1. ## Identity operator

$A$ 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 $A=\lambda I$.

2. Originally Posted by karkusha
$A$ 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.

3. Then there exists a unitary U.
Why? Can you write this operator?

4. Originally Posted by karkusha
Originally Posted by Opalg
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.