1 Show that α + β = β + α \alpha + \beta = \beta + \alpha α + β = β + α α , β ∈ C \alpha, \beta \in \mathbf{C} α , β ∈ C Solution C \mathbf{C} C α = a + b i \alpha = a + bi α = a + bi β = c + d i \beta = c + di β = c + d i a , b , c , d ∈ R a, b, c, d \in \R a , b , c , d ∈ R
α + β = ( a + b i ) + ( c + d i ) = ( a + c ) + ( b + d ) i = ( c + a ) + ( d + b ) i = ( c + d i ) + ( a + b i ) = β + α . □ \begin{aligned}
\alpha + \beta &= (a + bi) + (c + di) = (a + c) + (b + d)i \\
&= (c + a) + (d + b)i = (c + di) + (a + bi) \\
&= \beta + \alpha. \quad \square
\end{aligned} α + β = ( a + bi ) + ( c + d i ) = ( a + c ) + ( b + d ) i = ( c + a ) + ( d + b ) i = ( c + d i ) + ( a + bi ) = β + α . □
2 Show that ( α + β ) + λ = α + ( β + λ ) (\alpha + \beta) + \lambda = \alpha + (\beta + \lambda) ( α + β ) + λ = α + ( β + λ ) α , β , λ ∈ C \alpha, \beta, \lambda \in \mathbf{C} α , β , λ ∈ C
Solution
Let α = a + b i \alpha = a + bi α = a + bi β = c + d i \beta = c + di β = c + d i λ = e + f i \lambda = e + fi λ = e + f i
( α + β ) + λ = ( a + b i + c + d i ) + e + f i = ( ( a + c ) + ( b + d ) i ) + ( e + f i ) = ( a + c + e ) + ( b + d + f ) i = ( a + b i ) + ( ( c + e ) + ( d + f ) i ) = α + ( β + λ ) . □ (\alpha + \beta) + \lambda = (a + bi + c + di) + e + fi \\
= ((a + c) + (b + d)i) + (e + fi) = (a + c + e) + (b + d + f)i \\
= (a + bi) + ((c + e) + (d + f)i) = \alpha + (\beta + \lambda). \quad \square ( α + β ) + λ = ( a + bi + c + d i ) + e + f i = (( a + c ) + ( b + d ) i ) + ( e + f i ) = ( a + c + e ) + ( b + d + f ) i = ( a + bi ) + (( c + e ) + ( d + f ) i ) = α + ( β + λ ) . □
3 Show that ( α β ) λ = α ( β λ ) (\alpha \beta)\lambda = \alpha(\beta \lambda) ( α β ) λ = α ( β λ ) α , β , λ ∈ C \alpha, \beta, \lambda \in \mathbf{C} α , β , λ ∈ C
Solution
Let α = a + b i \alpha = a + bi α = a + bi β = c + d i \beta = c + di β = c + d i λ = e + f i \lambda = e + fi λ = e + f i
( α β ) λ = ( ( a + b i ) ( c + d i ) ) ( e + f i ) = ( ( a c − b d ) + ( a d + b c ) i ) ( e + f i ) = ( ( a c − b d ) e − ( a d + b c ) f ) + ( ( a c − b d ) f + ( a d + b c ) e ) i = ( a c e − b d e − a d f − b c f ) + ( a c f − b d f + a d e + b c e ) i . (\alpha \beta) \lambda = ((a + bi)(c + di)) (e + fi) \\
= ((ac - bd) + (ad + bc)i)(e + fi) \\
= ((ac - bd)e - (ad + bc)f) + ((ac - bd)f + (ad + bc)e)i \\
= (ace - bde - adf - bcf) + (acf - bdf + ade + bce)i. ( α β ) λ = (( a + bi ) ( c + d i )) ( e + f i ) = (( a c − b d ) + ( a d + b c ) i ) ( e + f i ) = (( a c − b d ) e − ( a d + b c ) f ) + (( a c − b d ) f + ( a d + b c ) e ) i = ( a ce − b d e − a df − b c f ) + ( a c f − b df + a d e + b ce ) i .
Calculate separately that
α ( β λ ) = ( a + b i ) ( ( c + d i ) ( e + f i ) ) = ( a + b i ) ( ( c e − d f ) + ( c f + d e ) i ) = ( a ( c e − d f ) − b ( c f + d e ) ) + ( a ( c f + d e ) + b ( c e − d f ) ) i = ( a c e − a d f − b c f − b d e ) + ( a c f + a d e + b c e − b d f ) i . \alpha (\beta \lambda) = (a + bi)((c + di)(e + fi)) \\
= (a + bi)((ce - df) + (cf + de)i) \\
= (a(ce - df) - b(cf + de)) + (a(cf+de) + b(ce-df))i \\
= (ace - adf - bcf - bde) + (acf + ade + bce - bdf)i. α ( β λ ) = ( a + bi ) (( c + d i ) ( e + f i )) = ( a + bi ) (( ce − df ) + ( c f + d e ) i ) = ( a ( ce − df ) − b ( c f + d e )) + ( a ( c f + d e ) + b ( ce − df )) i = ( a ce − a df − b c f − b d e ) + ( a c f + a d e + b ce − b df ) i .
Rearranging terms shows that
( α β ) λ = α ( β λ ) . □ (\alpha \beta) \lambda = \alpha (\beta \lambda). \quad \square ( α β ) λ = α ( β λ ) . □
4 Show that λ ( α + β ) = λ α + λ β \lambda (\alpha + \beta) = \lambda \alpha + \lambda \beta λ ( α + β ) = λ α + λ β λ , α , β ∈ C \lambda, \alpha, \beta \in \mathbf{C} λ , α , β ∈ C
Solution
Let α = a + b i \alpha = a + bi α = a + bi β = c + d i \beta = c + di β = c + d i λ = e + f i \lambda = e + fi λ = e + f i
λ ( α + β ) = ( e + f i ) ( ( a + b i ) + ( c + d i ) ) = ( e + f i ) ( ( a + c ) + ( b + d ) i ) = ( e ( a + c ) − f ( b + d ) ) + ( e ( b + d ) + f ( a + c ) ) i = ( e a + e c − f b − f d ) + ( e b + e d + f a + f c ) i . \lambda (\alpha + \beta) = (e + fi)((a + bi) + (c + di)) \\
= (e + fi)((a+c) + (b+d)i) \\
= (e(a+c) - f(b+d)) + (e(b+d) + f(a+c))i \\
= (ea + ec -fb - fd) + (eb + ed + fa + fc)i. λ ( α + β ) = ( e + f i ) (( a + bi ) + ( c + d i )) = ( e + f i ) (( a + c ) + ( b + d ) i ) = ( e ( a + c ) − f ( b + d )) + ( e ( b + d ) + f ( a + c )) i = ( e a + ec − f b − fd ) + ( e b + e d + f a + f c ) i .
Calculate separately that
λ α + λ β = ( e + f i ) ( a + b i ) + ( e + f i ) ( c + d i ) = ( ( e a − b f ) + ( e b + f a ) i ) + ( ( e c − f d ) + ( e d + f c ) i ) = ( e a − b f + e c − f d ) + ( e b + f a + e d + f c ) i . \lambda \alpha + \lambda \beta = (e + fi)(a + bi) + (e + fi)(c + di) \\
= ((ea - bf) + (eb + fa)i) + ((ec - fd) + (ed + fc)i) \\
= (ea - bf + ec - fd) + (eb + fa + ed + fc)i. λ α + λ β = ( e + f i ) ( a + bi ) + ( e + f i ) ( c + d i ) = (( e a − b f ) + ( e b + f a ) i ) + (( ec − fd ) + ( e d + f c ) i ) = ( e a − b f + ec − fd ) + ( e b + f a + e d + f c ) i .
Rearranging terms and the commutative property for reals gives
λ ( α + β ) = λ α + λ β . □ \lambda(\alpha + \beta) = \lambda \alpha + \lambda \beta. \quad \square λ ( α + β ) = λ α + λ β . □
5 Show that for every α ∈ C \alpha \in \mathbf{C} α ∈ C β ∈ C \beta \in \mathbf{C} β ∈ C α + β = 0 \alpha + \beta = 0 α + β = 0
Solution
Let α = a + b i \alpha = a + bi α = a + bi β = − a − b i \beta = -a - bi β = − a − bi α + β = 0 \alpha + \beta = 0 α + β = 0 β ′ = c + d i \beta' = c + di β ′ = c + d i α + β ′ = 0. \alpha + \beta' = 0. α + β ′ = 0.
Plug in,
( a + b i ) + ( c + d i ) = 0 ( a + c ) + ( b + d ) i = 0 c = − a , d = − b , (a + bi) + (c + di) = 0 \\
(a + c) + (b + d)i = 0 \\
c = -a, \quad d = -b, ( a + bi ) + ( c + d i ) = 0 ( a + c ) + ( b + d ) i = 0 c = − a , d = − b ,
so we have
β ′ = β \beta' = \beta β ′ = β
proving uniqueness. □ \quad \square □
6 Show that for every α ∈ C \alpha \in \mathbf{C} α ∈ C α ≠ 0 \alpha \neq 0 α = 0 β ∈ C \beta \in \mathbf{C} β ∈ C α β = 1 \alpha \beta = 1 α β = 1
Solution
Let α = a + b i \alpha = a + bi α = a + bi
α × α ‾ = ( a + b i ) ( a − b i ) = a 2 + b 2 \alpha \times \overline{\alpha} = (a + bi)(a - bi) = a^2 + b^2 α × α = ( a + bi ) ( a − bi ) = a 2 + b 2
Assume for the purpose of contradiction that a 2 + b 2 = 0 a^2 + b^2 = 0 a 2 + b 2 = 0 a = b = 0 a = b = 0 a = b = 0 α = 0 \alpha = 0 α = 0 α ≠ 0 \alpha \neq 0 α = 0 a 2 + b 2 ≠ 0 a^2 + b^2 \neq 0 a 2 + b 2 = 0
Let
β = a − b i a 2 + b 2 , \beta = \frac{a - bi}{a^2 + b^2}, β = a 2 + b 2 a − bi ,
which is well-defined because a 2 + b 2 ≠ 0 a^2 + b^2 \neq 0 a 2 + b 2 = 0 α β = 1 \alpha \beta = 1 α β = 1
Let β ′ = c + d i \beta' = c + di β ′ = c + d i
α β ′ = 1. \alpha \beta' = 1. α β ′ = 1.
Plug in to get
( a + b i ) ( c + d i ) = 1 ⟺ ( a c − b d ) + ( a d + b c ) i = 1 ⟺ a c − b d = 1 and a d + b c = 0 (a + bi)(c + di) = 1 \\
\iff (ac - bd) + (ad + bc)i = 1 \\
\iff ac - bd = 1 \quad \text{and} \quad ad + bc = 0 ( a + bi ) ( c + d i ) = 1 ⟺ ( a c − b d ) + ( a d + b c ) i = 1 ⟺ a c − b d = 1 and a d + b c = 0
a c = 1 + b d c = 1 + b d a 0 = a d + b c = a d + b + b 2 d a 0 = a 2 d + b + b 2 d a 2 d + b 2 d = − b d = − b a 2 + b 2 c = 1 + b d a = 1 − b 2 a 2 + b 2 a = a 2 + b 2 − b 2 a ( a 2 + b 2 ) = a a 2 + b 2 . □ ac = 1 + bd \\[4px]
c = \frac{1 + bd}{a} \\[4px]
0 = ad + bc = ad + \frac{b + b^2d}{a} \\
0 = a^2d + b + b^2d \\
a^2d + b^2d = -b \\
d = \frac{-b}{a^2+b^2} \\[4px]
c = \frac{1 + bd}{a} = \frac{1 - \frac{b^2}{a^2+b^2}}{a} = \frac{a^2 + b^2 - b^2}{a(a^2 + b^2)} \\
= \frac{a}{a^2 + b^2}. \quad \square a c = 1 + b d c = a 1 + b d 0 = a d + b c = a d + a b + b 2 d 0 = a 2 d + b + b 2 d a 2 d + b 2 d = − b d = a 2 + b 2 − b c = a 1 + b d = a 1 − a 2 + b 2 b 2 = a ( a 2 + b 2 ) a 2 + b 2 − b 2 = a 2 + b 2 a . □
7 Show that
− 1 + 3 i 2 \frac{-1 + \sqrt{3}i}{2} 2 − 1 + 3 i
is a cube root of 1 1 1 1 1 1
Solution x = − 1 + 3 i 2 x = \frac{-1 + \sqrt{3}i}{2} x = 2 − 1 + 3 i
( − 1 + 3 i ) ( − 1 + 3 i ) = ( 1 − 3 ) + ( − 3 − 3 ) i = − 2 − 2 3 i . \left(-1 + \sqrt{3}i\right) \left(-1 + \sqrt{3}i\right) = (1 - 3) + (-\sqrt{3}-\sqrt{3})i \\
= -2 - 2\sqrt{3}i. ( − 1 + 3 i ) ( − 1 + 3 i ) = ( 1 − 3 ) + ( − 3 − 3 ) i = − 2 − 2 3 i .
And so
( − 1 + 3 i ) 3 = ( − 2 − 2 3 i ) ( − 1 + 3 i ) = ( 2 + 2 × 3 ) + ( − 2 3 + 2 3 ) i = 8. \\[1px]
\left(-1 + \sqrt{3}i\right)^3 = \left(-2 - 2\sqrt{3}i\right)\left(-1 + \sqrt{3}i\right) \\
= (2 + 2 \times 3) + \left(-2\sqrt{3} + 2\sqrt{3}\right)i = 8. ( − 1 + 3 i ) 3 = ( − 2 − 2 3 i ) ( − 1 + 3 i ) = ( 2 + 2 × 3 ) + ( − 2 3 + 2 3 ) i = 8.
Therefore x 3 = 1 x^3 = 1 x 3 = 1 □ \quad \square □
8 Find two distinct square roots of i i i
Solution
( a + b i ) ( a + b i ) = i ⟺ ( a 2 − b 2 ) + ( a b + b a ) i = i ⟺ a 2 = b 2 and 2 a b = 1 (a+bi)(a+bi) = i \\
\iff (a^2-b^2) + (ab + ba)i = i \\
\iff a^2 = b^2 \quad \text{and} \quad 2ab = 1 ( a + bi ) ( a + bi ) = i ⟺ ( a 2 − b 2 ) + ( ab + ba ) i = i ⟺ a 2 = b 2 and 2 ab = 1
Consider 1 2 + 1 2 i \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i 2 1 + 2 1 i − 1 2 − 1 2 i -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i − 2 1 − 2 1 i
( 1 2 + 1 2 i ) ( 1 2 + 1 2 i ) = ( 1 2 − 1 2 ) + ( 1 2 + 1 2 ) i = i . □ \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i \right) \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i \right) \\[4px]
= \left(\frac{1}{2} - \frac{1}{2}\right) + \left(\frac{1}{2} + \frac{1}{2}\right)i = i. \quad \square ( 2 1 + 2 1 i ) ( 2 1 + 2 1 i ) = ( 2 1 − 2 1 ) + ( 2 1 + 2 1 ) i = i . □
9 Find x ∈ R 4 x \in \mathbf{R}^4 x ∈ R 4
( 4 , − 3 , 1 , 7 ) + 2 x = ( 5 , 9 , − 6 , 8 ) . (4,-3,1,7) + 2x = (5,9,-6,8). ( 4 , − 3 , 1 , 7 ) + 2 x = ( 5 , 9 , − 6 , 8 ) .
Solution
2 x = ( 1 , 12 , − 7 , 1 ) x = ( 0.5 , 6 , − 3.5 , 0.5 ) . □ 2x = (1, 12, -7, 1) \\
x = (0.5, 6, -3.5, 0.5). \quad \square 2 x = ( 1 , 12 , − 7 , 1 ) x = ( 0.5 , 6 , − 3.5 , 0.5 ) . □
10 Explain why there does not exist λ ∈ C \lambda \in \mathbf{C} λ ∈ C
λ ( 2 − 3 i , 5 + 4 i , − 6 + 7 i ) = ( 12 − 5 i , 7 + 22 i , − 32 − 9 i ) . \lambda(2-3i,5+4i,-6+7i)=(12-5i,7+22i,-32 -9i). λ ( 2 − 3 i , 5 + 4 i , − 6 + 7 i ) = ( 12 − 5 i , 7 + 22 i , − 32 − 9 i ) .
Solution
Let λ = a + b i \lambda = a + bi λ = a + bi
12 − 5 i = ( a + b i ) ( 2 − 3 i ) = ( 2 a + 3 b ) + ( − 3 a + 2 b ) i 12 = 2 a + 3 b − 5 = − 3 a + 2 b 12 - 5i = (a + bi)(2 - 3i) = (2a + 3b) + (-3a + 2b)i \\
12 = 2a + 3b \\
-5 = -3a + 2b 12 − 5 i = ( a + bi ) ( 2 − 3 i ) = ( 2 a + 3 b ) + ( − 3 a + 2 b ) i 12 = 2 a + 3 b − 5 = − 3 a + 2 b
Multiply the top equation by three and the bottom by two
36 = 6 a + 9 b − 10 = − 6 a + 4 b , 36 = 6a + 9b \\
-10 = -6a + 4b, 36 = 6 a + 9 b − 10 = − 6 a + 4 b ,
add both equations together to get
13 b = 26 b = 2 12 = 2 a + 3 × 2 12 = 2 a + 6 2 a = 6 a = 3 13b = 26 \\
b = 2 \\
12 = 2a + 3 \times 2 \\
12 = 2a + 6 \\
2a = 6 \\
a = 3 13 b = 26 b = 2 12 = 2 a + 3 × 2 12 = 2 a + 6 2 a = 6 a = 3
So we must have λ = 3 + 2 i \lambda = 3 + 2i λ = 3 + 2 i
( 3 + 2 i ) ( 2 − 3 i ) = ( 6 + 6 ) + ( − 9 + 4 ) i = 12 − 5 i . (3 + 2i)(2 - 3i) = (6 + 6) + (-9 + 4)i = 12 - 5i. ( 3 + 2 i ) ( 2 − 3 i ) = ( 6 + 6 ) + ( − 9 + 4 ) i = 12 − 5 i .
We also need to have
λ ( 5 + 4 i ) = 7 + 22 i . \lambda(5 + 4i) = 7 + 22i. λ ( 5 + 4 i ) = 7 + 22 i .
Calculate that
( 3 + 2 i ) ( 5 + 4 i ) = ( 15 − 8 ) + ( 12 + 10 ) i = 7 + 22 i . (3 + 2i)(5 + 4i) = (15 - 8) + (12 + 10)i = 7 + 22i. ( 3 + 2 i ) ( 5 + 4 i ) = ( 15 − 8 ) + ( 12 + 10 ) i = 7 + 22 i .
Okay, so that one works, but we still got one more to go! We still need
λ ( − 6 + 7 i ) = − 32 − 9 i . \lambda (-6 + 7i) = -32 - 9i. λ ( − 6 + 7 i ) = − 32 − 9 i .
Calculate that
( 3 + 2 i ) ( − 6 + 7 i ) = ( − 18 − 14 ) + ( 21 − 12 ) i = − 32 + 9 i ≠ − 32 − 9 i □ (3 + 2i) (-6 + 7i) = (-18 - 14) + (21 - 12)i \\
= -32 + 9i \neq -32 - 9i \quad \square ( 3 + 2 i ) ( − 6 + 7 i ) = ( − 18 − 14 ) + ( 21 − 12 ) i = − 32 + 9 i = − 32 − 9 i □
11 Show that ( x + y ) + z = x + ( y + z ) (x+y)+z = x+(y+z) ( x + y ) + z = x + ( y + z ) x , y , z ∈ F n x,y,z \in \mathbf{F}^n x , y , z ∈ F n
Solution
Let x = ( x 1 , … , x n ) x = (x_1, \dots, x_n) x = ( x 1 , … , x n ) y = ( y 1 , … , y n ) y = (y_1, \dots, y_n) y = ( y 1 , … , y n ) z = ( z 1 , … , z n ) z = (z_1, \dots, z_n) z = ( z 1 , … , z n )
( x + y ) + z = ( ( x 1 , … , x n ) + ( y 1 , … , y n ) ) + ( z 1 , … , z n ) = ( x 1 + y 1 , … , x n + y n ) + ( z 1 , … , z n ) = ( x 1 + y 1 + z 1 , … , x n + y n + z n ) = ( x 1 + ( y 1 + z 1 ) , … , x n + ( y n + z n ) ) = ( x 1 , … , x n ) + ( y 1 + z 1 , … , y n + z n ) = x + ( y + z ) . □ (x + y) + z = ((x_1, \dots, x_n) + (y_1, \dots, y_n)) + (z_1, \dots, z_n) \\
= (x_1 + y_1, \dots, x_n + y_n) + (z_1, \dots, z_n) \\
= (x_1 + y_1 + z_1, \dots, x_n + y_n + z_n) \\
= (x_1 + (y_1 + z_1), \dots, x_n + (y_n + z_n)) \\
= (x_1, \dots, x_n) + (y_1 + z_1, \dots, y_n + z_n) \\
= x + (y + z). \quad \square ( x + y ) + z = (( x 1 , … , x n ) + ( y 1 , … , y n )) + ( z 1 , … , z n ) = ( x 1 + y 1 , … , x n + y n ) + ( z 1 , … , z n ) = ( x 1 + y 1 + z 1 , … , x n + y n + z n ) = ( x 1 + ( y 1 + z 1 ) , … , x n + ( y n + z n )) = ( x 1 , … , x n ) + ( y 1 + z 1 , … , y n + z n ) = x + ( y + z ) . □
12 Show that ( a b ) x = a ( b x ) (ab)x = a(bx) ( ab ) x = a ( b x ) x ∈ F n x \in \mathbf{F}^n x ∈ F n a , b ∈ F a,b \in \mathbf{F} a , b ∈ F
Solution
Let x = ( x 1 , … , x n ) x = (x_1, \dots, x_n) x = ( x 1 , … , x n )
( a b ) x = ( a b ) ( x 1 , … , x n ) = ( ( a b ) x 1 , … , ( a b ) x n ) = ( a ( b x 1 ) , … , a ( b x n ) ) = a ( b x 1 , … , b x n ) = a ( b x ) . □ (ab)x = (ab)(x_1, \dots, x_n) = ((ab)x_1, \dots, (ab)x_n) \\
= (a(bx_1), \dots, a(bx_n)) = a(bx_1, \dots, bx_n) = a(bx). \quad \square ( ab ) x = ( ab ) ( x 1 , … , x n ) = (( ab ) x 1 , … , ( ab ) x n ) = ( a ( b x 1 ) , … , a ( b x n )) = a ( b x 1 , … , b x n ) = a ( b x ) . □
13 Show that 1 x = x 1x = x 1 x = x x ∈ F n x \in \mathbf{F}^n x ∈ F n
Solution
Let x = ( x 1 , … , x n ) x = (x_1, \dots, x_n) x = ( x 1 , … , x n )
1 x = ( 1 x 1 , … , 1 x n ) = ( x 1 , … , x n ) = x . □ 1x = (1x_1, \dots, 1x_n) = (x_1, \dots, x_n) = x. \quad \square 1 x = ( 1 x 1 , … , 1 x n ) = ( x 1 , … , x n ) = x . □
14 Show that λ ( x + y ) = λ x + λ y \lambda(x+y) = \lambda x + \lambda y λ ( x + y ) = λ x + λ y λ ∈ F \lambda \in \mathbf{F} λ ∈ F x , y ∈ F n x,y \in \mathbf{F}^n x , y ∈ F n
Solution
Let x = ( x 1 , … , x n ) x = (x_1, \dots, x_n) x = ( x 1 , … , x n ) y = ( y 1 , … , y n ) y = (y_1, \dots, y_n) y = ( y 1 , … , y n )
λ ( x + y ) = λ ( x 1 + y 1 , … , x n + y n ) = ( λ ( x 1 + y 1 ) , … , λ ( x n + y n ) ) = ( λ x 1 + λ y 1 , … , λ x n + λ y n ) = ( λ x 1 , … , λ x n ) + ( λ y 1 , … , λ y n ) = λ x + λ y . □ \lambda(x+y) = \lambda(x_1 + y_1, \dots, x_n + y_n) \\
= (\lambda(x_1 + y_1), \dots, \lambda(x_n + y_n)) = (\lambda x_1 + \lambda y_1, \dots, \lambda x_n + \lambda y_n) \\
= (\lambda x_1, \dots, \lambda x_n) + (\lambda y_1, \dots, \lambda y_n) = \lambda x + \lambda y. \quad \square λ ( x + y ) = λ ( x 1 + y 1 , … , x n + y n ) = ( λ ( x 1 + y 1 ) , … , λ ( x n + y n )) = ( λ x 1 + λ y 1 , … , λ x n + λ y n ) = ( λ x 1 , … , λ x n ) + ( λ y 1 , … , λ y n ) = λ x + λ y . □
15 Show that ( a + b ) x = a x + b x (a+b)x = ax + bx ( a + b ) x = a x + b x a , b ∈ F a,b \in \mathbf{F} a , b ∈ F x ∈ F n x \in \mathbf{F}^n x ∈ F n
Solution
Let x = ( x 1 , … , x n ) x = (x_1, \dots, x_n) x = ( x 1 , … , x n )
( a + b ) x = ( a + b ) ( x 1 , … , x n ) = ( ( a + b ) x 1 , … , ( a + b ) x n ) = ( a x 1 + b x 1 , … , a x n + b x n ) = ( a x 1 , … , a x n ) + ( b x 1 , … , b x n ) = a ( x 1 , … , x n ) + b ( x 1 , … , x n ) = a x + b x . □ (a+b)x = (a+b)(x_1, \dots, x_n) = ((a+b)x_1, \dots, (a+b)x_n) \\
= (ax_1 + bx_1, \dots, ax_n + bx_n) = (ax_1, \dots, ax_n) + (bx_1, \dots, bx_n) \\
= a(x_1, \dots, x_n) + b(x_1, \dots, x_n) = ax + bx. \quad \square ( a + b ) x = ( a + b ) ( x 1 , … , x n ) = (( a + b ) x 1 , … , ( a + b ) x n ) = ( a x 1 + b x 1 , … , a x n + b x n ) = ( a x 1 , … , a x n ) + ( b x 1 , … , b x n ) = a ( x 1 , … , x n ) + b ( x 1 , … , x n ) = a x + b x . □
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