Linear Algebra, Perpendicular to Perpendicular

Perpendicular to Perpendicular

Let S be a finite dimensional subspace of a larger vector space. Let P be the vectors that are perpendicular to S. In other words, s_x.p_y = 0 for every s_x in S and every p_y in P. (Note that P is indeed a vector space.) We will show that the space perpendicular to P is in fact S. For example, if S is the xy plane, then P is the z axis, and the vectors perpendicular to the z axis form the xy plane.

Using gram schmidt, build an orthogonal basis for S. Then build a basis for P. (Since P could have uncountably many dimensions, we can't say any more about its basis, other than the fact that it exists, and even that requires the axiom of choice.) Take the union to find a larger basis b, and suppose some vector v is not spanned by b. Use gram schmidt again to move v onto a vector w that is orthogonal to S. Now w is in P, yet w and v differ by vectors in S, hence v is spanned after all. Together, S and P span the entire space.

Suppose some vector v is perpendicular to P, but not in S. Write v as a linear combination of basis vectors, something like this.

c₁p₁ + c₂p₂ + c₃p₃ + c₄s₄ + c₅s₅

Take the dot product with any p_y in P and the last two terms drop out, hence c₁p₁+c₂p₂+c₃p₃ is perpendicular to P. This vector is perpendicular to p₁, and to p₂, and to p₃, hence it is perpendicular to any linear combination of these three vectors. In particular, it is perpendicular to itself, which is impossible for a nonzero vector. Therefore the space perpendicular to P is precisely S.