Definition: Null space. Let be linear. The null space of or the kernel of (denoted ). We know that as

Proposition: , linear is a linear subspace of . Proof: Pick TBA

Proposition: is linear then if and only if is one-one. Proof: Pick such that , as is a linear map So, Therefore, Hence, is one to one. Other side: Note, Pick Then, Since is one to one,

  • So essentially, we have a lot of vectors in the domain that map to zero
    • But we also have a lot of vectors that map to non-zero vectors, which gives the transform its non-triviality
  • Say some is linear is also linear, and a subspace
    • What follows is that is you have transform is linear, then is a subspace
  • In conclusion, the vectors in all map to , and the rest of the vectors span

Example: ,

Definition: Rank-nullity. Rank is the dimension of , nullity is the dimension of Let be a linear map. Then, Proof: Since is a subspace, let be a basis of . Pick is linearly independent. as is a basis, , Pick and complete it to a basis of Remains to show

Proof of rank nullity: Note that is a linear subspace of the codomain . Thus, there exists such that , which is a basis of . Now, we shall show that is linearly independent. Suppose . This implies that . We can apply the linear property such that . Therefore, Since is a subspace, say, is a basis of . Claim: is a basis of . We know this because the union of two linearly independent sets is linearly independent from our homework. Pick . Either , or This implies that , given . This implies that which implies that Which implies that So, is a basis of . P.

Example: as . Therefore, Therefore, Thus,