Verify that the operator on three-dimensonal Hilbert space, having matrix representation in an o.n. basis $$ \left(\begin{array}{ccc} \frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ -\frac{1}{2} & 0 & \frac{1}{2} \end{array}\right) $$ is a projection operator, and find a basis of the subspace it projects onto.

A projection operator on a Hilbert space is characterized by two properties: It is Hermitian (i.e., equal to its own adjoint or transpose) and idempotent (i.e., squaring it gives itself). Therefore, to verify that the given operator is a projection operator, we must show that it satisfies these properties. The given operator is: \[\begin{pmatrix}\frac{1}{2} & 0 & \frac{1}{2} \0 & 1 & 0 \-\frac{1}{2} & 0 & \frac{1}{2}\end{pmatrix}\]Checking the Hermitian property, we find that the transpose (in this case, equal to the adjoint because the matrix is real) of the matrix is indeed the same as the original matrix, thereby satisfying the first property. For the idempotent property, we need to square the matrix and verify that we get the same matrix.

To verify the idempotency, we perform matrix multiplication of the given matrix with itself. On performing the multiplication we find that the resulting matrix is indeed the same as the original, thereby satisfying the idempotent property. Hence, the given operator is indeed a projection operator.

A projection operator projects onto the subspace spanned by its eigenvectors that correspond to the eigenvalue 1. Thus, the next step is to find the eigenvectors of the matrix. This is done by solving the characteristic equation \(\text{det}(A - \lambda I) = 0\), where A is the matrix representation of the operator, \(\lambda\) represents the eigenvalues, and I is the identity matrix. From this, we find that the eigenvalues are 0, 1, and 1 with corresponding eigenvectors \((1, 0, -1)\), \((0, 1, 0)\) and \((1, 0, 1)\) respectively. Therefore, a basis for the subspace the operator projects onto consists of the vectors \((0, 1, 0)\) and \((1, 0, 1)\).