Given a non singular and symmetric $4 \times 4 $ matrix $G$ over the field of real numbers. Is it always possible to find another non singular $ 4 \times 4$ matrix $A$ such that
$\sum_j \sum_k G_{jk}A_{ij}A_{nk}=C_i\delta_i^n$.
where $G_{ij}$ is ith row and jth column entry of matrix $G$ and {$C_i$} are some constants.
$\delta_i^n=1$ if $i=n$ and $0$ otherwise.
It is always possible in the case $G$ is either positive or negative definite matrix. But I don't see how to generalize to the case where $G$ is not definite.
In other words, is there a matrix $A$ such that $A^TGA$ is diagonal?
All symmetric matrices are normal. I.e., $G^TG = GG^T$.
All normal matrices are diagonalizable. I.e., there is some $Q$ such that $Q^{-1}GQ$ is a diagonal matrix $C$ with diagonal entries $C_i$. Tjus $GQ = QC$, but the $i$-th column of $QC$ is $C_i$ times the $i$-th column of $Q$. So the columns of $Q$ are eigenvectors of $G$. In particular, the sum of the eigenspaces of $G$ is all of $\Bbb R^4$. And any basis of eigenvectors of $G$ for $\Bbb R^4$, can be combined to form the columns of such a matrix $Q$.
All normal matrices have orthogonal eigenspaces. I.e., if $Gv_1 = \lambda_1 v_1$ and $Gv_2 = \lambda_2 v_2$ with $\lambda_1\ne\lambda_2$, then $v_1 \perp v_2$.
Any inner product space, such as the eigenspace of an eigenvector, has orthonormal bases.
Thus it is possible to construct an orthonormal basis $\{a_i\}_i$ of $\Bbb R^4$ (or any other dimension) consisting of eigenvectors of $G$. If we then construct a matrix $A$ having these basis vectors as its columns, $A$ will be orthonormal. I.e., $A^T = A^{-1}$. Since the columns of $A$ are all eigenvectors, $A^{-1}GA$ is diagonal. And thus $A^TGA$ is diagonal.