Instructor's Solutions Manual, Section 1B Exercise 1 **1** Prove that - (-v) = v for every v $\in$ V. SOLUTION Let v $\in$ V. By the definition of additive inverse, we have v +（-v） = 0. The additive inverse of -v, which by definition is — (-v), is the unique vector that when added to -v gives 0. The equation above shows that v has this property. Thus -(-v) = v. COMMENT Using 1.32 twice leads to another proof that - (-v) = v. However, the proof given above uses only the additive structure of V, whereas a proof using 1.32 also uses the multiplicative structure. Edward Frenkel student comment: 1.32 (−1)𝑣 = −𝑣 for every 𝑣 ∈ 𝑉. is proven using association of multiplication --- Commentary: $\textbf{Exercise 1.}$ Prove that $-(−v) = v$ for every $v\in V$. $\textbf{Solution 1.}$ Let $v\in V$. By the definition of additive inverse, we have $v + (−v) = 0.$ The additive inverse of $−v$, which by definition is $−(−v)$, is the unique vector that when added to $−v$ gives $0$. The equation above shows that $v$ has this property. Thus $−(−v) = v$. $\textit{Commentary:}$ Using S.18 twice leads to another proof that $−(−v) = v$. However, the proof given above uses only the additive structure of $V$, whereas a proof using S.18 also uses the multiplicative structure. --- Commentary 2: $\textbf{Exercise 1.}$ Prove that $-(−v) = v$ for every $v\in V$. $\textbf{Solution 1.}$ Let $v\in V$. By the definition of additive inverse, we have $v + (−v) = 0.$ The additive inverse of $−v$, which by definition is $−(−v)$, is the unique vector that when added to $−v$ gives $0$. The equation above shows that $v$ has this property. Thus $−(−v) = v$. $\textit{Commentary:}$ This exercise demonstrates a fundamental property of additive inverses in a vector space: the additive inverse of the additive inverse of a vector is the vector itself. This is a consequence of the uniqueness of additive inverses and the fact that $0$ is the additive identity. This property holds in any mathematical structure where additive inverses are defined, such as groups, rings, and fields. It is a key step in proving that the set of additive inverses in a vector space (or any of these other structures) is closed under the operation of taking additive inverses. $\textit{Example:}$ In the vector space of real numbers $\mathbb{R}$, for any $x \in \mathbb{R}$, we have $-(-x) = x$. For instance, $-(-3) = 3$. In the vector space of $2 \times 2$ matrices over $\mathbb{R}$, denoted $M_{2\times 2}(\mathbb{R})$, for any matrix $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$, we have $-(-A) = A$. For example, if $A = \begin{pmatrix} 1 & -2 \ 3 & 4 \end{pmatrix}$, then $-A = \begin{pmatrix} -1 & 2 \ -3 & -4 \end{pmatrix}$ and $-(-A) = \begin{pmatrix} 1 & -2 \ 3 & 4 \end{pmatrix} = A$. In the vector space of polynomials with real coefficients, denoted $\mathbb{R}[x]$, for any polynomial $p(x)$, we have $-(-p(x)) = p(x)$. For instance, if $p(x) = x^2 - 3x + 1$, then $-p(x) = -x^2 + 3x - 1$ and $-(-p(x)) = x^2 - 3x + 1 = p(x)$. ---