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
Closed under addition.

Let a \in \textbf{F} and (p,q,r) \in U
Then
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
Closed under scalar multiplication.

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