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Math Help - Zero Operator

  1. #1
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    Zero Operator

    Dear Colleagues,

    Could you please help me in solving the following problem:
    Let T:X\longrightarrow X be a bounded linear operator on a complex inner product space X. If  \langle Tx,x \rangle =0 for all x\in X, show that T=0.
    Moreover, this is does not hold in the case of real inner product space.


    Regards,

    Raed.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by raed View Post
    Dear Colleagues,

    Could someone solve the problem.


    Regards.
    I'll help you with the problem. Note first that if \langle u,x\rangle=\langle v,x\rangle for all x\in V (where [tex]V{/math] is our inner product space) then u=v (indeed, take x=u-v and subtract). From this we get that if \langle T(x),y\ranlge=0 for all x,y\in V then T=0 (indeed, note that \langle T(x),y\rangle=0=\langle 0,y\rangle...conclude from the first part). Now, to our actual problem let x=zv+u\;\; z\in\mathbb{C} and compute that our assumption gives 0=z\langle T(u),v\rangle+\overline{z}\langle T(v),u\rangle. Pick z cleverly (don't think too hard) to create a system of equations that will reduce this problem to the last one (note that you'll have to pick one of the values of z to be non-real otherwise you'll have disproved the second part). For the second part take your inner product space to be \mathbb{R}^2 with the usual inner product and recall that the inner product of orthogonal vectors is orthogonal...so what if a transformation sent a vector to its orthogonal vector.
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  3. #3
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by raed View Post
    Dear Colleagues,

    Could you please help me in solving the following problem:
    Let T:X\longrightarrow X be a bounded linear operator on a complex inner product space X. If  \langle Tx,x \rangle =0 for all x\in X, show that T=0.
    Moreover, this is does not hold in the case of real inner product space.
    Hint: Apply the polarisation identity \displaystyle \langle Tx,y\rangle = \frac14\sum_{k=0}^3 i^k\langle T(x+i^ky),x+i^ky\rangle.

    Note that this is an intrinsically complex relation. In the real case the result fails. For example, in two-dimensional space a rotation through a right angle takes every vector to an orthogonal vector.

    Edit. Sorry, didn't see drexel28's comment.
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