1. ## 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

2. ## Re: basis

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 pj is the linear functional that projects v onto its j-th coordinate, i.e., pj(v) = vj,

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

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

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# standard basis of annihilator

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