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I was reading a proof of the theorem that the range of a linear map $T$ is always a subspace of the target space, and when the author was showing that the $0$ vector was included in the range, he made an appeal to a previous theorem which says that the null space of $T$ is always a subspace of $T$.

In other words, he says that because the nullspace is a subspace, $0$ is always in the nullspace, and therefore since $T(0) = 0$, then $0$ is in the range of $T$.

That makes sense, but is it possible that the nullspace is empty? My feeling is no. Because $T$ acts on a vector space $V$, then $V$ must include $0$, and since we showed that the nullspace is a subspace, then $0$ is always in the nullspace of a linear map, so therefore the nullspace of a linear map can never be empty as it must always include at least one element, namely $0$.

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  • $\begingroup$ "That makes sense, but is it possible that the nullspace is empty?" What do you mean by null space? $\endgroup$ – Git Gud Jan 31 '15 at 22:10
  • $\begingroup$ The set {v in V: Tv = 0} $\endgroup$ – user1236 Jan 31 '15 at 22:12
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    $\begingroup$ Does $0$ satisfy $T0=0$? $\endgroup$ – Git Gud Jan 31 '15 at 22:13
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    $\begingroup$ It's worth pointing out that a vector space, by definition, cannot be empty. Specifically, the vector space axioms require that for any vector space $V$, there exist a $0 \in V$ such that for any $v \in V$ $$0+v = v$$ so we must have $0 \in V$, and so $V\not=\emptyset$. $\endgroup$ – Strants Jan 31 '15 at 22:16
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Let $T:V\to W$ be a linear map. Then $$ T(\mathbf{0})=T(\mathbf{0}-\mathbf{0})=T(\mathbf{0})-T(\mathbf{0})=\mathbf{0} $$ This proves that $\mathbf 0$ is always in the nullspace of $T$. Hence the nullspace of $T$ cannot be empty.

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  • $\begingroup$ I don't think there is anything wrong with your post. This is the way it was done in the text book too actually. I didn't realize this at first when I was doing my first reading. $\endgroup$ – user1236 Feb 9 '15 at 0:17
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It is not possible that the nullspace is empty. The element $\overline 0$ is always contained in the domain and as $T$ is linear we have $T(\overline 0) = T(0\cdot\overline 0) = 0\cdot T(\overline 0) = \overline 0$.

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