Worksheet 2

Solution.

We have two directions to prove.

"$\implies$"

Assume that $\set{u, v}$ is linearly dependent. Then there exist $a, b \in \F$ not both zero such that $au + bv = 0$. If $a \neq 0$, then we may multiply both sides by $a^{-1}$ to get

Otherwise, $b \neq 0$, and we analogously get

so in both cases, $u$ and $v$ are multiples of each.

"$\impliedby$"

Suppose $u$ and $v$ are multiples of each other. Without loss of generality, we assume that $u$ is a multiple of $v$, i.e., there exists a scalar $c \in \F$ such that $u = cv$. In the other case, we can swap $u$ and $v$ in the following proof. Rearranging $u = cv$, we get

Thus, the zero vector is a non-trivial linear combination of $u$ and $v$ since $1 \neq 0$, so $\set{u, v}$ is linearly dependent.

Solution.

1. "$\implies$"

   Assume that $\set{u, v}$ is linearly independent. Let's show that $\set{u + v, u - v}$ is linearly independent. Suppose we have scalars $a, b \in \F$ such that

   $$a\p{u + v} + b\p{u - v} = 0.$$

   We need to show that $a = b = 0$. Distributing and refactoring, we get

   $$\p{a + b}u + \p{a - b}v = 0,$$

   but $\set{u, v}$ is linearly independent, so this forces

   $$\begin{cases} a + b = 0, \\ a - b = 0. \end{cases}$$

   Adding the equations together and subtracting them, we get the system

   $$\begin{cases} 2a = 0, \\ 2b = 0. \end{cases}$$

   Since $\F$ has characteristic different from $2$, we know $2 \neq 0$, so $\frac{1}{2} \in \F$. Thus, multiplying the equations by $\frac{1}{2}$, we get $a = b = 0$.

   "$\impliedby$"

   Assume that $\set{u + v, u - v}$ is linearly independent. Now assume $a, b \in \F$ are such that $au + bv = 0$. We need to show that $a = b = 0$.

   In order to use the fact that $\set{u + v, u - v}$, we need to turn $au + bv$ into a linear combination of $u + v$ and $u - v$. Let's work backwards: assume there were scalars $c, d \in \F$ such that

   $$\begin{aligned} au + bv &= c\p{u + v} + d\p{u - v} \\ &= \p{c + d}u + \p{c - d}v. \end{aligned}$$

   By comparing coefficients, we know that if we could solve

   $$\begin{cases} a = c + d, \\ b = c - d \end{cases}$$

   for $c$ and $d$, then our idea works. Adding and subtracting, this system becomes

   $$\begin{cases} a + b = 2c, \\ a - b = 2d. \end{cases}$$

   Like before, because $\F$ has characteristic not equal to $2$, we can divide by $2$, so

   $$\begin{cases} c = \frac{a + b}{2}, \\ d = \frac{a - b}{2}. \end{cases}$$

   With these choices of $c$ and $d$, we have

   $$0 = au + bv = \p{\frac{a+b}{2}}\p{u + v} + \p{\frac{a-b}{2}}\p{u - v}.$$

   So because $\set{u + v, u - v}$ is linearly independent, this forces the coefficients to be $0$:

   $$\begin{cases} \frac{a + b}{2} = 0, \\ \frac{a - b}{2} = 0. \end{cases}$$

   Solving like before, we get $a = b = 0$.

   If $\F$ has characteristic $2$, then $1 + 1 = 2 = 0$, so $1 = -1$. This means that $u - v = u + \p{-1}v = u + v$, and so $\set{u + v, u - v}$ can never be linearly dependent. As a concrete example, let $\F = \Z_2$ and $V = \F^2$. Then if $u = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $v = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, then $\set{u, v}$ is linearly independent, but

   $$u + v = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \quad\text{and}\quad u - v = \begin{pmatrix} \phantom{-}1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}.$$

2. This is almost identical to part 1, so I'll omit it.