Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 6 / ii / a

For $B{x}_i={\lambda }_i{x}_i$, Hermitian symmetry givesDistinct eigenvalues therefore imply $x_i^{†}{x}_j=0$ for $i\ne j$. Together with the unit normalizations,The $n$ mutually orthogonal nonzero vectors are linearly independent and span the $n$-dimensional space, so they form an orthonormal basis. Here ${\delta }_{ij}$ is the Kronecker delta.