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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2First, we prove that if \( \mathrm{{\left(,\right)}} \) is a scalar product on \( \mathrm{{V}} \), then \( \mathrm{{f}} \) is one-to-one.Suppose that \( \mathrm{{\left(,\right)}} \) is a scalar product on \( \mathrm{{V},} \) is not one-to-one. Then there exist vectors \( \mathrm{{u}_{{1}}{u}_{{2}}} \) in V such that . Then we have\( \mathrm{{\left({u}_{{1}}-{u}_{{2}},{u}_{{1}}-{u}_{{2}}\right)}={\left({u}_{{1}},{u}_{{1}}\right)}-{2}{\left({u}_{{1}},{u}_{{2}},\right)}+{\left({u}_{{2}},{u}_{{2}}\right)}>{0}} \)Since \( \mathrm{{\left(,\right)}} \) is a positive definite scalar product. On the other hand, we have\( \mathrm{{\left({f{{\left({u}_{{1}}\right)}}}-{f{{\left({u}_{{2}}\right)}}},{f{{\left({u}_{{1}}\right)}}}-{f{{\left({u}_{{2}}\right)}}}\right)}={\left({f{{\left({u}_{{1}}\right)}}},{f{{\left({u}_{{2}}\right)}}}\right)}-{2}{f{{\left({\left({u}_{{1}}\right)},{f{{\left({u}_{{2}}\right)}}}\right)}}}+{\left({f{{\left({u}_{{2}}\right)}}},{f{{\left({u}_{{2}}\right)}}}\right)}} \)\( \mathrm{={\left({u}_{{1}},{u}_{{1}}\right)}-{2}{\left({u}_{{1}},{u}_{{2}}\right)}+{\left({u}_{{2}},{u}_{{2}}\right)}} \)since \( \mathrm{{\left(,\right)}} \) is induced by f. ExplanationTherefore, we obtain a contradiction, which implies that f must be one-to-one.Next, we prove that if f is one-to-one, then \( \mathrm{{\left(,\right) ... See the full answer