# Linear Algebra Done Right 2nd Edition – Chapter 1 Exercises 2, 3, 4, 5(a)

Chapter 1

2. Show that $\frac{-1 + \sqrt{3}i}{2}$ is a cube root of 1 (meaning that its cube equals 1).

Solution:

$\frac{(-1 + \sqrt{3}i)(-1 + \sqrt{3}i)(-1 + \sqrt{3}i)}{8} =$
$\frac{(1 + (3)(-1) - 2\sqrt{3}i)(-1 + \sqrt{3}i)}{8} =$
$\frac{(-2 + -2\sqrt{3}i)(-1 + \sqrt{3}i)}{8} =$
$\frac{2 -2(3)(-1) + 2\sqrt{3}i - 2\sqrt{3}i)}{8} =$
$\frac{2 + 6}{8} = 1$

3. Prove that $-(-v) = v$ for every $v \in V$.

Solution:

$-(-v) = -1(-v)$ by proposition 1.6
$-1(-v) = -1( (-1) v)$ by proposition 1.6
$-1( (-1) v) = 1v = v$ $\blacksquare$

This is the alternate proof suggested in the solutions manual.

4. Prove that if $a \in F, v \in V$ and $av = 0$, then $a = 0$ or $v = 0$.

Solution:

$\text{Suppose } a \ne 0$
$av = 0 = a0$ by proposition 1.5
$av - a0 = 0$
$a(v - 0) = 0$
$(v - 0) = 0 \text{ since } a \ne 0$
$v = 0$

Now let $a = 0$, then done. $\blacksquare$

Check: The solution manual provides a different proof
$\frac{1}{a}(av) = \frac{1}{a}0 \text{ since } a \ne 0$
$1v = 0$
$v = 0$

5.(a) Determine if ${(x_1,x_2,x_3) \in \textbf{F}^3:x_1 + 2x_2 + 3x_3 = 0}$ is a subspace of $\textbf{F}^3$.

Solution:

Let $U = {(x_1,x_2,x_3) \in \textbf{F}^3:x_1 + 2x_2 + 3x_3 = 0}$

$0 + 20 + 30 = 0 \Rightarrow (0,0,0) \in U$
Additive identity in $U$.

Let $(a,b,c) \in U$ and $(p,q,r) \in U$
Then
$a + 2b + 3c = 0$ and
$p + 2q + 3r = 0$
$\Rightarrow$
$(a,b,c) + (p,q,r) =$
$a + p + 2b + 2q + 3c + 3r =$
$(a + 2b + 3c) + (p + 2q + 3r) = 0$
$\Rightarrow (a,b,c) + (p,q,r) \in U$
Let $a \in \textbf{F}$ and $(p,q,r) \in U$
$a(p,q,r) = a(p + 2q + 3r) = 0$
$\Rightarrow$
$ap + a2q + a3r = 0$
$\Rightarrow$
$(ap, aq, ar) \in U$
$\Rightarrow$
$a(p, q, r) \in U$