Proposition 4.7 on Lang's semisimple rings part. In page 654 on Lang's Algebra 3rd revised edition, proposition 4.7 and its proof are written.

[Proposition 4.7]
  Let $k$ be a field, $E$ be a finite dimensional $k$-vector space, $R$
  be a sub-algebra of $End_{k}(E)$. Then, $R$ is semisimple $\iff$ $E$
  is semisimple $R$-module.

And proof is below. 

If $R$ is semisimple, then $E$ is semisimple by proposition 4.1. Suppose $E$ be semisimple. Then, $E = \oplus_{i=1}^{n}E_i$ for some $E_i$ which are simple ring. Thus, $\exists x_i \in E_i$ such that $Rx_{i} = E_i$. Consider $\phi : R \to E$ by $r \mapsto (rx_1, \cdots , rx_n)$. Then,$\phi$ is $R$-linear and injective. Thus, $R \cong Im \phi \subset E$. Since $Im \phi$, a submodule of $E$ is semisimple by proposition 2.2, so does $R$.

My question is, I don't know why the map $\phi$ is injective. What if $\{x_{1}, \cdots, x_{n} \}$ doesn't form a $k$-basis? This case is possible when you think $E = \mathbb{R}^{3}$, and consider $E_{1} = \mathbb{R}(1,1,1)$ and $E_{2} = E_{1}^{\perp}$. Obviously, $E_2$ is simple, so it can be generated by one element, say $\bar{y}$, of $\mathbb{R}^{3}$, however, we cannot say that $(1,1,1)$ and $\bar{y}$ is $\mathbb{R}$-basis of $E$. How can I show that this map is injective?
Also I read comment on this article from @m_t_ but I don't understand why it would be injection from $R$ into $E^{\oplus \dim E}$. Could anyone explain this?
 A: As pointed out by m_t_ in his comment the map $R \to E$, $r \mapsto (r x_1, \dotsc, r x_n)$ is not necessarily injective.
Taking for example $R = \operatorname{End}_K(E)$ we find that $E$ is simple as an $R$-module, and we can take any nonzero $x \in E$ is an $R$-generator.
But the map $R \to E$, $r \mapsto r x$ is not injective if $\dim E > 1$.
Instead we can embedd $R$ into $E^{\oplus \dim E}$ as follows:
For $n = \dim E$ we have a decomposition $\operatorname{Mat}_n(K) \cong (K^n)^{\oplus n}$ as $\operatorname{Mat}_n(K)$-modules, and therefore a decomposition $\operatorname{End}_K(E) \cong E^{\oplus \dim E}$ as $\operatorname{End}_K(E)$-modules (if $x_1, \dotsc, x_n \in E$ is a $K$-basis then such an isomorphism is given by $\operatorname{End}_K(E) \to E^{\oplus \dim E}$, $r \mapsto (r x_1, \dotsc, r x_n)$).
Because the canonical inclusion $R \to \operatorname{End}_K(E)$ is an embedding of $R$-modules we find that $R$ is isomorphic (as an $R$-module) to a submodule of $E^{\oplus \dim E}$.
Since direct sums and submodules of semisimple modules are again semisimple it then follows that $R$ is semisimple as an $R$-module.
