 Solution

ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2015/3/a/iii/solution

u = \sum_{i} u_{i} e_{i}

and

v = \sum_{i} v_{i} e_{i}

, evaluation against the dual basis gives

ϕ (u, v) = i \sum u_{i} \overline{v_{i}} .

This has codomain $C$ , not the printed $R$ . For a nonzero vector space, take

u = i e_{1}

and

v = e_{1}

to obtain

ϕ (u, v) = i

. Only the diagonal values

ϕ (v, v)

must be real. The displayed formula defines the intended complex inner product.

It is linear in the first argument and conjugate-linear in the second:

ϕ (a u + b w, v) = a ϕ (u, v) + b ϕ (w, v), ϕ (u, a v + b w) = \overline{a} ϕ (u, v) + \overline{b} ϕ (u, w) .

ϕ (v, u) = \overline{ϕ (u, v)}

. Finally,

ϕ (v, v) = \sum_{i} ∣ v_{i} ∣^{2} \geq 0,

with equality exactly when every coordinate vanishes, equivalently

v = 0

. This proves positive definiteness. Since

ϕ (e_{i}, e_{j}) = δ_{ij}

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