 Solution

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

The restriction of a positive definite form to each

V_{m}

P

such that

P^{T} A_{m} P = diag (d_{1}, \dots, d_{m})

Each

d_{i} = φ (f_{i}, f_{i}) > 0

, since a basis vector is nonzero. Taking determinants in part (i) gives

det (P^{T} A_{m} P) = (det P)^{2} det A_{m} = det A_{m},

\pm 1

. Consequently

det A_{m} = i = 1 \prod m d_{i} > 0 (1 \leq m \leq n) .

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