Consider the mechanical system
$$\boldsymbol{M\ddot{q}}(t)+\boldsymbol{Kq}(t)=\boldsymbol{0}$$
with the mass matrix $\boldsymbol{M}$ and the stiffness matrix $\boldsymbol{K}$. The displacements of the individual masses are given by the vector $\boldsymbol{q}(t)\in \mathbb{R}^{n}$.
If we already know the eigenvectors $\hat{\boldsymbol{q}}_i$ and eigenvalues (eigenfrequencies = imaginary part of complex conjugate eigenvalues) $\omega_i$, is there a quick method in deriving the diagonal modal mass matrix $\widetilde{\boldsymbol{M}}$ and the diagonal modal stiffness matrix $\widetilde{\boldsymbol{K}}$ which are given by the following relationships
$$\widetilde{\boldsymbol{M}}=\boldsymbol{Q}^T\boldsymbol{MQ} \qquad (1)$$ $$\widetilde{\boldsymbol{K}}=\boldsymbol{Q}^T\boldsymbol{KQ} \qquad \,\,\,(2),$$
in which the eigenvector matrix is given by $$\boldsymbol{Q}=\begin{bmatrix}\hat{\boldsymbol{q}}_1 & \hat{\boldsymbol{q}}_2 & \ldots \hat{\boldsymbol{q}}_n \end{bmatrix}.$$
It is clear that the substitution $\boldsymbol{q}=\boldsymbol{Qp}$ transforms the initial ODE into the following form
$$\widetilde{m}_i\ddot{p}_i(t)+\widetilde{k}_ip_i(t)=0 \qquad \text{for } i=1,\ldots,n.$$
The entries $\widetilde{m}_i$ and $\widetilde{k}_i$ are the corresponding entries in $i^{\text{th}}$ row of the diagonal modal matrices.
By dividing by $\widetilde{m}_i\neq 0 $ we obtain:
$$\ddot{p}_i(t)+\dfrac{\widetilde{k}_i}{\widetilde{m}_i}p_i(t)=0$$
because we know that
$$\omega_i^2 = \dfrac{\widetilde{k}_i}{\widetilde{m}_i}$$ it is possible to write down the decoupled equation for the normalized case, as we already know the eigenfrequencies. But what I actually want is the modal mass matrix and the modal stiffness matrix.
Is it possible to deduce the modal masses $\widetilde{m}_i$ and modal stiffnesses $\widetilde{k}_i$ directly from the structure of the eigenvectors without multiplying the eigenvector matrix with the mass matrix (see equation $(1)$) and the stiffness matrix (see equation $(2)$)?