Exercise 6.A.24

Answers

Proof. For positivity, we have v,v1 = Sv,Sv 0.

For definiteness, suppose v,v1 = 0. Then Sv,Sv = 0, which implies that Sv = 0. However S is injective, thus v = 0. Conversely, if v = 0, then v,v1 = Sv,Sv = 0,0 = 0.

For additivity in the first slot, we have u + v,w1 = Su + Sv,Sw = Su,Sw + Sv,Sw = u,w1 + v,w1.

For homogeneity, λu,v1 = λSu,Sv = λSu,Sv = λu,v1.

For conjugate symmetry, we have u,v1 = Su,Sv = Sv,Su¯ = v,u1¯. □

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2017-10-06 00:00
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