Use the position equation
1 answer:
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Answer:
Step-by-step explanation:
From the given information.
R3 is a vector space over the field R, where R is the set of real numbers.
Where;
The set W = {(0, x2, x3): x2 and x3 are real numbers} is the subspace of ³
To proof:
(0,0,0) ∈ W ⇒ W ≠ ∅
Suppose u and v is an element of W;
i.e.
u,v ∈ W (which implies that) ⇒ u (0,x2,x3) and v = (0,y2,y3) are real numbers.
Then
u+v = (0,x2,x3) +(0, y2, y3)
u+v = (0,x2+y2, x2+y3) ∈ W
⇒ u+v ∈ W ----- (1)
Now, if we take any integer to be an element of the real number
i.e
∝ ∈
∝* = ∝(0,x2,x3)
∝* = (0,x2,x3) ∈ W
⇒ ∝*u ∈ W ------(2)
Thus from (1) and (2), W is a subspace of ³