J
Equation
Solution Space
0
∑ q = 0 0 β q γ ( k , q ) = β 0 γ ( k , 0 ) = 0
⇒ β 0 = 0 i.e. {0} the point
1
∑ q = 0 1 β q γ ( k , q ) = β 0 γ ( k , 0 ) + β 1 γ ( k , 1 ) = 0
⇒ ϕ ( β 0 , β 1 ) = 0 i.e. a subspace of ℝ 2
2
∑ q = 0 2 β q γ ( k , q ) = β 0 γ ( k , 0 ) + β 1 γ ( k , 1 ) + β 2 γ ( k , 1 ) = 0
⇒ ϕ ( β 0 , β 1 , β 2 ) = 0 i.e. a subspace of ℝ 3
⋮
n
∑ q = 0 n β q γ ( k , q ) = 0
⇒ ϕ ( β 0 , β 1 , ⋯ , β n ) = 0 i.e. a subspace of ℝ n + 1