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Thread: Identity operator

  1. #1
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    Identity operator

    Hello. Could you please help me to solve this problem:
    A is a linear unbounded operator on Hilbert space H. If AU=UA for all U- unitary operator on H, then A is bounded operator and A=\lambda I.
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  2. #2
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    Quote Originally Posted by karkusha View Post
    Hello. Could you please help me to solve this problem:
    A is a linear unbounded operator on Hilbert space H. If AU=UA for all U- unitary operator on H, then A is bounded operator and A=\lambda I.
    Fix a unit vector x_0\in H. Let y_0\in H be another unit vector, orthogonal to both x_0 and Ax_0. Let x and y be an arbitrary pair of unit vectors orthogonal to each other. Then there exists a unitary U with Ux_0=x and Uy_0=y. Therefore \langle Ax,y\rangle = \langle AUx_0,Uy_0\rangle = \langle UAx_0,Uy_0\rangle = \langle Ax_0,y_0\rangle = 0. Similarly, \langle Ax,x\rangle = \langle Ax_0,x_0\rangle.

    Thus Ax is orthogonal to everything in the orthogonal complement of x, so that Ax must be a multiple of x. What's more, that multiple must be the same for x as it is for x_0. Conclusion: A is a scalar multiple of the identity operator.
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    Then there exists a unitary U.
    Why? Can you write this operator?
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    Quote Originally Posted by karkusha View Post
    Quote Originally Posted by Opalg View Post
    Then there exists a unitary U.
    Why? Can you write this operator?
    Let V be the two-dimensional subspace spanned by x_0 and y_0, and let W be the two-dimensional subspace spanned by x and y. Then U, as defined above, maps V isometrically onto W. Also, the subspaces H\ominus V and H\ominus W have the same dimension, and so they are isometrically isomorphic to each other. Define U to be a mapping implementing this isomorphism. Then U, defined in that way, is an isometric isomorphism from H onto H, hence a unitary operator.
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