What is an easiest way to check does matrix $A$ corresponds to a bilinear form $\mathcal{A}$ in some basis?
For example let’s say that in a standard basis bilinear form $\mathcal{A}:\mathbb{C}^{3}\times\mathbb{C}^{3}\to\mathbb{C}$ is given by matrix
\begin{equation} A_{0} = \begin{pmatrix}0&-1&-2\\-5&-8&6\\ -8&-8&19\end{pmatrix}. \end{equation}
Are there any $\mathbb{C}^{3}$ in which matrix $A_{1}$ or $A_{2}$ are its Gram’s matrices?
\begin{align} A_{1} & = \begin{pmatrix} 3&4&6\\5&7&10\\8&11&17\end{pmatrix} & A_{2} = \begin{pmatrix}-1&4&10\\-2&12&25\\4&-3&-24\end{pmatrix} \end{align}
As I understand this is the same question as asking whether or not there are some matrices $U_{1}$ and $U_{2}$ such that $A_{1} = U_{1} A_{0} U_{1}^{\top}$ and $A_{2} = U_{2} A_{0} U_{2}^{\top}$.
There is a following possible solution:
Let’s note that $A_{1}^{\top} = \left(U_{1}^{\top}A_{0}U_{1}\right)^{\top} = U_{1}^{\top}\left(U_{1}^{\top}A_{0}\right)^{\top} = U_{1}^{\top} A_{0}^{\top} {U_{1}^{\top}}^{\top} = U_{1}^{\top}A_{0}U_{1}$ and $A_{1}^{-1} = \left(U_{1}^{\top}A_{0}U_{1}\right)^{- 1} = U_{1}^{-1}A_{0}^{-1}{U_{1}^{\top}}^{-1}$. Denote $M_{0} = A_{0}^{- 1}A_{0}^{\top}$ and $M_{1} = A_{1}^{-1}A_{1}^{\top}$. Then
\begin{equation} M_{1} = A_{1}^{-1}A_{1}^{\top} = U_{1}^{-1} A_{0}^{-1}{U_{1}^{\top}}^{-1} U_{1}^{\top}A_{0}^{\top}U_{1} = U_{1}^{-1}A_{0}^{-1}A_{0}^{\top}U_{1} = U_{1}^{-1}M_{0}U_{1}. \end{equation}
Thus $M_{1}$ is similar to $M_{0}$ ($M_{1}\sim M_{0}$), and know we have to check that $M_{1}$ and $M_{2}$ have the same eigenvalues. If indeed these two matrices are similar, then there is a basis in which $A_{1}$ is a Gram matrix of $\mathcal{A}$, and $U$ is a basis-changing matrix.
In our case \begin{gather} M_{0} = A_{0}^{-1}A_{0}^{\top} = \begin{pmatrix}0&-17&-2\\-5&-8&6\\,-8&-8&19 \end{pmatrix} \sim \begin{pmatrix}1&0&0\\ 0&-3-2\sqrt{2}&0\\0&0&-3+2\sqrt{2}\end{pmatrix}\\ M_{1} = A_{1}^{-1}A_{1}^{\top} = \begin{pmatrix}7&11&16\\-3&-4&-7\\ -1&-1&-\end{pmatrix} \sim \begin{pmatrix}-i&0&0\\ 0&i&0\\0&0&1\end{pmatrix} \\ M_{2} = A_{2}^{-1}A_{2}^{\top} = \begin{pmatrix}277&718&-570\\-66&-171&136\\54&140&-111\end{pmatrix}\sim \begin{pmatrix}1&0&0\\ 0&-3-2\sqrt{2}&0\\0&0&-3+2\sqrt{2}\end{pmatrix} \end{gather} Therefore $A_{2}$ does represent the same bilinear form as $A_{0}$, and $A_{1}$ does not.
As far as it does the trick maybe anyone knows a better solution, which for example does not include calculating a matrix inverse.