Exercise 2C Problem 1

Problem Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$ such that $\dim U = \dim V$. Prove that $U = V$. Solution $\dim U = \dim V$ implies that a basis of $U$ consist of $\dim V$ linearly independent vectors $u_1, \dots, u_{\dim V}$, but as $U \subseteq V$ all the $u_i$’s are in $V$. The list $u_1, \dots, u_{\dim V}$ is therefore a linearly independent list in $V$ of the right length i....

June 26, 2022 · 1 min · 84 words · Me

Exercise 2B Problem 8

Problem Suppose $U$ and $W$ are subspaces of $V$ such that $V = U \oplus W$. Suppose also that $u_1, \dots, u_m$ is a basis of $U$ and $w_1, \dots, w_n$ is a basis of $W$. Prove that $$u_1, \dots, u_m, w_1, \dots, w_n$$ is a basis of $V$ Solution Every vector in $V$ can be written uniquely as a vector $u \in U$ plus a vector $w \in W$ as per the fact that $V = U \oplus W$....

June 26, 2022 · 1 min · 172 words · Me

Exercise 2B Problem 7

Problem Prove or give counterexample: If $v_1, v_2, v_3, v_4$ is a basis of $V$ and $U$ is a subspace of $V$ such that $v_1,v_2 \in U$ and $v_3$ $\cancel{\in}$ $U$ and $v_4$ $\cancel{\in}$ $U$, then $v_1,v_2$ is a basis of $U$. Solution The list $v_1, v_2$ is linearly independent because the list $v_1, v_2, v_3, v_4$ is a basis. Suppose now that $v_1, v_2$ does not span $U$, then there exists $u \in U$ such that $u$ $\cancel{\in}$ $\textrm{span}(v_1,v_2)$....

June 26, 2022 · 1 min · 150 words · Me

Exercise 2B Problem 6

Problem Suppose $v_1, v_2, v_3, v_4$ is a basis of $V$. Prove that $$v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4$$ is also a basis of $V$. Solution Let $a_1,a_2,a_3,a_4 \in \mathbb{F}$ with: $$0 = a_1(v_1 + v_2) + a_2(v_2 + v_3) + a_3(v_3 + v_4) + a_4v_4$$ $$= a_1v_1 + (a_1 + a_2)v_2 + (a_2 + a_3)v_3 + (a_3 + a_4)v_4$$ Implying that $a_1 = a_1 + a_2 = a_2 + a_3 = a_3 + a_4 = 0$, by the linear independence of $v_1, v_2, v_3, v_4$, which in turn implies $a_1 = a_2 = a_3 = a_4 = 0$....

June 26, 2022 · 1 min · 137 words · Me

Exercise 2B Problem 5

Problem Prove or disprove: there exists a basis $p_0, p_1, p_2, p_3$ of $\mathcal{P}_3(\mathbb{F})$ such that none of the polynomials $p_0, p_1, p_3$ has degree 2. Solution The list $z^2, z^3, z, 1$ is a basis for $\mathcal{P}_3(\mathbb{F})$ by example 2.28 (g), and therefore, by P2A7, so is the list $5z^2 - 4z^3, z^3, z, 1$ in which none of the polynomials has degree 2.

June 26, 2022 · 1 min · 65 words · Me