can anyone point me in the right direction? 
 A: A one dimensional subspace is the span of a vector. So let $\vec{w} = \langle a,b,c \rangle \in W$. It suffices to show that the span $S$ of $\vec{w}$ is not a subrepresentation.
If it were, then $\langle 1, \frac{b}{a}, \frac{c}{a}\rangle$ would be in $S$ (we can assume $a\neq 0$). Note that $W$, hence $S$, is invariant under permuting the coordinates (since this doesn't change their sum). Hence $\langle \frac{b}{a}, 1, \frac{c}{a}\rangle$ is also in $S$. We will also use the fact that the difference $\vec{d} = \langle 1- \frac{a}{b}, \frac{a}{b} - 1, 0\rangle$ must too be in $S$.
If $a=b$ then $\frac{c}{a} = -2$ so that $\vec{v_{1}} = \langle 1, 1, -2 \rangle \in S$. Again using permutation invariance, $\vec{v_{2}} = \langle 1, -2, 1\rangle \in S$. Therefore $\vec{v_{1}} - \vec{v_{2}}$ and $\vec{v_{2}} - \vec{v_{1}}$ are both in $S$, which reveals that $\langle 1, -1, 0\rangle$ and $\langle 0, 1, -1\rangle$ are in $S$. Since these two vectors form a basis for $W$ (check this), we must have $S = W$.
If $a \neq b$ then $\frac{a}{b} \neq 1$ and so $\frac{1}{1-\frac{a}{b}}\vec{d} = \langle 1, -1, 0\rangle \in S$. By permuting coordinates we again see $\langle 0,1, -1\rangle \in S$, so by the same logic above we must have $S=W$. 
A: Can I suggest using character theory? This may not be the smartest approach here, but the method will serve you well for more complicated problems in the future.
Here is a very useful fact: If $\rho : G \to {\rm Gl}(V)$ is a representation of $G$, and if $\chi : G \to \mathbb C$ is the character of this representation, then
$$ \frac 1 {|G|} \sum_{g \in G} | \chi(g) |^2$$
is the number of irreducible invariant subspaces in $V$ (i.e. it is the number of irreps you get when you decompose $\rho$ into irreps). If you don't already know this fact, then I would encourage you to derive it from the orthogonality of characters.
I would now invite you to work out the character of your representation $\rho$, and use it to show that $V$ has two irreducible invariant subspaces.
Clearly,
$$ V = W \oplus {\rm Span}(1,1,1),$$
and $W$ and ${\rm Span}(1,1,1)$ are both invariant subspaces of $V$.
And we know that $V$ only has two irreducible invariant subspaces, from character theory! So actually, $W$ and ${\rm Span}(1,1,1)$ are irreducible invariant subspaces of $V$. In particular, $W$ is irreducible, which means that you can't find a one-dimensional invariant subspace within $W$!
A: You haven't specified your field, so I'll take it to be arbitrary (that way we can see what goes wrong in positive characteristic). 
Note that $W$ is 2-dimensional with basis $\alpha_1=e_1-e_2$ and $\alpha_2=e_2-e_3$. It is useful to observe the following formulas:
$$
\begin{array}{ll} \rho_{(12)}(\alpha_1)=-\alpha_1&\rho_{(23)}(\alpha_1)=\alpha_1+\alpha_2\\
\rho_{(12)}(\alpha_2)=\alpha_1+\alpha_2&\rho_{(23)}(\alpha_2)=-\alpha_2
\end{array}
$$
If $X\subset W$ is a 1-dimensional subspace, then $X=\mathrm{span}\{a\alpha_1+b\alpha_2\}$ for some $a,b\in F$. Now, since $X$ is $S_3$-invariant, any $\sigma\in S_3$ must satisfy $\rho_\sigma(a\alpha_1+b\alpha_2)=\lambda_\sigma(a\alpha_1+b\alpha_2)$ for some $\lambda_\sigma\in F$.
Next, using the formulas above, we have
$$\lambda_{(12)}(a\alpha_1+b\alpha_2)=\rho_{(12)}(a\alpha_1+b\alpha_2)=(-a+b)\alpha_1+b\alpha_2.$$
Comparing the coefficients of $\alpha_2$, we deduce that $\lambda_{(12)}=1$ and, therefore, $-a+b=a$.
Using the formulas above again, we compute
$$\lambda_{(23)}(a\alpha_1+b\alpha_2)=\rho_{(23)}(a\alpha_1+b\alpha_2)=a\alpha_1+(a-b)\alpha_2.$$
Comparing the coefficients of $\alpha_1$, we deduce that $\lambda_{(23)}=1$ and $a-b=b$.
We now have two equations:
\begin{align}
-a+b&=a\\
a-b&=b
\end{align}
Hence, $2a=b$ and $2b=a$. It follows that $4b=b$ and $4a=a$. We now deduce that either $4=1\in F$ (so $\mathrm{char}(F)=3$) or $a=b=0$.
Notice that in characteristic 3, $2=-1$ and $\alpha_1-\alpha_2$ does span a 1-dimensional invariant subspace. On the other hand,
$$
\alpha_1-\alpha_2=(e_1-e_2)-(e_2-e_3)=e_1-2e_2+e_3=e_1+e_2+e_3.
$$
Weird, eh.
