Injection of a vector space into its tensor algebra Let $V$ be a vector space and $TV$ the tensor algebra of $V$, defined as the quotient of the free algebra over $V$ by the (two-sided) ideal $I$ generated by all elements of the type $(v+v')-(v)-(v')$ or $(\lambda v)- \lambda (v)$, for vectors $v,v'$ and scalar $\lambda$. 
Using this definition, how do we prove that the space $V$ injects into $TV$, 
or in other words, why is it that if $(v)\in I$, then $v$ has to be zero. 
 A: That is a somewhat unusual definition of the tensor algebra, which does not make obvious the natural grading that this universal algebra possesses. What you should do is show that the traditional definition of the tensor algebra of $A$ module $M$ as:
$$\mathrm{T}_{A}(M)=\bigoplus_{n \in \mathbb{N}}\ \bigotimes_{A}^{n}M$$
satisfies the same universal property as the object you described, canonically attached to $M$. Thus, denoting the canonical maps from $M$ to the attached structure by greek minuscules, $(\mathrm{T}_{A}(M), \iota)$ and $(A\langle M\rangle, \gamma)$ are both initial objects in the comma category $M \downarrow (A-\mathrm{Alg})$ and are thus isomorphic. As $\iota$ is by definition easily seen to be injective ($M$ is canonically isomorphic to the tensor product of just one copy of itself, which then embeds in the direct sum), so will $\gamma$ be, as the composition between the injection $\iota$ and a bijection.
A: You can show this using just the universal property. To show that $V \to T(V)$ is an injection it suffices to show that $V$ injects into any algebra whatsoever, because any such injection $V \hookrightarrow A$ must, by the universal property, factor as a composite $V \hookrightarrow T(V) \to A$.
So here's an algebra which canonically works but is easier to construct than the tensor algebra: the square-zero extension $k \oplus V$, where multiplication is defined so that $k$ is spanned by the unit but all products of elements of $V$ vanish.
(Once you construct the tensor algebra as in ΑΘΩ's answer, this algebra appears as the quotient $T(V)/V^{\otimes 2}$.)
Non-canonically it suffices to show that there exists some algebra of dimension at least $\dim V$, and for example $\text{End}(V)$ is such an algebra.
