Is the following statement true? An injective linear operator on a vector space is surjective

Counterexample

Is this true?

An injective linear operator on a vector space is surjective

No — it is false.

The counterexample

The right shift on $\ell^{2}: (x_{1}, x_{2}, \dots) \mapsto (0, x_{1}, x_{2}, \dots)$

Why it fails

Injective but misses everything with a non-zero first coordinate. Rank–nullity needs finite dimension.

Keep practicing

Execution slip

The idea was right. The computation was not.

Published 23 Aug 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.