Thread: Subspaces and orthogonality

Hi everyone

Stuck on this problem, hoping someone here could point me in the right direction.

We have two subspaces of an inner product space. The dimension of one is larger than the other (and both finite). The question is if there is at least one nonzero vector in the larger subspace that is orthogonal to all vectors of the smaller one.

Clearly this seems to be true for R² and R³ but I don't know how to generalize this and in general I just suck at proving things

in an inner product space, for any subspace W of a vector space V, you can define W^⊥ = {v in V: <v,w> = 0, for all w in W}, the orthogonal complement of W.

moreover, V = W⊕W^⊥.

if we have another subspace U, with dim(U) > dim(W), then there must exist some vector u in U, that is not in W (otherwise U would be a subspace of W, and thus dim(U) ≤ dim(W)).

since u is not in W, it cannot be 0 (because 0 is in every subspace), and therefore it is a non-zero element of W^⊥. hence <u,w> = 0 for all w in W.

*****

here is an example from R⁴, with the standard dot product:

let W = span({(1,0,0,1),(1,1,0,0)}), and let U = span({1,0,0,0),(0,1,0,0),(0,0,1,0)}).

an element of W is of the form: (a+b,b,0,a). first let's find a basis for W^⊥:

writing a typical element of R⁴ as (x,y,z,w) we have:

(a+b,b,0,a).(x,y,z,w) = 0, so: ax+by+aw = 0. letting b = 0, a ≠ 0, we get:

ax+aw = 0, so w = -x.

letting a = 0, b ≠ 0, we get: by = 0, so y = 0. so every element of W^⊥ is of the form:

(x,0,z,-x) = x(1,0,0,-1) + z(0,0,1,0), so {(1,0,0,-1),(0,0,1,0)} is a basis for W^⊥.

now we just need to find an element of U in W^⊥, which is easy: (0,0,1,0) clearly fits the bill. as a check:

(a+b,b,0,a).(0,0,1,0) = (a+b)(0) + b(0) + (0)(1) + a(0) = 0 + 0 + 0 + 0 = 0, and clearly (0,0,1,0) is NOT in W, for every element of W has a 0 in the 3rd coordinate.