 Solution

The spectral theorem gives a real orthonormal eigenbasis $v_1,\dots ,v_n$ for a real symmetric matrix $A$. Distinct eigenspaces are orthogonal; within a repeated eigenspace, choose an orthonormal basis, for example by Gram-Schmidt orthogonalization. Put these eigenvectors in the columns of $T$. Then $T^{-1}=T^T$ andFor a change of variables $y=Sx$, choose $\boxed{S=T^T}$. Since $x=Ty$, the quadratic form becomes $x^{T}Ax={\sum }_i{\lambda }_i{y}_i^2$; the ${\lambda }_i$ are precisely the eigenvalues of $A$.

In the numerical case, the off-diagonal entries are half the cross-term coefficients, soA direct eigenvector calculation gives the unit eigenvectors $(1,1,0)/\sqrt{2}$, $(1,-1,1)/\sqrt{3}$, $(1,-1,-2)/\sqrt{6}$, with eigenvalues two, three and minus six. They are pairwise orthogonal. Therefore the required orthogonal diagonalization isand the transformed quadratic form is $\boxed{2y_1^2+3y_2^2-6y_3^2}$.