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 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 3

n

q = 0 n β q γ ( k , q ) = 0

ϕ ( β 0 , β 1 , , β n ) = 0 i.e. a subspace of n + 1