I am asked to show that if $\mathfrak{g}$ is a semisimple Lie Algebra with root system of type $G_2$, then it has a unique, $7$-dimensional faithful representation.
To start, let $\omega_1, \omega_2$ be the fundamental weights of the root system $G_2$ and suppose that $\lambda = m_1\omega_1 + m_2\omega_2$ is a dominant weight. Then by using the Weyl Dimension Formula, we see that:
$$\dim(V(\lambda)) = \frac{1}{120}(m_1+1)(m_2+1)(m_1+m_2+2)(m_1+2m_2+3)(m_1+3m_2+4)(2m_1 + 3m_2+5).$$
Using this formula we see that $V(\omega_1)$ is the only irreducible representation of $\mathfrak{g}$ and there are no irreducible representations of dimensions $2$, $3$, $4$, $5$, or $6$. This in turn means there are no representations of dimensions $2$, $3$, $4$, $5$, or $6$ not of the form $V(0) ^{\otimes n}$
Finally, we note $V(0)^{\otimes 7}$ is not faithful, and so we must be required to show that the representation $V(\omega_1)$ is faithful.
It is here that I am stuck since I do not know how to do this.
My immediate thought is that if we assume we do not have injectivity then we can find a subrepresentation, though I am unsure of how to do this, or how to use the fact that the representation is specifically $V(\omega_1)$ and not just any $7$-dimensional vector space. Below is my attempt
Let $\phi$ be the representation $\phi : \mathfrak{g} \rightarrow \mathfrak{gl}(V(\omega_1))$ and suppose that for some $x \neq y, x,y \in \mathfrak{g}$ we have that $\phi(x) = \phi(y)$
If $x = ky$ for some $k\in \mathbb C$ then by linearity of $\phi$, we have $\phi(x) = 0$
Let $v_1, \dotsc, v_n$ be a basis for $\mathfrak{g}$ with $v_1 = x$. Then as $\phi$ is a homomorphism, we must have that $\dim(\phi(\mathfrak{g})) < \dim(\mathfrak{g})$.
[I am unable to pursue this approach any further].
In the case that $x$ and $y$ are linearly independent, then I have even fewer ideas.
I think ultimately, treating $\mathfrak{gl}(V(\omega_1))$ as a vector space of matrices, then I would like to show that in $\phi(\mathfrak{g})$ there is an entire row or column that is not in use. Therefore, we may ignore the corresponding basis vector(s) of $V(\omega_1)$ to get a subrepresentation and thus a contradiction.
However, I am unsure of how to achieve this and would appreciate any help.
It is clear that the simple module of smallest dimension, different from the trivial one, is $L(\omega_1)$, which has dimension $7$. Since ${\mathfrak g}_2$ is a simple Lie algebra, the module is faithful. Every ${\mathfrak g}_2$-module is semisimple, i.e., direct sum of simple modules.