Reduction of a quadratic form by eigenvalues

Matrix. $A=\begin{pmatrix}3&1\\1&3\end{pmatrix}$.

Eigenvalues. $\det(A-\lambda I)=(3-\lambda)^2-1=\lambda^2-6\lambda+8=(\lambda-2)(\lambda-4)$. So $\lambda_1=2$, $\lambda_2=4$.

Orthonormal eigenvectors. For $\lambda_1=2$: $u_1=(1,-1)/\sqrt2$. For $\lambda_2=4$: $u_2=(1,1)/\sqrt2$.

Orthogonal change of variables. Let $X=QU$ with $Q=\frac{1}{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}$. Then
$$Q(x,y)=2u^2+4v^2.$$

Conclusion. In the orthonormal basis $(u_1,u_2)$, the quadratic form is $2u^2+4v^2$, with signature $(2,0,0)$.