MAT2611
ASSIGNMENT 3
2025
, PROBLEM 1
U = {(x, y, z, 0) ∈ ℝ4 ∶ x = y − z}
Let ∶ u ∈ U
u = (x, y, z, 0) ∴x=y−z
u = (y − z, y, z, 0)
Choosing y = 0 and z = 0
u = (y − z, y, z, 0)
u = (0 − 0,0,0,0)
u = (0,0,0,0) (Contains the zero vector)
Let u1 , u2 ∈ U such that
u1 = (y1 − z1 , y1 , z1 , 0) and u2 = (y2 − z2 , y2 , z2 , 0)
u1 + u2 = (y1 − z1 , y1 , z1 , 0) + (y2 − z2 , y2 , z2 , 0)
= (y1 − z1 + y2 − z2 , y1 + y2 , z1 + z2 , 0)
= ((y1 + y2 ) − (z1 + z2 ), y1 + y2 , z1 + z2 , 0) ∈ U (Closed under addition)
Let ∶ λ ∈ F ∶
λu = λ(y − z, y, z, 0)
= (λy − λz, λy, λz, λ0)
= (λy − λz, λy, λz, 0) ∈ U (Closed under multiplication)
Hence U is a subspace of ℝ4
ASSIGNMENT 3
2025
, PROBLEM 1
U = {(x, y, z, 0) ∈ ℝ4 ∶ x = y − z}
Let ∶ u ∈ U
u = (x, y, z, 0) ∴x=y−z
u = (y − z, y, z, 0)
Choosing y = 0 and z = 0
u = (y − z, y, z, 0)
u = (0 − 0,0,0,0)
u = (0,0,0,0) (Contains the zero vector)
Let u1 , u2 ∈ U such that
u1 = (y1 − z1 , y1 , z1 , 0) and u2 = (y2 − z2 , y2 , z2 , 0)
u1 + u2 = (y1 − z1 , y1 , z1 , 0) + (y2 − z2 , y2 , z2 , 0)
= (y1 − z1 + y2 − z2 , y1 + y2 , z1 + z2 , 0)
= ((y1 + y2 ) − (z1 + z2 ), y1 + y2 , z1 + z2 , 0) ∈ U (Closed under addition)
Let ∶ λ ∈ F ∶
λu = λ(y − z, y, z, 0)
= (λy − λz, λy, λz, λ0)
= (λy − λz, λy, λz, 0) ∈ U (Closed under multiplication)
Hence U is a subspace of ℝ4