Theorem

Suppose ${latex.inlineT \in L(V, W)}, then null T is a subspace of V.

Proof

Because T is a linear map, T(0) = 0 per 1753751116 - Axler 3.10 Linear maps take 0 to 0|3.10. Thus, ${latex.inline0 \in null\ T}. We now need to check if the null space is closed under addition and closed under scalar multiplication.

Suppose \({latex.inline[u, v \in null\ T](u, v \in null\ T)}. Then, by definition of linear map, we know that \){latex.inlineT(v + v) = Tu + Tv = 0 + 0 = 0}. Thus the null space is closed under addition.

Suppose \({latex.inline[u \in V](u \in V)} and \){latex.inline\lambda \in F}. Then ${latex.inlineT(\lambda\ v) = \lambda T v = \lambda * 0 = 0}.

Thus, we have shown that the null space is a vector space.