if W is subspace of V_{4}(R) spanned by x_{1}=(1,2,-3,4) and x_{2}=(0,1,4,-1). find the basis of annihilator of W?
can we start like, f _{1} (x_{1})=0 and f_{1}(x_{2})=0 & f_{2}(x_{1})=0 and f_{2}(x_{2})=0
well, sure f_{1} and f_{2} will have to annihilate the basis of W. bt that doesn't reveal "which" linear functionals f_{1} and f_{2} 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:
v_{1} + 2v_{2} - 3v_{3} + 4v_{4} = 0
v_{2} + 4v_{3} - v_{4} = 0
subtracting 2 times the second equation from the first, turns the first equation into:
v_{1} - 11v_{3} + 6v_{4} = 0
making it clear that if we choose any two values for v_{3} and v_{4}, say s and t, we obtain values for v_{1} and v_{2}:
v_{1} = 11s - 6t
v_{2} = -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 f_{1}(v) = <(11,-4,1,0),v> and f_{2}(v) = <(-6,1,0,1),v>
it should be clear that both f_{1}, f_{2} are in ann(W). suppose that af_{1} + bf_{2} = 0,
that is: (af_{1} + bf_{2})(v) = 0, for all v in R^{4}. this means:
(11a - 6b)v_{1} + (-4a + b)v_{2} + av_{3} + bv_{4} = 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 {f_{1},f_{2}} is a linearly independent set.
finally, since dim(ann(W)) = dim(R^{4}) - dim(W) = 4 - 2 = 2, we see {f_{1},f_{2}} 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 f_{1} = (11,-4,1,0) in the basis {p_{j}} for (R^{4})*, and f_{2} = (-6,1,0,1) in that basis,
which might make clearer why we looked at W^{⊥} in the first place.