Suppose $X$ is a Banach space and $T: X \to X$ is a bounded linear operator. If $T$ is onto, is it necessarily 1-1?
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No. This is true in the finite dimensional case, where a linear transformation is one-to-one iff it is onto iff it is invertible. Some counterexamples for your question are the "backward shifts" on sequence spaces like $c_0$ or $\ell^p$ with $1\leq p\leq \infty$. Let $X$ be one of these spaces, and let $T(a_0,a_1,a_2,\ldots)=(a_1,a_2,a_3,\ldots)$.