Instructor's Solutions Manual, Section 1B Exercise 3 **3** Suppose v, w $\in$ V. Explain why there exists a unique x $\in$ V such that v + 3x = w. SOLUTION There exists a unique vector -v $\in$ V such that v + (-v) = 0. v + 3x = w. Adding -v to both sides of the equation above, we see that the equation above is equivalent to the equation 3x＝ w - v which is equivalent to the equation x = $\frac{1}{3}(w - v)$ which shows that our original equation has a unique solution. Edward Frenkel student note: uses unique vector -v , unique vector w to form x --- Commentary: $\textbf{Exercise 3.}$ Suppose $v, w\in V$. Explain why there exists a unique $u\in V$ such that $v + u = w$. $\textbf{Solution 3.}$ There exists a unique vector $−v\in V$ such that $v + (−v) = 0$. Adding $−v$ to both sides of the equation above, we see that the equation above is equivalent to the equation $u = w − v,$ which is equivalent to the equation $u = 1(w − v),$ which shows that our original equation has a unique solution. --- $\textbf{Exercise 3.}$ Suppose $v, w\in V$. Explain why there exists a unique $u\in V$ such that $v + u = w$. $\textbf{Solution 3.}$ There exists a unique vector $−v\in V$ such that $v + (−v) = 0$. Adding $−v$ to both sides of the equation $v + u = w$, we see that the equation is equivalent to $u = w − v,$ which is equivalent to $u = 1(w − v),$ which shows that our original equation has a unique solution. $\textit{Commentary:}$ This exercise is about the solvability of the equation $v + u = w$ in a vector space $V$. It shows that for any vectors $v$ and $w$, there is a unique vector $u$ that satisfies this equation. This is a fundamental property of vector spaces that follows from the existence of additive inverses and the properties of vector addition. In fact, this property characterizes vector spaces among more general algebraic structures. A set with an addition operation where this property holds is called an "affine space". Every vector space is an affine space, but not every affine space is a vector space (affine spaces don't necessarily have a scalar multiplication operation). This property is the basis for many practical applications of vector spaces, such as solving systems of linear equations, which can be interpreted as finding a vector $u$ that satisfies an equation of the form $Au = w$ for a matrix $A$ and a vector $w$. $\textit{Example:}$ In $\mathbb{R}^2$, for any vectors $\mathbf{v} = (v_1, v_2)$ and $\mathbf{w} = (w_1, w_2)$, there is a unique vector $\mathbf{u} = (u_1, u_2)$ such that $\mathbf{v} + \mathbf{u} = \mathbf{w}$. Specifically, $\mathbf{u} = \mathbf{w} - \mathbf{v} = (w_1 - v_1, w_2 - v_2)$. In $M_{2\times 2}(\mathbb{R})$, for any matrices $A = \begin{pmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{pmatrix}$ and $B = \begin{pmatrix} b_{11} & b_{12} \ b_{21} & b_{22} \end{pmatrix}$, there is a unique matrix $X = \begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \end{pmatrix}$ such that $A + X = B$. Specifically, $X = B - A = \begin{pmatrix} b_{11} - a_{11} & b_{12} - a_{12} \ b_{21} - a_{21} & b_{22} - a_{22} \end{pmatrix}$. In $\mathbb{R}[x]$, for any polynomials $p(x)$ and $q(x)$, there is a unique polynomial $r(x)$ such that $p(x) + r(x) = q(x)$. Specifically, $r(x) = q(x) - p(x)$.