Problem (a) Let $U$ be the subspace of $\mathbb{C}^5$ defined by $$U = \lbrace (z_1,z_2,z_3,z_4,z_5) \in \mathbb{C}^5 | 6z_1 = z_2 \thinspace \textrm{span} \thinspace z_3 + 2z_4 + 3z_5 = 0 \rbrace$$ Find a basis of $U$. (b) Extend the basis in part (a) to a basis of $\mathbb{C}^5$ (c) Find a subspace $W$ of $\mathbb{C}^5$ such that $\mathbb{C}^5 = U \oplus W$ Solution (a) Any vector in $U$ can be written as $(x,6x,-2y - 3z,y,z)$, therefore a basis for $U$ is the list $(1,6,0,0,0), (0,0,-2,1,0), (0,0,-3,0,1)$ as any vector $u = (x,6x,-2y - 3z,y,z) \in U$ can be written uniquely as the linear combination $(x,6x,-2y - 3z,y,z) = x(1,6,0,0,0) + y(0,0,-2,1,0) + z(0,0,-3,0,1)$, therefore by theorem 2....

Problem (a) Let $U$ be the subspace of $\mathbb{R}^5$ defined by $$U = \lbrace (x_1,x_2,x_3,x_4,x_5) \in \mathbb{R}^5 | x_1 = 3x_2 \thinspace \textrm{and} \thinspace x_3 = 7x_4 \rbrace$$ Find a basis for $U$ (b) Extend the basis in part (a) to a basis of $\mathbb{R}^5$ (c) Find a subspace $W$ of $\mathbb{R}^5$ such that $\mathbb{R}^5 = U \oplus W$ Solution (a) Any vector in $U$ can be written as $(x,x,y,y,z)$, therefore by theorem 2....

Problem Verify all the assertions in Example 2.28 Solution Any vector $(x_1, x_2, \dots, x_n) \in \mathbb{F}^n$ can be written as a linear combination of the given list as such $(x_1, x_2, \dots, x_n) = x_1(1,0,\dots,0) + x_2(0,1,0,\dots,0) + \dots + x_n(0,\dots,0,1)$, therefore the list spans $\mathbb{F}^n$. The list is also clearly linearly independent and therefore a basis. The list $(1,2), (3,5)$ is a basis of $\mathbb{F}^2$ because any vector $(x,y)$ can be written as a linear combination of $(1,2), (3,5)$ as such $(x,y) = (3y - 5x)(1,2) + (2x - y)(3,5)$, also $(1,0),(0,1)$ is a basis of $\mathbb{F}^2$ by (1), therefore by theorem 2....

Problem Find all vector spaces that have exactly one basis Solution The only vector space that has exactly one basis is $\lbrace 0 \rbrace$, proof: Let $v_1, v_2, \dots, v_n$ be a basis of $V$, then for all non-zero $\lambda$ $\lambda v_1, \lambda v_2, \dots, \lambda v_n$ is also a basis as per P2A8. Therefore only the empty list basis can be a unique basis, the empty list is a basis for $\lbrace 0 \rbrace$....

Problem Suppose $p_0, p_1, \dots, p_m$ are polynomials in $\mathcal{P}_m(\mathbb{F})$ such that $p_j(2) = 0$ for each $j$. Prove that $p_0, p_1, \dots, p_m$ is not linearly independent in $\mathcal{P}_m(\mathbb{F})$ Solution We know that $z-2 | p_i$ for all $i$, therefore we now denote $q_i$ as the polynomial $p_i/(z-2)$. Therefore: $$0 = a_0p_0 + a_1p_1 + \dots + a_mp_m$$ Is only true in general if and only if: $$0 = a_0q_0 + a_1q_1 + \dots + a_mq_m$$ But $q_i$ has degree at most $m-1$ and the linearly independent list $1,z,\dots,z^{m-1}$ spans $\mathcal{P}_{m-1}(\mathbb{F})$, therefore by theorem 2....