MAT2611 ASSIGNMENT 3 2025
Problem 1
(a)
𝑈 = {(𝑥, 𝑦, 𝑧, 0) ∈ ℝ4 ∶ 𝑥 = 𝑦 − 𝑧}
𝑈 = {(𝑥, 𝑦, 𝑧, 0) ∈ ℝ4 ∶ 𝑧 = 𝑦 − 𝑥 }
𝑈 = {(𝑥, 𝑦, 𝑦 − 𝑥, 0) ∈ ℝ4 }
Let 𝑢1 = ( 𝑥1 ,𝑦1 , 𝑦1 − 𝑥1 ,0) and 𝑢2 = (𝑥2 , 𝑦2 , 𝑦2 − 𝑥2 , 0) ∈ 𝑈
𝑘 ∈ ℝ be a scalar.
(i)
By definition, 𝑈 is a subset of ℝ4
(ii)
(0, 0, 0 − 0, 0) ∈ 𝑈
∴ (0, 0, 0, 0) ∈ 𝑈
∴ 𝟎ℝ𝟒 ∈ 𝑈
∴ 𝑈 is a nonempty subset of ℝ4
(iii)
𝑢1 + 𝑢2 = ( 𝑥1 ,𝑦1 , 𝑦1 − 𝑥1 ,0) + ( 𝑥2 , 𝑦2 , 𝑦2 − 𝑥2 , 0)
𝑢1 + 𝑢2 = ( 𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , 𝑦1 − 𝑥1 + 𝑦2 − 𝑥2 , 0 + 0)
𝑢1 + 𝑢2 = ( 𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , (𝑦1 + 𝑦2 ) − (𝑥1 + 𝑥2 ), 0) ∈ 𝑈
So, 𝑢1 + 𝑢2 ∈ 𝑈
Problem 1
(a)
𝑈 = {(𝑥, 𝑦, 𝑧, 0) ∈ ℝ4 ∶ 𝑥 = 𝑦 − 𝑧}
𝑈 = {(𝑥, 𝑦, 𝑧, 0) ∈ ℝ4 ∶ 𝑧 = 𝑦 − 𝑥 }
𝑈 = {(𝑥, 𝑦, 𝑦 − 𝑥, 0) ∈ ℝ4 }
Let 𝑢1 = ( 𝑥1 ,𝑦1 , 𝑦1 − 𝑥1 ,0) and 𝑢2 = (𝑥2 , 𝑦2 , 𝑦2 − 𝑥2 , 0) ∈ 𝑈
𝑘 ∈ ℝ be a scalar.
(i)
By definition, 𝑈 is a subset of ℝ4
(ii)
(0, 0, 0 − 0, 0) ∈ 𝑈
∴ (0, 0, 0, 0) ∈ 𝑈
∴ 𝟎ℝ𝟒 ∈ 𝑈
∴ 𝑈 is a nonempty subset of ℝ4
(iii)
𝑢1 + 𝑢2 = ( 𝑥1 ,𝑦1 , 𝑦1 − 𝑥1 ,0) + ( 𝑥2 , 𝑦2 , 𝑦2 − 𝑥2 , 0)
𝑢1 + 𝑢2 = ( 𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , 𝑦1 − 𝑥1 + 𝑦2 − 𝑥2 , 0 + 0)
𝑢1 + 𝑢2 = ( 𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , (𝑦1 + 𝑦2 ) − (𝑥1 + 𝑥2 ), 0) ∈ 𝑈
So, 𝑢1 + 𝑢2 ∈ 𝑈
