Instructor's Solutions Manual, Section 1B Exercise 5 **5** Show that in the definition of a vector space (1.20), the additive inverse condition can be replaced with the condition that 0v = 0 for all v $\in$ V. Here the 0 on the left side is the number 0, and the 0 on the right side is the additive identity of V. The phrase a "condition can be replaced" in a definition means that the collection of objects satisfying the definition is unchanged if the original condition is replaced with the new condition. SOLUTION Suppose the additive inverse condition in 1.20 is replaced with the condition that 0v = 0 for all v $\in$ V. Let v $\in$ E V. Then $ \begin{align} 0 &= 0v\\ &= (1 + (-1) )v\\ &= 1v + (-1)v\\ &＝ \; v +（-1)v. \end{align} $ Thus (-1)v is an additive inverse of v. Hence the additive inverse condition is satisfied. Edward Frenkel student note: The additive inverse property or condition says : **For every 𝑣 ∈ 𝑉, there exists 𝑤 ∈ 𝑉 such that 𝑣 + 𝑤 = 0** . This new condition that 0v = 0 for all v $\in$ V. doesn't say "exists"; it shows how to get a 0 vector for all v $\in$ V. Then it shows that 0v existing is the same as v + (-1)v, so (-1)v is the additive inverse. If you have one form is the same as having the other form. --- Commentary: $\textbf{Exercise 5.}$ Show that in the definition of a vector space (S.20), the additive inverse condition can be replaced with the condition that $0v = 0 \text{ for all } v\in V.$ Here the $0$ on the left side is the number $0$, and the $0$ on the right side is the additive identity of $V$. $\textit{Commentary:}$ The phrase "a condition can be replaced" in a definition means that the collection of objects satisfying the definition is unchanged if the original condition is replaced with the new condition. $\textbf{Solution 5.}$ Suppose the additive inverse condition in S.20 is replaced with the condition that $0v = 0 \text{ for all } v\in V.$ Let $v\in V$. Then $\begin{align*} 0 &= 0v \\ &= (1 + (−1))v \\ &= 1v + (−1)v \\ &= v + (−1)v. \end{align*}$ This $(−1)v$ is an additive inverse of $v$. Hence the additive inverse condition is satisfied. --- $\textbf{Exercise 5.}$ Show that in the definition of a vector space (S.20), the additive inverse condition can be replaced with the condition that $0v = 0 \text{ for all } v\in V.$ Here the $0$ on the left side is the scalar $0$, and the $0$ on the right side is the zero vector in $V$. $\textit{Commentary:}$ The phrase "a condition can be replaced" in a definition means that the collection of objects satisfying the definition is unchanged if the original condition is replaced with the new condition. $\textbf{Solution 5.}$ Suppose the additive inverse condition in S.20 is replaced with the condition that $0v = 0$ for all $v\in V$. Let $v\in V$. Then $\begin{align*} 0 &= 0v \\ &= (1 + (−1))v \\ &= 1v + (−1)v \\ &= v + (−1)v. \end{align*}$ This $(−1)v$ is an additive inverse of $v$. Hence the additive inverse condition is satisfied. $\textit{Commentary:}$ This exercise shows that in the definition of a vector space, the condition that every vector has an additive inverse can be replaced by the condition that the scalar multiple of any vector by $0$ is the zero vector. This is a non-trivial result because it relates a condition about the existence of certain vectors (additive inverses) to a condition about the interaction between scalars and vectors (scalar multiplication by $0$). The proof uses the properties of the field of scalars, particularly the fact that $1 + (-1) = 0$, and the properties of scalar multiplication in a vector space, particularly the distributive property. This exercise illustrates the interplay between the vector space axioms and how they relate to each other. It also shows that there can be different equivalent formulations of the definition of a vector space. $\textit{Example:}$ In $\mathbb{R}^n$, for any vector $\mathbf{v} = (v_1, v_2, \ldots, v_n)$, we have $0\mathbf{v} = (0, 0, \ldots, 0)$, the zero vector in $\mathbb{R}^n$. In $M_{m\times n}(\mathbb{R})$, for any matrix $A = (a_{ij})$, we have $0A = (0)$, the $m \times n$ zero matrix. In the vector space of continuous functions from $[0,1]$ to $\mathbb{R}$, denoted $C([0,1], \mathbb{R})$, for any function $f \in C([0,1], \mathbb{R})$, we have $(0f)(x) = 0$ for all $x \in [0,1]$, where $0f$ is the zero function.