Exercise 9.B.3

Answers

Proof. For u1,u2,v1,v2 ∈ V we have

⟨(u1 + iv1) + (u1 + iv2),x + iy⟩ = ⟨(u1 + u2) + i(v1 + v2),x + iy⟩ = ⟨u1 + u2,x⟩ + ⟨v1 + v2,y⟩ + (⟨v1 + v2,x⟩−⟨u1 + u2,y⟩ )i = ⟨u1,x⟩ + ⟨u2,x⟩ + ⟨v1,y⟩ + ⟨v2,x⟩ + (⟨v1,x⟩−⟨u1,y⟩ )i + (⟨v2,x⟩−⟨u2,y⟩ )i = ⟨u1 + iv1,x + iy⟩ + ⟨u2 + iv2,x + iy⟩.

Therefore it satisfies additivity in the first slot. For a,b ∈ ℝ, we have

⟨(a + bi)(u + iv),x + iy⟩ = ⟨(au − bv) + i(av + bu),x + iy⟩ = ⟨au − bv,x⟩ + ⟨av + bu,y⟩ + (⟨av + bu,x⟩−⟨au − bv,y⟩ )i = ⟨au,x⟩−⟨bv,x⟩ + ⟨av,y⟩ + ⟨bu,y⟩ + (⟨av,x⟩−⟨au,y⟩ )i + (⟨bu,x⟩ + ⟨bv,y⟩)i = ⟨au,x⟩ + ⟨av,y⟩ + (⟨av,x⟩−⟨au,y⟩ )i −⟨bv,x⟩ + ⟨bu,y⟩ + (⟨bu,x⟩ + ⟨bv,y⟩)i = a⟨u + iv,x + iy⟩ + bi (i⟨v,x⟩− i⟨u,y⟩ + ⟨u,x⟩ + ⟨v,y⟩ ) = a⟨u + iv,x + iy⟩ + bi⟨u + iv,x + iy⟩ = (a + bi)⟨u + iv,x + iy⟩

Hence homogeneity in the first slot is satisfied. For positivity, we have

⟨u + iv,u + iv⟩ = ⟨u,u⟩ + ⟨v,v⟩ + (⟨v,u⟩−⟨u,v⟩ )i = ⟨u,u⟩ + ⟨v,v⟩≥ 0,

where the second equality follows because V is a real inner product space. The equation above also displays definiteness, because the left side equals 0 if and only if u = v = 0, in other words, if u + iv = 0. For conjugate symmetry, we have

⟨u + iv,x + iy⟩ = ⟨u,x⟩ + ⟨v,y⟩− (⟨v,x⟩−⟨u,y⟩)i¯ = ⟨x,u⟩ + ⟨y,v⟩− (⟨x,v⟩−⟨y,u⟩)i¯ = ⟨x + iy,u + iv⟩¯.

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2017-10-06 00:00
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