The kernel/null space of a linear transformation is a subspace

Created almost 9 years ago, updated about 8 hours ago

The kernel contains zero and is closed under combinations: if $T(u)=T(v)=0$, then $T(au+bv)=0$. Hence it is a subspace of the domain. This is also the inverse-image-of-a-subspace result with the codomain subspace $\{0\}$.

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