Is the tensor algebra a Lie algebra? I found that the reverse is true. To show that it is a Lie algebra I just have to show that the Jacobi identity holds, so I was wontering if it could be true either for the tensor algebra of a vector space or for a subalgebra of that space.

As Torsten says in the comments, every associative algebra is a Lie algebra with bracket the commutator bracket $[a, b] = ab - ba$.
The tensor algebra $T(V)$, as a Lie algebra, has a subalgebra generated by $V$ (which is not all of $T(V)$), and $T(V)$ has a special relationship to this subalgebra: at least in characteristic $0$, it is precisely the universal enveloping algebra of this subalgebra, and moreover this subalgebra is the free Lie algebra $L(V)$ on $V$. This mostly follows from the fact that the two have the same universal property, except that there's a little work to be done explaining why the natural map $L(V) \to U(L(V))$ is injective. But this follows from the Poincare-Birkhoff-Witt theorem.
