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ご回答誠に有難うございます。


>>> ところが, C^3 のある正規直交基底 u_1, u_2, u_3 に対して,
>>> x = k u_2Λu_3 (k \neq 0) と表せますから,
>>> (C(u_2Λu_3), u_2Λu_3) > 0 を示せば良い.
>>> ここまでは理解されましたか.
>> ここは xは任意なのだから,x=ku_i∧u_j 但し,i,j∈{1,2,3},i<j.
>> ではないのでしょうか?
> 正規直交基底はその並び方の順序にはよりませんから,
> x = k u_2Λu_3 となるように, u_1, u_2, u_3 を
> 取ることができます.

つまり, U:={{u_2,u_3}∈2^{C^3};{u_1,u_2,u_3}は∧^2C^3の正規直交基底}とすると,
∀x∈∧^2C^3に対して,{u_2,u_3}∈Uと複素数k(≠0)が存在して,x=ku_2∧u_3となるという事ですね?


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