Thread: basis

if W is subspace of V₄(R) spanned by x₁=(1,2,-3,4) and x₂=(0,1,4,-1). find the basis of annihilator of W?



can we start like, f ₁ (x₁)=0 and f₁(x₂)=0 & f₂(x₁)=0 and f₂(x₂)=0

well, sure f₁ and f₂ will have to annihilate the basis of W. bt that doesn't reveal "which" linear functionals f₁ and f₂ might be.

a more profitable approach is to find a basis first for W^⊥, the orthogonal complement of W.

this gives us the following two equations:

v₁ + 2v₂ - 3v₃ + 4v₄ = 0
v₂ + 4v₃ - v₄ = 0

subtracting 2 times the second equation from the first, turns the first equation into:

v₁ - 11v₃ + 6v₄ = 0

making it clear that if we choose any two values for v₃ and v₄, say s and t, we obtain values for v₁ and v₂:

v₁ = 11s - 6t
v₂ = -4s + t

so that any solution is of the form s(11,-4,1,0) + t(-6,1,0,1), so a basis for W^⊥ is {(11,-4,1,0),(-6,1,0,1)}.

now define f₁(v) = <(11,-4,1,0),v> and f₂(v) = <(-6,1,0,1),v>

it should be clear that both f₁, f₂ are in ann(W). suppose that af₁ + bf₂ = 0,

that is: (af₁ + bf₂)(v) = 0, for all v in R⁴. this means:

(11a - 6b)v₁ + (-4a + b)v₂ + av₃ + bv₄ = 0, no matter how we choose v.

letting v = (0,0,1,0), we see a = 0, and letting v = (0,0,0,1), we see b = 0, thus {f₁,f₂} is a linearly independent set.

finally, since dim(ann(W)) = dim(R⁴) - dim(W) = 4 - 2 = 2, we see {f₁,f₂} is a basis for ann(W).

note that if p_j is the linear functional that projects v onto its j-th coordinate, i.e., p_j(v) = v_j,

we can write f₁ = (11,-4,1,0) in the basis {p_j} for (R⁴)*, and f₂ = (-6,1,0,1) in that basis,

which might make clearer why we looked at W^⊥ in the first place.