 Solution

The matrix is real symmetric, so the spectral theorem guarantees an orthonormal basis of eigenvectors. Its characteristic determinant factors asSolving the three corresponding equations for the null space gives the normalized eigenvectorsTheir pairwise inner products vanish. With these vectors as columns,Since $Q^{T}Q=I$, this also gives $A=Q\mathrm{diag}(6,-3,1)Q^T$.