linear algebra

Problem (a) Let $U = \lbrace p \in \mathcal{P}_4(\mathbb{F}) | p(2) = p(5) \rbrace$. 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) As $z$ $\cancel{\in}$ $U$, we must have $\dim U < \dim \mathcal{P}_4(\mathbb{F}) = 5$, and as the linearly independent (linear independence follows from applying the procedure in example 2....

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....

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....

Problem Show that the subspaces of $\mathbb{R}^3$ are precisely $\lbrace 0 \rbrace$, $\mathbb{R}^3$, all lines in $\mathbb{R}^3$ through the origin, and all planes in $\mathbb{R}^3$ through the origin. Solution As per theorem 2.38, we must have that any subspace $U$ of $R^3$ has a basis of length $0$, $1$, $2$ or $3$. The cases $\dim U = 0$ and $\dim U = 3$ are precisely the cases when $U = \lbrace 0 \rbrace$ and $U = \mathbb{R}^3$ respectively - as per P2B1 and P2C1....

Problem Show that the subspaces of $\mathbb{R}^2$ are precisely $\lbrace 0 \rbrace, \mathbb{R}^2$, and all lines through the origin. Solution As per theorem 2.38, we must have that any subspace $U$ of $R^2$ has a basis of length $0$, $1$ or $2$. The case when a basis of $U$ has length $0$ is exactly when $U = \lbrace 0 \rbrace$, when $\dim U = 2 = \dim \mathbb{R}^2$ we must have $U = \mathbb{R}^2$ by P2C1....