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.