Instructor's Solutions Manual, Section 1B Exercise 2 **2** Suppose a $\in$ F, v $\in$V, and av = 0. Prove that a = 0 or v = 0. SOLUTION We want to prove that a = 0 or v = 0. If a = 0, then we are done. So suppose a $\ne$ 0. av = 0 Multiplying both sides of the equation above by 1/a gives $ \frac{1}{a}(av)= \frac{1}{a}0 $ The associative property shows that the left side of the equation above equals 1v, which equals v. The right side of the equation above equals 0 (by 1.31). Thus v = 0, completing the proof. Edward Frenkel student note: 1.31 is 𝑎0 = 0 for every 𝑎 ∈ 𝐅. 0 is a vector --- Commentary: $\textbf{Exercise 2.}$ Suppose $a\in \mathbb{F}$, $v\in V$, and $av = 0$. Prove that $a = 0$ or $v = 0$. $\textbf{Solution 2.}$ We want to prove that $a = 0$ or $v = 0$. If $a = 0$, then we are done. So suppose $a \neq 0$. Multiplying both sides of the equation above by $\frac{1}{a}$ gives $\frac{1}{a}(av) = \frac{1}{a}0.$ The associative property shows that the left side of the equation above equals $1v$, which equals $v$. The right side of the equation above equals $0$ (by S.13). Thus $v = 0$, completing the proof. --- $\textbf{Exercise 2.}$ Suppose $a\in \mathbb{F}$, $v\in V$, and $av = 0$. Prove that $a = 0$ or $v = 0$. $\textbf{Solution 2.}$ We want to prove that $a = 0$ or $v = 0$. If $a = 0$, then we are done. So suppose $a \neq 0$. Multiplying both sides of the equation $av = 0$ by $\frac{1}{a}$ gives $\frac{1}{a}(av) = \frac{1}{a}0.$ The associative property shows that the left side of the equation above equals $1v$, which equals $v$. The right side of the equation above equals $0$ (by the property of the zero vector). Thus $v = 0$, completing the proof. $\textit{Commentary:}$ This exercise proves a basic but important property in vector spaces: if the product of a scalar and a vector is the zero vector, then either the scalar is zero or the vector is the zero vector. This is sometimes called the "zero product property" and it holds in any vector space. This property is a consequence of the field axioms (particularly the existence of multiplicative inverses for non-zero elements) and the definition of scalar multiplication in a vector space. It is a key property that distinguishes vector spaces from more general algebraic structures like modules, where this property may not hold. $\textit{Example:}$ In $\mathbb{R}^3$, if $a(x, y, z) = (0, 0, 0)$ for some $a \in \mathbb{R}$ and $(x, y, z) \in \mathbb{R}^3$, then either $a = 0$ or $(x, y, z) = (0, 0, 0)$. In $M_{2\times 2}(\mathbb{R})$, if $aA = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix}$ for some $a \in \mathbb{R}$ and $A \in M_{2\times 2}(\mathbb{R})$, then either $a = 0$ or $A = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix}$. In $\mathbb{R}[x]$, if $ap(x) = 0$ for some $a \in \mathbb{R}$ and $p(x) \in \mathbb{R}[x]$, then either $a = 0$ or $p(x) = 0$ (the zero polynomial).