MAT2611 ASSIGNMENT 3 2023
Problem 7
𝑈, 𝑉 are subsets of ℝ4 defined by:
𝑈 = {(𝑥, 𝑦, 𝑧, 𝑢) ∈ ℝ4 ∶ 𝑥 2 − 𝑦𝑧 = 𝑢}
𝑉 = {(𝑥, 𝑦, 𝑧, 𝑣) ∈ ℝ4 ∶ 𝑥 = 2𝑧 and 𝑥 − 3𝑦 = 𝑣}
A subset 𝐻 of a vector space 𝐺 is a subspace of 𝐺 if:
1. 𝐻 is a non-empty subset of 𝐺. (Easier to just check if the zero vector is in 𝐻)
2. For ℎ1 , ℎ2 ∈ 𝐻, we have ℎ1 + ℎ2 ∈ 𝐻 (Closed under addition)
3. For all 𝑘 ∈ ℝ and ℎ ∈ 𝐻, we have that 𝑘ℎ ∈ 𝐻 (Closed under scalar multiplication)
Checking 𝑈
Let 𝑥 = 1, 𝑦 = 2, 𝑧 = −1
Therefore 𝑢 = (1)2 − (2)(−1)
𝑢=3
𝒖𝟏 = (1, 2, −1, 3)
Let 𝑥 = 2, 𝑦 = 1, 𝑧 = 3
Therefore 𝑢 = (2)2 − (1)(3)
𝑢=1
𝒖𝟐 = (2, 1, 3, 1)
We now have two vectors that are in 𝑈 which are 𝒖𝟏 and 𝒖𝟐
𝒖𝟏 + 𝒖𝟐 = (1, 2, −1, 3) + (2, 1, 3, 1)
𝒖𝟏 + 𝒖𝟐 = (3, 3, 2, 4)
To test if 𝒖𝟏 + 𝒖𝟐 ∈ 𝑈
𝑥 = 3, 𝑦 = 3, 𝑧 = 2, 𝑢 = 4
In order for 𝒖𝟏 + 𝒖𝟐 to be in 𝑈, the following equation must be satisfied:
𝑥 2 − 𝑦𝑧 = 𝑢
Problem 7
𝑈, 𝑉 are subsets of ℝ4 defined by:
𝑈 = {(𝑥, 𝑦, 𝑧, 𝑢) ∈ ℝ4 ∶ 𝑥 2 − 𝑦𝑧 = 𝑢}
𝑉 = {(𝑥, 𝑦, 𝑧, 𝑣) ∈ ℝ4 ∶ 𝑥 = 2𝑧 and 𝑥 − 3𝑦 = 𝑣}
A subset 𝐻 of a vector space 𝐺 is a subspace of 𝐺 if:
1. 𝐻 is a non-empty subset of 𝐺. (Easier to just check if the zero vector is in 𝐻)
2. For ℎ1 , ℎ2 ∈ 𝐻, we have ℎ1 + ℎ2 ∈ 𝐻 (Closed under addition)
3. For all 𝑘 ∈ ℝ and ℎ ∈ 𝐻, we have that 𝑘ℎ ∈ 𝐻 (Closed under scalar multiplication)
Checking 𝑈
Let 𝑥 = 1, 𝑦 = 2, 𝑧 = −1
Therefore 𝑢 = (1)2 − (2)(−1)
𝑢=3
𝒖𝟏 = (1, 2, −1, 3)
Let 𝑥 = 2, 𝑦 = 1, 𝑧 = 3
Therefore 𝑢 = (2)2 − (1)(3)
𝑢=1
𝒖𝟐 = (2, 1, 3, 1)
We now have two vectors that are in 𝑈 which are 𝒖𝟏 and 𝒖𝟐
𝒖𝟏 + 𝒖𝟐 = (1, 2, −1, 3) + (2, 1, 3, 1)
𝒖𝟏 + 𝒖𝟐 = (3, 3, 2, 4)
To test if 𝒖𝟏 + 𝒖𝟐 ∈ 𝑈
𝑥 = 3, 𝑦 = 3, 𝑧 = 2, 𝑢 = 4
In order for 𝒖𝟏 + 𝒖𝟐 to be in 𝑈, the following equation must be satisfied:
𝑥 2 − 𝑦𝑧 = 𝑢
