 Solution

By the spectral theorem for real symmetric matrices, there is an orthogonal matrix $Q$ such thatFor clarity, the diagonalization can also be justified recursively. The characteristic polynomial has a complex root of a polynomial by the fundamental theorem of algebra. By ii its eigenvalue is real; taking a nonzero real or imaginary part of an associated eigenvector in ${\mathbb{C}}^n$ gives a real eigenvector, which can be normalized. Its orthogonal complement is invariant, because $⟨Ax,z⟩=⟨x,Az⟩=\lambda ⟨x,z⟩=0$. Restriction to this linear subspace is again symmetric, and induction on dimension gives an orthonormal basis of eigenvectors.

All ${\lambda }_i\ge 0$ by ii. The permitted invariance of determinant and matrix trace under change of basis, applied to the diagonal matrix, yields