Proof of Elementary Properties of a Complex Inner Product

Proof

Conditions 1-4 below are those of Complex Inner Product Space, and we use the properties of complex conjugation recorded in Properties of Complex Conjugation and Modulus, in particular that conjugation preserves sums and products (claim 1 there).

Claim 1. By condition 1, condition 2, and the additivity of conjugation,

\langle u+v,w\rangle=\overline{\langle w,u+v\rangle}=\overline{\langle w,u\rangle+\langle w,v\rangle}=\overline{\langle w,u\rangle}+\overline{\langle w,v\rangle}=\langle u,w\rangle+\langle v,w\rangle .

Claim 2. By condition 1, condition 3, and the multiplicativity of conjugation,

\langle\lambda u,v\rangle=\overline{\langle v,\lambda u\rangle}=\overline{\lambda\langle v,u\rangle}=\overline{\lambda}\,\overline{\langle v,u\rangle}=\overline{\lambda}\,\langle u,v\rangle .

Claim 3. By claim 3 of Elementary Identities in a Vector Space applied to the vector $0_{V}$ and the scalar $0$ , we have $0\,0_{V}=0_{V}$ . Hence, by condition 3 and the field identity $0\cdot z=0$ in $\mathbb{C}$ ,

\langle v,0_{V}\rangle=\langle v,0\,0_{V}\rangle=0\,\langle v,0_{V}\rangle=0 .

Applying condition 1 and $\overline{0}=0$ , which holds because $0$ has real part $0$ and imaginary part $0$ by Real and Imaginary Parts of a Complex Number and Complex Conjugate, we get $\langle 0_{V},v\rangle=\overline{\langle v,0_{V}\rangle}=\overline{0}=0$ .

Claim 4. If $v=0_{V}$ , then $\langle v,v\rangle=0$ by claim 3. Conversely, if $\langle v,v\rangle=0$ , then $v=0_{V}$ by condition 4.

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lemmaElementary Properties of a Complex Inner Productlem:inner-product-elementary-properties-2026a