Schauder basis for Banach spaces

Let $X$ and $Y$ be Banach spaces that are known to have Schauder bases.

If $x_i$ is a Schauder basis for $X$, and $T:X \to Y$ is a linear homeomorphism, is it true that $Tx_i$ is a Schauder basis for $Y$?

It seems obvious that the answer is yes but I am only confused about the role of "unconditional convergence" here...

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Take $X = Y = \mathbb{R}$ and let $T(x) = x - 1$, then $T$ is a homeomorphism that sends the basis $\{1\}$ to $\{0\}$ which is not a basis. I suspect you meant to impose stronger conditions on $T$. – Rob Arthan Sep 4 '13 at 17:15
He probably means that $T$ is an isomorphism of Banach spaces. – Christopher A. Wong Sep 4 '13 at 17:16
@Christopher A. Wong: well isomorphism of Banach spaces would make it trivial. My guess is that the OP meant is isomorphism of $X$ and $Y$ as vector spaces, but why not let the OP tell us. – Rob Arthan Sep 4 '13 at 17:18
Yes, it's true. If we transport the norm of $Y$ to $X$ via $T$, the assumption that $T$ is an isomorphism of topological vector spaces (or even of Banachable spaces) says you got an equivalent norm. So the question is: "Is being a Schauder basis a topological property or norm-dependent?" The answer is: topological. – Daniel Fischer Sep 4 '13 at 19:37
Daniel is right. And BigUser should try to prove it. – GEdgar Sep 4 '13 at 20:17