Theorem

Let T be a linear map from U to V. Then T is injective if and only if null T = {0}.

${latex.inline\rightarrow}

Suppose T is injective. We already know that \({latex.inline[\{0\} \subseteq null\ T](\{0\} \subseteq null\ T)} by [1753751116 - Axler 3.10 Linear maps take 0 to 0|3.10](1753751116 - Axler 3.10 Linear maps take 0 to 0|3.10). We need to prove inclusion in the other direction, namely that \){latex.inlinenull\ T \subseteq {0}}. Suppose \({latex.inline[v \in null\ T](v \in null\ T)}. Then we know \){latex.inlineTv = 0 = T(0)}. But we know T is injective, which means that this implies v = 0, and we are done.

${latex.inline\leftarrow}

Suppose null T = {0}. We want to prove that T is injective. Suppose \({latex.inline[u,v \in V](u,v \in V)} and Tu = Tv. Then, because T is a linear map, \){latex.inlineTu - Tv = T(u - v) = 0}. But this means \({latex.inline[u - v \in null\ T](u - v \in null\ T)} which means \){latex.inlineu - v = 0} which implies that ${latex.inlineu = v}. This shows that T is injective.