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線性代數的問題,煩請高手解答~
Let T be a linear operator on an inner product space V and suppose that ||T(x)||=||x|| for
all x.Prove that T is one-to-one.
題目我懂~但不知如何下手~
Update:
||x||= 的平方根~
順道一提~很急~拜託各位了~
2 Answers
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- 2 decades agoFavorite Answer
Since T be a linear operator , we have T(x1-x2)=T(x1)-T(x2),
and then ||T(x1-x2)||=||T(x1)-T(x2)||.
On the other hand, ||T(x1-x2)||=||x1-x2||, then we have ||x1-x2||=||T(x1)-T(x2)||.
Now, if T(x1)=T(x2) then ||x1-x2||=||T(x1)-T(x2)||=0 , that is x1=x2
So,T is one-to-one.
Q.E.D.
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