Suppose u; v is the element of V . Prove that || au + bv || = || bu + av || for all a, b is the element of R if and only if || u || = || v || .
Suppose that "V={\\mathbb{R}}^n". Then, we have: "u=(u_1,...u_n)" and "v=(v_1,...,v_n)". We have: "(au+bv)=(au_1+bv_1,...,au_n+bv_n)" and "(bu+av)=(bu_1+av_1,...,bu_n+av_n)". We must obtain the equality: "\\sqrt{(au_1+bv_1)^2+...+(au_n+bv_n)^2}=\\sqrt{(bu_1+av_1)^2+...+(bu_n+av_n)^2}". Simplify the left side: "\\sqrt{(a^2u^2_1+2abu_1v_1+b^2v^2_1+...+a^2u^2_n+2abu_nv_n+b^2v^2_n}." Right side has the form: "\\sqrt{(bu_1+av_1)^2+...+(bu_n+av_n)^2}=\\sqrt{b^2u^2_1+2abu_1v_1+a^2v_1^2+...+b^2u^2_n+2abu_nv_n+a^2v_n^2}"
Since by assumption we have: "\\sqrt{u_1^2+...+u_n^2}=\\sqrt{v_1^2+...+v_n^2}" . Using the latter we observe that left and right side are equal. Thus, the equality "||au+bv||=||bu+av||" holds. We showed that if "||u||=||v||", then the equality "||au+bv||=||bu+av||" holds. Otherwise, we can set "a=1,b=0" and obtain "||u||=||v||."
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