 Solution

ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2014/3/a/iii/solution

y

lies in

(im Φ ∣_{U})^{\circ}

exactly when

j (y) (φ_{x}) = φ_{x} (y) = φ (x, y) = 0 for all x \in U .

That is precisely membership of the orthogonal complement for a bilinear form. Therefore

U^{⊥} = (im Φ ∣_{U})^{\circ} .

The dimension formula in part (ii) gives

dim U^{⊥} = n - rank (Φ ∣_{U}) \geq n - dim U,

dim U + dim U^{⊥} \geq n

. More precisely, the rank-nullity theorem gives

dim U + dim U^{⊥} = n + dim (U \cap ker Φ) .

This is the orthogonal complement dimension for a degenerate bilinear form. The possible excess comes from vectors annihilated by the form against all of

V

 New to topics? Read the docs here!