# 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.

Update:

||x||= 的平方根~

Rating

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.