 Solution

The spectral theorem for a real symmetric matrix gives real eigenvalues and an orthonormal eigenbasis. Compute the roots of $\det (A-\lambda I)=0$, find a basis for each eigenspace, and orthonormalize within any repeated-eigenvalue eigenspace. Eigenvectors belonging to distinct eigenvalues are already orthogonal: if $Av=\lambda v$ and $Aw=\mu w$, symmetry gives $(\lambda -\mu )v^{T}w=0$.

Putting the normalized basis vectors in the columns of an orthogonal matrix $Q$ yieldsFor repeated eigenvalues, the choice of axes inside their eigenspace is not unique.