Problem
(a) Let $U = \lbrace p \in \mathcal{P}_4(\mathbb{F}) | p^{\prime\prime}(6) = 0 \rbrace$. Find a basis of $U$
(b) Extend the basis in part (a) to a basis of $\mathcal{P}_4(\mathbb{F})$
(c) Find a subspace $W$ of $\mathcal{P}_4(\mathbb{F})$ such that $\mathcal{P}_4(\mathbb{F}) = U \oplus W$
Solution
(a) Since the linearly independent list $z^2$ is not in $U$, we must have that $\dim U < \dim \mathcal{P}_4(\mathbb{F}) = 5$, and as the linearly independent (linear independence follows from applying the procedure in example 2.41) list of length 4: $1,z,(z-6)^3,(z-6)^4$ consists of vectors that are all in $U$, it must be a basis for $U$ by theorem 2.39.
(b) We can extend $U$ to a basis for $\mathcal{P}_4(\mathbb{F})$ by appending $z^2$ to any basis of $U$, as $z^2 \thinspace \cancel{\in} \thinspace U$ and $\dim U = 4$, by P2A11 and theorem 2.39.
(c) Let $W = \textrm{span}(z^2)$, then problem (b) implies $\mathcal{P}_4(\mathbb{F}) = U \oplus W$.