Problem
(a) Let $U = \lbrace p \in \mathcal{P}_4(\mathbb{F}) | p(6) = 0 \rbrace$. 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) It’s clear that $U$ is a proper subspace of $V$ i.e. $\dim U < \dim \mathcal{P}_4(\mathbb{F})$, as there are vectors in $\mathcal{P}_4(\mathbb{F})$ that are not in $U$ e.g. $p(z) = z$. Moreover, following the process described in example 2.41, we see that the vectors $z-6,(z - 6)^2, (z - 6)^3, (z - 6)^4$ are all linearly independent and since they’re all in $U$. Therefore (by using $\dim U < \dim \mathcal{P}_4(\mathbb{F}) = 5$) we must have that the list $z-6,(z - 6)^2, (z - 6)^3, (z - 6)^4$ is a basis of $U$ as per theorem 2.29.
(b) As $p(z) = 1$ $\cancel{\in}$ $U$, the list $1,z-6,(z - 6)^2, (z - 6)^3, (z - 6)^4$ is therefore a linearly independent list, by P2A11, of length 5. Theorem 2.39 implies that it is a basis for $\mathcal{P}_4(\mathbb{F})$.
(c) Let $W = \textrm{span}(1)$, then by (b) we must have $U \oplus W = \mathcal{P}_4(\mathbb{F})$