# Can the null space be empty?

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 $0$.

• "That makes sense, but is it possible that the nullspace is empty?" What do you mean by null space? – Git Gud Jan 31 '15 at 22:10
• The set {v in V: Tv = 0} – user1236 Jan 31 '15 at 22:12
• Does $0$ satisfy $T0=0$? – Git Gud Jan 31 '15 at 22:13
• 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 $V\not=\emptyset$. – Strants Jan 31 '15 at 22:16

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.
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 $T(\overline 0) = T(0\cdot\overline 0) = 0\cdot T(\overline 0) = \overline 0$.