Exercises — Lesson 1: Vectors and Linear Combinations
Exercise 1.1. [Hand] Let \(v = (2,-1)\) and \(w = (1,3)\). Compute \(3v + 2w\), \(v - w\), and find scalars \(a, b\) with \(av + bw = (0,7)\).
Solution.
\[ \begin{aligned} 3v + 2w &= (6,-3) + (2,6) = (8,3) \\[2pt] v - w &= (1,-4) \end{aligned} \]
For the scalars, \(av + bw = (0,7)\) gives \((2a + b,\; -a + 3b) = (0,7)\), a system solved by elimination:
\[ \begin{aligned} \left[\begin{array}{cc|c} 2 & 1 & 0 \\ -1 & 3 & 7 \end{array}\right] &\to \left[\begin{array}{cc|c} 1 & 4 & 7 \\ -1 & 3 & 7 \end{array}\right] && \text{[$R_1 \leftarrow R_1 + R_2$]} \\[2pt] &\to \left[\begin{array}{cc|c} 1 & 4 & 7 \\ 0 & 7 & 14 \end{array}\right] && \text{[$R_2 \leftarrow R_2 + R_1$]} \\[2pt] &\to \left[\begin{array}{cc|c} 1 & 4 & 7 \\ 0 & 1 & 2 \end{array}\right] && \text{[$R_2 \leftarrow \tfrac{1}{7} R_2$]} \\[2pt] &\to \left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & 2 \end{array}\right] && \text{[$R_1 \leftarrow R_1 - 4 R_2$]} \end{aligned} \]
\[ \boxed{\;a = -1, \qquad b = 2\;} \]
Exercise 1.2. [Hand] Which vectors in \(\mathbb{R}^3\) are linear combinations of \(u = (1,1,0)\) and \(v = (0,1,1)\)? Describe the set geometrically, and exhibit one specific vector in \(\mathbb{R}^3\) that is not a combination of \(u\) and \(v\), with justification.
Solution.
\[ \begin{aligned} \{\, t \in \mathbb{R}^3 : t = a(1,1,0) + b(0,1,1);\ a, b \in \mathbb{R} \,\} &= (a,\; a + b,\; b). \end{aligned} \]
Note \(z = (1,0,1) \neq (a,\, a+b,\, b)\) for all \(a, b \in \mathbb{R}\).
Exercise 1.3. [Proof, \(\star\)] Prove that in any vector space, \((-1)v = -v\) for every \(v\); that is, \((-1)v\) is the additive inverse of \(v\). (Use the proposition \(0v = 0\) and the distributive property. This is LADR 1.B territory.)
Proof. Prove \((-1)v = -v\) for all \(v \in V\), i.e. \((-1)v\) is the additive inverse of \(v\).
\[ \begin{aligned} (-1)v &= (-1 + 0)v \\ &= (-1)v + 0v && \text{[distributivity]} \\ &= (-1)v + 0 && \text{[$0v = 0$]} \\ &= (-1)v + \bigl(v + (-v)\bigr) && \text{[$0 = v + (-v)$]} \\ &= \bigl((-1)v + (1)v\bigr) + (-v) && \text{[$v = 1v$; associativity]} \\ &= (-1 + 1)v + (-v) && \text{[distributivity]} \\ &= 0v + (-v) \\ &= 0 + (-v) \\ &= -v. \qquad\square \end{aligned} \]
Exercise 1.4. [Proof] Prove that the additive inverse of each vector is unique (mirror the proof of the unique-additive-identity proposition).
Proof. Prove \(-v\) is unique for all \(v \in V\).
Suppose \(-v\) and \(-v'\) are both additive inverses of \(v \in V\). Then
\[ \begin{aligned} -v &= -v + 0 \\ &= -v + (-v' + v) && \text{[$-v'$ is an inverse of $v$]} \\ &= -v' + (-v + v) && \text{[commutativity, associativity]} \\ &= -v' + 0 && \text{[$-v$ is an inverse of $v$]} \\ &= -v'. \qquad\square \end{aligned} \]