Two basic sequences $(x_n)$ and $(y_n)_n$ in Banach spaces $X$ and $Y$ are said to be equivalent if there is an isomorphism $T$ mapping the closed linear span of $(x_n)_n$ onto the closed linear span of $(y_n)_n$ mapping $Tx_n=y_n$. Is the idea that this is more than an isomorphism between the closed linear spans? In finite dimensions, you can always take two bases for two isomorphic spaces, and define an isomorphism taking one basis to another. But in infinite dimensions, even if you know the span of $(x_n)_n$ and the span of $(y_n)_n$ are isomorphic, you can't necessarily find an isomorphism mapping $Tx_n$ to $y_n$?
Also, this definition is equivalent to saying that there are constants $0<A<B<\infty$ such that for all $N\in\mathbb{N}$ and all scalars $\alpha_1,...,\alpha_N$, $A||\sum_{n=1}^N\alpha_nx_n||\leq||\sum_{n=1}^N\alpha_ny_n||\leq B||\sum_{n=1}^N\alpha_nx_n||$.
I've also seen this stated for any infinite sequence of scalars $(\alpha)_n$, so is the equivalence between these two definitions, for finite sequences of scalars and infinite sequences of scalars, and application of the bounded linear transformation theorem?
There are two non-equivalent bases of the Banach space $c_0$.
For the first one, take the canonical basis $(e_n)$ where $e_n = (0,\dots,0,1,0,\dots)$ with $1$ on the nth place.
For the second one, take the so-called summing basis $(f_n)$ where $f_n = (1,\dots,1,0,\dots)$ with $1$ on the first $n$ places and $0$ elsewhere.
These two are both bases of $c_0$ but there is no isomorphism $T:c_0 \rightarrow c_0$ which maps $e_n$ to $f_n$ (for example, the canonical basis is unconditional and the summing basis is not).
So, two bases being equivalent is indeed stronger than their closed spans being isomorphic.