What is the conventional way to write a basis for infinite dimension vector space. Suppose $\mathrm{R}$ a field and let $V$ be vectorspace over $\mathrm{R}$, let $(.)$ be an arbitrary inner product over $V$.
Suppose $T:V\to V$ a linear transformation, $\lambda \in \mathrm{R}$ an eigenvalue of $T$, prove that $V_\lambda$ (eigenspace) is an invariant subspace of $V$.
The question is rather simple to solve in finite (and infinite dimensions), my question is how would I write the notations for infinite one.
Let $V_\lambda = span \{v_1,v_2,\dots\}$, then for any $u\in V$, $u$ can be represented as a sum of the vector basis, $u=\alpha_1v_1+\alpha_2v_2+\dots$
Then it follows that $T(u)=T(\alpha_1v_1+\alpha_2v_2+\dots)=T(\alpha_1v_1)+T(\alpha_2v_2)+\dots$
All $v_i$'s are eigenvectors of $T$, therefore exists $\beta_i$ such that $T(v_i)=\beta v_i$, it follows that
$T(u)=\alpha_1\beta_1v_1+\alpha_2\beta_2v_2+\dots\in V_\lambda$
Two questions, is the notation correct for infinite vectors? do I replace $\dots$ with anything else? and how do I even know basis for $V_\lambda$ exist? for infinite dimensions sure, can I ensure it for infinite ones?
 A: Not an answer to the question (about how to write a basis), but the additional comment about this problem:

Suppose $T:V\to V$ a linear transformation, $\lambda \in \mathrm{R}$ an eigenvalue of $T$, prove that $V_\lambda$ (eigenspace) is an invariant subspace of $V$.

Proof.  Definition:
$$
V_\lambda = \big\{u \in V\;:\; T(u) = \lambda u\big\} .
$$
To show $V_\lambda$ is invariant we must show: if $u \in V_\lambda$ then $T(u) \in V_\lambda$.   
Let $u \in V_\lambda$.  Then $T(u) = \lambda u$.  Compute (using linearity of $T$):
$$
T(T(u)) = T(\lambda u) = \lambda T(u)
$$
and that means $T(u) \in V_\lambda$.  QED.   
(As I said in my comment, using a basis to prove this is not the best way to do it.)
A: As commented, this problem is much easier if you do not use bases. But let me comment on what you have done:
The notation $V_\lambda = span \{v_1,v_2,\dots\}$ is not correct as it suggests that $V_\lambda$ has a countable basis, which needs not be the case.
The notation $u=\alpha_1v_1+\alpha_2v_2+\cdots$ is not incorrect, but it looks old-fashioned. To make sure that you're talking about a finite sum (even if the basis is infinite), write $u=\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_m v_m$.
Note that by definition $span(S)$ is the set of all finite linear combinations of elements of $S$.
The proof you offer is not correct because $v_i \in V_\lambda$ and so $T(v_i)=\lambda v_i$, that is, the $\beta_i$ are all the same and this is what makes the proof work.
The existence of basis for infinite-dimensional vector spaces follows from the axiom of choice (and is actually equivalent to it), typically in the form of Zorn's lemma.
