1. Let U={p∈P4(F):p(6)=0}U = \{p \in \mathcal P_4(\mathbf F) : p(6) = 0\}. Find a basis for UU
  2. Extend the basis to a basis of P4(F)\mathcal P_4(\mathbf F)
  3. Find a subspace WW of P4(F)\mathcal P_4(\mathbf F) such that P4(F)=U⊕W\mathcal P_4(\mathbf F) = U \oplus W

This has the same structure as 2b/3 with the subspace from 2a/17. except now we have access to 2.39 so the proofs are much shorter.

Note dim⁡P4(F)=5\dim \mathcal P_4(\mathbf F) = 5 and dim⁡U≤5\dim U \le 5 because it's a subspace.

  1. Examine (x−6),…,(x−6)4(x-6),\dots,(x-6)^4. they're independent by the same logic as in 2a/2. they're the right size since 1∉U1 \notin U implies dim⁡U<5\dim U < 5, and our list of 4 independent vectors in UU implies dim⁡U≥4\dim U \ge 4. Thus dim⁡U=4\dim U = 4 and applying 2.39 allows us to conclude we have a basis.
  2. (x−6),…,(x−6)4,1(x-6),\dots,(x-6)^4,1 is independent in P4(F)\mathcal P_4(\mathbf F) since there's no way to write 11 using previous terms, (see 2.21 a), and the right size (see 2.39) as dim⁡P4(F)=5\dim \mathcal P_4(\mathbf F) = 5.
  3. Let W=span(1)W = \text{span}(1). Clearly W∩U={0}W \cap U = \{0\} and W+U=VW+U = V by (2).