 Solution

For a self-adjoint operator on a finite-dimensional complex inner product space, every eigenvalue is real and eigenspaces for distinct eigenvalues are orthogonal. The spectral theorem gives an orthonormal eigenbasis. Accordinglyand the $E_i$ are orthogonal projections satisfying $E_i^{\ast }=E_i$. Since $T$ is nonsingular, none of its eigenvalues is zero. Its representing matrix in the orthonormal eigenbasis is real diagonal.