Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 2 / b

Solution

w

has the coordinate expansion

w = \sum_{i} ⟨ w, e_{i} ⟩ e_{i}

. If

⟨ S u, v ⟩ = ⟨ u, R v ⟩

for all

u, v

, apply it to

u = e_{i}

v = e_{j}

⟨ R e_{j}, e_{i} ⟩ = ⟨ e_{j}, S e_{i} ⟩ .

Consequently

R e_{j} = i = 1 \sum n ⟨ e_{j}, S e_{i} ⟩ e_{i} .

Conversely, this formula gives

⟨ e_{i}, R e_{j} ⟩ = \overline{⟨ e_{j}, S e_{i} ⟩} = ⟨ S e_{i}, e_{j} ⟩ .

Linearity in the first argument and conjugate-linearity in the second extend the equality from basis vectors to all

u, v

. This proves the equivalence.

The displayed formula specifies exactly one image for each basis vector, so it defines a unique linear transformation

S^{*}

⟨ S^{*} u, v ⟩ = \overline{⟨ v, S^{*} u ⟩} = \overline{⟨ S v, u ⟩} = ⟨ u, S v ⟩ .

(S^{*})^{*} = S

. Also

⟨ ST u, v ⟩ = ⟨ T u, S^{*} v ⟩ = ⟨ u, T^{*} S^{*} v ⟩,

(ST)^{*} = T^{*} S^{*}

. The reversal of order follows directly from moving the two factors across the inner product one at a time.

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