Simultaneous congruence diagonalization with a positive form ($P^{T}AP=I,P^{T}BP=\mathrm{diag}({\lambda }_j)$)

If $A$ is real symmetric positive definite and $B$ is real symmetric, let $S=A^{-1/2}$. The matrix $SBS$ is symmetric and has an orthonormal eigenbasis, represented by $Q$. Taking $P=SQ$ gives the two displayed congruences. The diagonal entries are determined by the generalized eigenvalue problem $\det (B-\lambda A)=0$, since $P^T(B-\lambda A)P$ is diagonal with entries ${\lambda }_j-\lambda$. Positive definiteness of the first form is essential to this construction.