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By orthogonal diagonalization of the positive-definite symmetric matrix $A$, form its positive symmetric inverse square root $S=A^{-1/2}$. Then $S^{T}AS=I$, and $C=SBS$ is real symmetric. Let $Q$ have an orthonormal eigenbasis of $C$ as its columns, so $Q^{T}CQ=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. Taking $P=SQ$ givesBoth factors of $P$ are invertible. This proves the simultaneous congruence diagonalization with a positive form by an explicit construction; it is congruence rather than simultaneous similarity.

To find the entries without constructing $P$, take determinants inSince $\det P\ne 0$, the entries are precisely the roots, with multiplicity, of $\det (B-\lambda A)=0$. Equivalently they are the eigenvalues of $A^{-1}B$, although that matrix need not itself be symmetric.

For the given $A$, its leading principal minors are $2,3,4$, so it is positive definite. Expanding the determinant givesThus the diagonal entries, in any order, are