Are all linear operators always surjective?

If TT is a linear operator, i.e: T:V⟶VT: V \longrightarrow V. Then, in our case, TT is surjective if rangeT=VT = V. With rangeTT := {TvTv: v∈Vv \in V} and Tv=wTv = w for some w∈Vw \in V, doesn’t it follow that rangeT=VT = V??

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No, an extreme example is the null function, T(v)=0T(v)=0 for all v∈Vv \in V. It is indeed a lineal transformation, but if V≠{0}V \neq \{0\} then ranT={0}⊊VT = \{0\} \subsetneq V so TT is not surjective.
– Antioquia3943
Oct 21 at 1:19

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